InternLMXComposer_GeoQA.txt

ID:0
picturePath:image/0.png
answer:80°
answer:80
correct answer:140
total:1  correct:0  acc:0.0
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ID:1
picturePath:image/1.png
answer:This is a geometry problem involving parallel lines and angles. Given that AB is parallel to CD, we can use the property of parallel lines that corresponding angles are congruent. Therefore, ∠1 = ∠3 = 50°. Since EG is an angle bisector of ∠BEF, we can use the property of angle bisectors that they cut off congruent angles. Therefore, ∠2 = ∠4 = (180° - ∠3) / 2 = (180° - 50°) / 2 = 65°. Hence, the answer is D.
answer:65
correct answer:65
total:2  correct:1  acc:0.5
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ID:2
picturePath:image/2.png
answer:12\sqrt{2}
answer:2
correct answer:6
total:3  correct:1  acc:0.3333333333333333
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ID:3
picturePath:image/3.png
answer:由于OA=OB=OC，∠ACB=30°，因此△AOB是等边三角形。在等边三角形中，所有的角度都相等，因此∠AOB=30°。
answer:30
correct answer:60
total:4  correct:1  acc:0.25
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ID:4
picturePath:image/4.png
answer:∠2=180°-90°-∠1=180°-90°-70°=20°
answer:20
correct answer:70
total:5  correct:1  acc:0.2
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ID:5
picturePath:image/5.png
answer:根据平行线的性质，两直线平行，同位角相等。因此，∠2=∠1=120°。故选B。
answer:120
correct answer:120
total:6  correct:2  acc:0.3333333333333333
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ID:6
picturePath:image/6.png
answer:45
answer:45
correct answer:45
total:7  correct:3  acc:0.42857142857142855
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ID:7
picturePath:image/8.png
answer:解:∵四边形ABCD为等腰梯形,∴AB∥CD,∴四边形EFGH为矩形,∴EH=FG,∵各边的中点分别是E,F,G,H,∴EH=\frac{1}{2}AC=\frac{1}{2}×10=5,∴矩形EFGH的周长=2(EH+FG)=2×5×2=20(米)．故选:B．
answer:20
correct answer:20
total:8  correct:4  acc:0.5
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ID:8
picturePath:image/9.png
answer:7米
answer:7
correct answer:6
total:9  correct:4  acc:0.4444444444444444
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ID:9
picturePath:image/11.png
answer:14
answer:14
correct answer:22
total:10  correct:4  acc:0.4
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ID:10
picturePath:image/13.png
answer:解:∵OC⊥OD,∴∠COD=90°,∵∠COA=36°,∴∠DOB=90°-36°=54°．故选:B．
answer:54
correct answer:54
total:11  correct:5  acc:0.45454545454545453
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ID:11
picturePath:image/14.png
answer:根据平行四边形的性质，平行四边形的对角线互相平分，因此OE是BD的中垂线，所以BE=DE。又因为OE⊥BD，所以AE=BE=DE。所以△ABE的周长=AB+AE+BE=AB+AE+DE=AB+AD=20cm。
answer:20
correct answer:10
total:12  correct:5  acc:0.4166666666666667
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ID:12
picturePath:image/17.png
answer:The figure shows a triangle ABC with its sides labeled as AB, AC, and BC. The lengths of these sides are given as AB = AC and BC = 5. The point D is the midpoint of AB, and the point E is the midpoint of AC. The line DE is the perpendicular bisector of AB. Given that the perimeter of triangle ABC is 21, we can find the length of side BC by subtracting the lengths of AB and AC from the perimeter: BC = 21 - (AB + AC) = 21 - (AB + AB) = 21 - 2AB. Since AB = AC, we can substitute AB for AC in the equation: BC = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 2AB = 21 - 
answer:21
correct answer:13
total:13  correct:5  acc:0.38461538461538464
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ID:13
picturePath:image/21.png
answer:20
answer:20
correct answer:40
total:14  correct:5  acc:0.35714285714285715
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ID:14
picturePath:image/22.png
answer:20
answer:20
correct answer:125
total:15  correct:5  acc:0.3333333333333333
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ID:15
picturePath:image/23.png
answer:Since arc BC is congruent to arc BD, we know that the angles intercepted by these arcs are also congruent. Therefore, ∠BOD = 2∠A. Given that ∠A = 25°, we can calculate ∠BOD as follows: ∠BOD = 2∠A = 2 * 25° = 50°. Thus, the angle ∠BOD is 50 degrees.
answer:50
correct answer:50
total:16  correct:6  acc:0.375
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ID:16
picturePath:image/24.png
answer:100°
answer:100
correct answer:100
total:17  correct:7  acc:0.4117647058823529
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ID:17
picturePath:image/25.png
answer:∠A+∠B=180°-∠AOB=180°-100°=80°
answer:80
correct answer:50
total:18  correct:7  acc:0.3888888888888889
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ID:18
picturePath:image/26.png
answer:3.75m.
answer:3.75
correct answer:5
total:19  correct:7  acc:0.3684210526315789
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ID:19
picturePath:image/27.png
answer:解:∵AD是⊙O的直径,∴∠ACD=90°,∵∠AOC=2∠ABC,∠ABC=25°,∴∠AOC=50°,∴∠CAD=90°-∠AOC=90°-50°=40°．故选:B．
answer:40
correct answer:65
total:20  correct:7  acc:0.35
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ID:20
picturePath:image/29.png
answer:解:如图,∵AB∥CD,∴∠1=∠3=62°,∵∠2+∠3=90°,∴∠2=90°-∠3=90°-62°=28°．故选:B．
answer:28
correct answer:56
total:21  correct:7  acc:0.3333333333333333
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ID:21
picturePath:image/30.png
answer:15
answer:15
correct answer:25
total:22  correct:7  acc:0.3181818181818182
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ID:22
picturePath:image/32.png
answer:100
answer:100
correct answer:50
total:23  correct:7  acc:0.30434782608695654
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ID:23
picturePath:image/35.png
answer:The answer is 20.
answer:20
correct answer:40
total:24  correct:7  acc:0.2916666666666667
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ID:24
picturePath:image/37.png
answer:解:∵AB=AC,∠A=36°,∴∠ABC=∠C=72°,∵BD是AC边上的高,∴∠CBD=90°,∴∠DBC=180°-90°-72°=18°．故选:A．
answer:18
correct answer:18
total:25  correct:8  acc:0.32
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ID:25
picturePath:image/38.png
answer:解:∵OC⊥AB,在Rt△OCD中,CD=8,OC=5,∴OD=√{CD²-OC²}=√{8²-5²}=5√{3},∴AB=2OD=2×5√{3}=10√{3}．故选:C．
answer:3
correct answer:8
total:26  correct:8  acc:0.3076923076923077
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ID:26
picturePath:image/39.png
answer:78° ÷ 2 = 39°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°, ∠BAC = 180° - ∠BOC = 180° - 78° = 102°
answer:102
correct answer:39
total:27  correct:8  acc:0.2962962962962963
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ID:27
picturePath:image/40.png
answer:解:∵AB是⊙O的直径,∴∠ADB=90°,∵∠ABD=58°,∴∠A=90°-∠ABD=32°,∴∠BCD=∠A=32°．故选:B．
answer:32
correct answer:32
total:28  correct:9  acc:0.32142857142857145
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ID:28
picturePath:image/42.png
answer:100
answer:100
correct answer:30
total:29  correct:9  acc:0.3103448275862069
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ID:29
picturePath:image/43.png
answer:5
answer:5
correct answer:25
total:30  correct:9  acc:0.3
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ID:30
picturePath:image/44.png
answer:The answer is 45.
answer:45
correct answer:112.5
total:31  correct:9  acc:0.2903225806451613
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ID:31
picturePath:image/45.png
answer:8cm
answer:8
correct answer:8
total:32  correct:10  acc:0.3125
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ID:32
picturePath:image/46.png
answer:40
answer:40
correct answer:40
total:33  correct:11  acc:0.3333333333333333
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ID:33
picturePath:image/48.png
answer:根据题意，直线a∥b，则∠1=∠3（平行线的性质）。又因为∠1=55°，所以∠3=55°。在△ABC中，∠2+∠3=180°，所以∠2=180°-∠3=180°-55°=125°。故选：B。
answer:125
correct answer:55
total:34  correct:11  acc:0.3235294117647059
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ID:34
picturePath:image/49.png
answer:解:连接OD、OE,∵⊙O与AB、AC相切,切点分别为D、E,∴OD⊥AB,OE⊥AC,∴∠DOB=∠DOE=90°,∵点O为BC的中点,∴OD⊥BC,∴∠ODM=∠ONE=90°,∴∠MND=∠DOE+∠ONE=180°,∴∠MND=180°．故选:A．
answer:180
correct answer:22.5
total:35  correct:11  acc:0.3142857142857143
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ID:35
picturePath:image/50.png
answer:40°
answer:40
correct answer:40
total:36  correct:12  acc:0.3333333333333333
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ID:36
picturePath:image/51.png
answer:∠C = 180° - ∠B = 180° - 20° = 160°.
answer:160
correct answer:20
total:37  correct:12  acc:0.32432432432432434
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ID:37
picturePath:image/52.png
answer:60
answer:60
correct answer:30
total:38  correct:12  acc:0.3157894736842105
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ID:38
picturePath:image/53.png
answer:100°
answer:100
correct answer:100
total:39  correct:13  acc:0.3333333333333333
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ID:39
picturePath:image/54.png
answer:15°
answer:15
correct answer:15
total:40  correct:14  acc:0.35
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ID:40
picturePath:image/55.png
answer:∠BEC=100°,ED平分∠BEC,所以∠BED=\frac{1}{2}∠BEC=\frac{1}{2}×100°=50°.又因为AB∥CD,所以∠D=∠BED=50°.故选B.
answer:50
correct answer:60
total:41  correct:14  acc:0.34146341463414637
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ID:41
picturePath:image/56.png
answer:由AB∥CD,∠CED=90°,根据两直线平行,同旁内角互补,可得∠AEC+∠CED=180°,∴∠AEC=180°-90°=90°,∵∠AEC=∠D,∴∠D=90°,故选B.
answer:90
correct answer:55
total:42  correct:14  acc:0.3333333333333333
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ID:42
picturePath:image/57.png
answer:The answer is A.
ID:43
picturePath:image/58.png
answer:根据题意，AB平行于CD，所以∠BAD=∠DCB=70°。又因为AD平分∠BAC，所以∠ADC=\frac{1}{2}∠BAC=\frac{1}{2}(180°-∠BAD)=65°。所以∠ACD=180°-∠ADC-∠DCB=180°-65°-70°=45°。故选B。
answer:45
correct answer:40
total:44  correct:14  acc:0.3181818181818182
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ID:44
picturePath:image/59.png
answer:Since \(AB \parallel CD\), we can use the corresponding angles postulate, which states that if two parallel lines are cut by a transversal, then the corresponding angles are congruent. Therefore, \(∠1 = ∠2\). Given that \(∠2 = 135°\), we can conclude that \(∠1 = 135°\).
answer:135
correct answer:45
total:45  correct:14  acc:0.3111111111111111
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ID:45
picturePath:image/60.png
answer:解:∵∠OBC=50°,∴∠A=\frac{1}{2}∠OBC=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:40
total:46  correct:14  acc:0.30434782608695654
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ID:46
picturePath:image/61.png
answer:60
answer:60
correct answer:25
total:47  correct:14  acc:0.2978723404255319
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ID:47
picturePath:image/63.png
answer:40
answer:40
correct answer:130
total:48  correct:14  acc:0.2916666666666667
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ID:48
picturePath:image/64.png
answer:解:∵∠C=90°,AC=4㎝,AB=7㎝,∴BC=√{AC^{2}+AB^{2}}=√{4^{2}+7^{2}}=5(厘米),∵AD平分∠BAC,∴∠BAD=∠DAE,∵DE⊥AB,∴∠BAE=90°,∴∠DAE=∠BAE-∠BAD=90°-∠BAD,∴∠DAE=∠BAD,∴AD=BD,∴EB=BC-BD=BC-AD=5-4=1(厘米)．故选:A．
answer:1
correct answer:3
total:49  correct:14  acc:0.2857142857142857
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ID:49
picturePath:image/65.png
answer:解:∵A、C、B是⊙O上三点,∠AOC=40°,∴∠ABC=\frac{1}{2}∠AOC=\frac{1}{2}×40°=20°．故选:B．
answer:20
correct answer:20
total:50  correct:15  acc:0.3
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ID:50
picturePath:image/66.png
answer:根据圆周角定理，同弧所对的圆周角相等，因此∠ACB=∠ABO=30°。故选A。
answer:30
correct answer:60
total:51  correct:15  acc:0.29411764705882354
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ID:51
picturePath:image/69.png
answer:解:∵∠BAC=50°,∴∠D=\frac{1}{2}∠BAC=\frac{1}{2}×50°=25°,∵AB为⊙O的直径,∴∠ADB=90°,∴∠ADC=90°-∠D=90°-25°=65°．故选:A．
answer:65
correct answer:40
total:52  correct:15  acc:0.28846153846153844
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ID:52
picturePath:image/70.png
answer:根据三角形全等，我们有∠OAD=∠OBC。已知∠O=70°，∠C=25°，所以∠OBC=∠O-∠C=70°-25°=45°。因此，∠OAD=45°。
answer:45
correct answer:85
total:53  correct:15  acc:0.2830188679245283
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ID:53
picturePath:image/72.png
answer:由于∠ABC和∠AOC是弧AB所对的角，根据圆周角定理，同弧所对的圆周角相等，因此∠AOC=2∠ABC。将∠ABC=40°代入，得到∠AOC=2×40°=80°。所以，∠AOC的度数为80°。
answer:80
correct answer:80
total:54  correct:16  acc:0.2962962962962963
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ID:54
picturePath:image/75.png
answer:解:∵BO、CO分别平分∠ABC与∠ACB,∴∠ABO=∠BCO,∠ACO=∠ACB,∵MN∥BC,∴∠ABO=∠BCO,∠ACO=∠ACB,∴△ABO∽△BCO,△ACO∽△ACB,∴\frac{AB}{BC}=\frac{BO}{CO},\frac{AC}{CB}=\frac{AO}{CO},∵AB=36,AC=24,∴\frac{36}{BC}=\frac{BO}{CO},\frac{24}{CB}=\frac{AO}{CO},解得:AO:CO=2:3,BO:CO=3:2,∴AO:BO:CO=6:4:3,设AO=6x,B0=4x,CO=3x,∵MN∥BC,∴△AMN∽△ABC,∴\frac{AM}{AB}=\frac{AN}{AC},∴\frac{AM}{36}=\frac{AN}{24},解得:AM:AN=2:3,∴△AMN的周长=AM+AN+MN=6x+4x+3x=13x,当x取最大值6时,△AMN的周长最大为13×6=78．故选:B．
answer:78
correct answer:60
total:55  correct:16  acc:0.2909090909090909
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ID:55
picturePath:image/78.png
answer:5
answer:5
correct answer:130
total:56  correct:16  acc:0.2857142857142857
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ID:56
picturePath:image/79.png
answer:解:∵DE∥AB,∠ADE=46°,∴∠ADC=∠ADE=46°,∵∠C=90°,∴∠ADC+∠DAC=90°,∴∠DAC=90°-∠ADC=44°,∴∠B=∠DAC=44°．故选:B．
answer:44
correct answer:44
total:57  correct:17  acc:0.2982456140350877
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ID:57
picturePath:image/80.png
answer:解:∵△BDE∽△ACE,∴\frac{BD}{AB}=\frac{DC}{AC},∵BD=120m,DC=60m,EC=50m,∴\frac{120}{AB}=\frac{60}{60+50},解得AB=100m．故选:A．
answer:100
correct answer:100
total:58  correct:18  acc:0.3103448275862069
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ID:58
picturePath:image/81.png
answer:Since \(AB \parallel CD\), we know that the corresponding angles are congruent. Therefore, \(∠B = ∠ECD\). Given that \(CE = CD\) and \(∠D = 74°\), we can deduce that \(∠ECD = ∠D = 74°\). Hence, the correct answer is B, 74°.
answer:74
correct answer:32
total:59  correct:18  acc:0.3050847457627119
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ID:59
picturePath:image/82.png
answer:∠B = 180° - ∠1 - ∠A = 180° - 145° - 90° = 45°.
answer:45
correct answer:55
total:60  correct:18  acc:0.3
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ID:60
picturePath:image/83.png
answer:∠D = 180° - ∠B = 180° - 40° = 140°.
answer:140
correct answer:140
total:61  correct:19  acc:0.3114754098360656
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ID:61
picturePath:image/84.png
answer:根据题意，∠1和∠2是对顶角，∠1=120°，∠2=80°，所以∠1=∠2=80°。又因为a∥b，所以∠3=∠1=120°。所以∠3的度数是120°。
answer:120
correct answer:40
total:62  correct:19  acc:0.3064516129032258
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ID:62
picturePath:image/86.png
answer:∠ACD=∠BAE=40°.
answer:40
correct answer:140
total:63  correct:19  acc:0.30158730158730157
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ID:63
picturePath:image/87.png
answer:根据题意,AB∥CD,则∠A+∠EOD=180°,∠A=34°,则∠EOD=180°-34°=146°,又∵OF平分∠EOD,则∠EOF=\frac{1}{2}∠EOD=73°,又∵AB∥CD,则∠EOF=∠DOF,则∠DOF=73°,则∠COD=180°-∠DOF-∠DOF=180°-73°-73°=34°,故选:A.
answer:34
correct answer:68
total:64  correct:19  acc:0.296875
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ID:64
picturePath:image/88.png
answer:1200cm
answer:1200
correct answer:1200
total:65  correct:20  acc:0.3076923076923077
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ID:65
picturePath:image/89.png
answer:解:∵四边形ABCD为⊙O的内接四边形,∴∠A=∠BCD=110°,
由圆周角定理得,∠BAD=180°-∠A=70°,
故选:A．
answer:70
correct answer:70
total:66  correct:21  acc:0.3181818181818182
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ID:66
picturePath:image/90.png
answer:根据题意，AB∥CD，∠BED=40°，则∠BDC=∠BED=40°。又因为EA是∠CEB的平分线，所以∠CEB=2∠EAB。由三角形内角和定理得，∠A+∠EAB+∠BDC=180°，即∠A+2∠EAB+40°=180°，解得∠A=100°。故选B.
answer:100
correct answer:70
total:67  correct:21  acc:0.31343283582089554
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ID:67
picturePath:image/91.png
answer:15
answer:15
correct answer:37
total:68  correct:21  acc:0.3088235294117647
------------------------------------------------------------------------------------
ID:68
picturePath:image/92.png
answer:100°
answer:100
correct answer:40
total:69  correct:21  acc:0.30434782608695654
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ID:69
picturePath:image/93.png
answer:70°
answer:70
correct answer:110
total:70  correct:21  acc:0.3
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ID:70
picturePath:image/95.png
answer:解:∵AB∥CD,AB=160cm,∴△OAB∽△OCD,∴\frac{AB}{CD}=\frac{OA}{OC},即\frac{160}{CD}=\frac{200}{OC},解得:OC=80cm,即油的最大深度为80cm．故选:A．
answer:80
correct answer:40
total:71  correct:21  acc:0.29577464788732394
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ID:71
picturePath:image/96.png
answer:解:∵∠1=40°,∴∠3=∠1=40°,∵∠3+∠4=90°,∴∠4=90°-∠3=50°,∵直尺的两边互相平行,∴∠2=∠4=50°．故选:B．
answer:50
correct answer:50
total:72  correct:22  acc:0.3055555555555556
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ID:72
picturePath:image/97.png
answer:解:如图,连接AC,∵AD∥BC,∴∠A+∠ACB=180°,∵∠A=120°,∠B=150°,∴∠ACB=150°-120°=30°,∵AD∥BC,∴∠C=∠ACB=30°．故选:B．
answer:30
correct answer:150
total:73  correct:22  acc:0.3013698630136986
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ID:73
picturePath:image/98.png
answer:圆锥模具的母线长为10cm,底面半径为5cm,则这个圆锥模具的侧面积是10π×5=50πcm².
answer:50
correct answer:50
total:74  correct:23  acc:0.3108108108108108
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ID:74
picturePath:image/99.png
answer:46米
answer:46
correct answer:46
total:75  correct:24  acc:0.32
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ID:75
picturePath:image/100.png
answer:20
answer:20
correct answer:40
total:76  correct:24  acc:0.3157894736842105
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ID:76
picturePath:image/101.png
answer:100
answer:100
correct answer:55
total:77  correct:24  acc:0.3116883116883117
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ID:77
picturePath:image/102.png
answer:76°
answer:76
correct answer:76
total:78  correct:25  acc:0.32051282051282054
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ID:78
picturePath:image/103.png
answer:解:∵AB与⊙O相切于点B,∴∠OBA=90°,∵∠A=36°,∴∠AOB=90°-36°=54°,∴∠C=∠AOB=54°．故选:B．
answer:54
correct answer:27
total:79  correct:25  acc:0.31645569620253167
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ID:79
picturePath:image/104.png
answer:解:∵四边形ABCD是圆内接四边形,
∴∠A+∠C=180°,
又∵∠C=80°,
∴∠A=180°-80°=100°．
故选:B．
answer:100
correct answer:100
total:80  correct:26  acc:0.325
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ID:80
picturePath:image/105.png
answer:20°
answer:20
correct answer:115
total:81  correct:26  acc:0.32098765432098764
------------------------------------------------------------------------------------
ID:81
picturePath:image/106.png
answer:25
answer:25
correct answer:25
total:82  correct:27  acc:0.32926829268292684
------------------------------------------------------------------------------------
ID:82
picturePath:image/107.png
answer:解:∵AB∥CD,∠1=56°,∴∠2=∠1=56°．故选:A．
answer:56
correct answer:56
total:83  correct:28  acc:0.3373493975903614
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ID:83
picturePath:image/108.png
answer:解:∵AD⊥BC,∴⁀{AB}=⁀{AC},∵∠D=36°,∴∠BAD=\frac{1}{2}∠D=18°．故选:A．
answer:18
correct answer:54
total:84  correct:28  acc:0.3333333333333333
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ID:84
picturePath:image/109.png
answer:70°
answer:70
correct answer:110
total:85  correct:28  acc:0.32941176470588235
------------------------------------------------------------------------------------
ID:85
picturePath:image/110.png
answer:∠2等于50°,因为∠1和∠2是邻补角,所以∠1+∠2=180°,所以∠2=180°-∠1=180°-50°=130°.
answer:130
correct answer:50
total:86  correct:28  acc:0.32558139534883723
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ID:86
picturePath:image/111.png
answer:解:∵AB是△ABC外接圆的直径,∴∠ACB=90°,∵∠A=35°,∴∠B=90°-∠A=55°．故选:A．
answer:55
correct answer:55
total:87  correct:29  acc:0.3333333333333333
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ID:87
picturePath:image/112.png
answer:如图,∵AB//CD,∠1=50°,∴∠3=∠1=50°,∵∠2+∠3=90°,∴∠2=90°-50°=40°.故选B.
answer:40
correct answer:130
total:88  correct:29  acc:0.32954545454545453
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ID:88
picturePath:image/113.png
answer:∠MNB的度数等于四边形ABCD的内角和减去∠B和∠C的度数，即360° - 70° - 80° = 210° - 70° - 80° = 60°。因此，∠MNB的度数为60°。
answer:60
correct answer:95
total:89  correct:29  acc:0.3258426966292135
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ID:89
picturePath:image/118.png
answer:解:∵∠ABC的平分线与∠ACB的外角平分线相交于D点,∴∠1=\frac{1}{2}∠ABC,∠2=\frac{1}{2}∠ACB,∴∠1+∠2=50°,∴∠D=25°．故选:A．
answer:25
correct answer:25
total:90  correct:30  acc:0.3333333333333333
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ID:90
picturePath:image/119.png
answer:解:∵△ABC≌△DCB,∴BD=AC=9cm．故选:C．
answer:9
correct answer:9
total:91  correct:31  acc:0.34065934065934067
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ID:91
picturePath:image/120.png
answer:由三角形全等可得∠B=∠D,∠C=∠D,即∠D=70°,∠D=30°,∠DAC=35°,根据三角形内角和定理可得∠EAC=180°-70°-30°-35°=65°.故选B
answer:65
correct answer:45
total:92  correct:31  acc:0.33695652173913043
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ID:92
picturePath:image/121.png
answer:5
answer:5
correct answer:3
total:93  correct:31  acc:0.3333333333333333
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ID:93
picturePath:image/122.png
answer:360
answer:360
correct answer:360
total:94  correct:32  acc:0.3404255319148936
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ID:94
picturePath:image/123.png
answer:根据题意,∠C=90°,AC=BC,AD平分∠CAB交BC于D,DE⊥AB于E,若AB=6cm,则AD=DB=\frac{1}{2}AB=\frac{1}{2}×6=3cm,DE=BE,所以△DBE的周长=DB+BE+DE=3+3+3=9cm.故选B.
answer:9
correct answer:6
total:95  correct:32  acc:0.3368421052631579
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ID:95
picturePath:image/124.png
answer:60
answer:60
correct answer:10
total:96  correct:32  acc:0.3333333333333333
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ID:96
picturePath:image/125.png
answer:根据三角形全等,有AB=BD,AD=CD,则BC=BD-AD=5-4=1(厘米)．故选:A．
answer:1
correct answer:6
total:97  correct:32  acc:0.32989690721649484
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ID:97
picturePath:image/127.png
answer:根据题意,AB是⊙O的弦,半径OC⊥AB于点D,且AB=6cm,OD=4cm,则CD=\frac{1}{2}AB=\frac{1}{2}×6=3cm．故选:A．
answer:3
correct answer:1
total:98  correct:32  acc:0.32653061224489793
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ID:98
picturePath:image/128.png
answer:根据圆周角定理，同弧所对的圆周角的度数等于它所对的圆心角度数的一半。因此，∠CBD的度数为30°÷2=15°。
answer:15
correct answer:60
total:99  correct:32  acc:0.32323232323232326
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ID:99
picturePath:image/129.png
answer:5√2cm
answer:2
correct answer:10
total:100  correct:32  acc:0.32
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ID:100
picturePath:image/130.png
answer:根据三角形的一个外角等于与它不相邻的两个内角的和，可得∠BOC=\frac{1}{2}∠A=\frac{1}{2}×120°=60°．故选B．
answer:60
correct answer:150
total:101  correct:32  acc:0.31683168316831684
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ID:101
picturePath:image/131.png
answer:8cm
answer:8
correct answer:8
total:102  correct:33  acc:0.3235294117647059
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ID:102
picturePath:image/132.png
answer:20
answer:20
correct answer:70
total:103  correct:33  acc:0.32038834951456313
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ID:103
picturePath:image/133.png
answer:Since CD is perpendicular to AB, we have ∠ADC = 90°. Since ∠DAB = 65°, we have ∠BAC = 180° - ∠DAB - ∠ADC = 180° - 65° - 90° = 25°. Since AB is a diameter, we have ∠AOC = 90°. Therefore, ∠AOC = 90° - ∠BAC = 90° - 25° = 65°.The answer is: 65
answer:65
correct answer:50
total:104  correct:33  acc:0.3173076923076923
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ID:104
picturePath:image/134.png
answer:70
answer:70
correct answer:70
total:105  correct:34  acc:0.3238095238095238
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ID:105
picturePath:image/136.png
answer:根据题意，∠A+∠B=200°，由于∠ADC和∠DCB的平分线相交于点O，因此∠AOC和∠BOC的和等于∠ADC和∠DCB的一半，即∠AOC+∠BOC=\frac{1}{2}(∠ADC+∠DCB)=\frac{1}{2}(∠A+∠B)=100°。又因为四边形ABCD是四边形，所以∠AOC+∠BOC+∠COD=360°，将∠AOC+∠BOC=100°代入，得到∠COD=360°-100°=260°。所以，∠COD的度数是260°。
answer:260
correct answer:100
total:106  correct:34  acc:0.32075471698113206
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ID:106
picturePath:image/137.png
answer:100°-20°-20°=60°,∠ADB′=60°.
answer:60
correct answer:40
total:107  correct:34  acc:0.3177570093457944
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ID:107
picturePath:image/138.png
answer:根据题意,∠C=70°,∠ABC=48°,∵AD为△ABC的角平分线,∴∠BAD=\frac{1}{2}∠C=35°,∠ABD=\frac{1}{2}∠ABC=24°,∵BE为△ABC的高,∴∠ABE=90°,∴∠3=∠ABE-∠ABD=90°-24°=66°.故选B.
answer:66
correct answer:59
total:108  correct:34  acc:0.3148148148148148
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ID:108
picturePath:image/139.png
answer:根据题意，DE是边AC的垂直平分线，所以AD=CD。在△ABC中，BC=18cm，AB=10cm，所以AC=BC+AB=18+10=28cm。由于AD=CD，所以△ABD的周长=AB+AD+BD=AB+AD+CD=AB+AC=10+28=38cm。所以，△ABD的周长为38cm。
answer:38
correct answer:28
total:109  correct:34  acc:0.3119266055045872
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ID:109
picturePath:image/140.png
answer:30
answer:30
correct answer:75
total:110  correct:34  acc:0.3090909090909091
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ID:110
picturePath:image/141.png
answer:The answer is 60.
answer:60
correct answer:120
total:111  correct:34  acc:0.3063063063063063
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ID:111
picturePath:image/142.png
answer:12.5°
answer:12.5
correct answer:115
total:112  correct:34  acc:0.30357142857142855
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ID:112
picturePath:image/148.png
answer:90
answer:90
correct answer:70
total:113  correct:34  acc:0.3008849557522124
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ID:113
picturePath:image/150.png
answer:36°
answer:36
correct answer:72
total:114  correct:34  acc:0.2982456140350877
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ID:114
picturePath:image/151.png
answer:The answer is A.
ID:115
picturePath:image/152.png
answer:根据三角形的一个外角等于与它不相邻的两个内角的和，可得∠D=∠1+∠2=90°+25°=115°.
answer:115
correct answer:25
total:116  correct:34  acc:0.29310344827586204
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ID:116
picturePath:image/153.png
answer:36°
answer:36
correct answer:40
total:117  correct:34  acc:0.2905982905982906
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ID:117
picturePath:image/154.png
answer:解:连接OC,∵PC是切线,∴∠OCP=90°,∵∠AOC=35°,∴∠P=90°-35°=55°．故选:B．
answer:55
correct answer:20
total:118  correct:34  acc:0.288135593220339
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ID:118
picturePath:image/155.png
answer:40°
answer:40
correct answer:40
total:119  correct:35  acc:0.29411764705882354
------------------------------------------------------------------------------------
ID:119
picturePath:image/156.png
answer:30
answer:30
correct answer:110
total:120  correct:35  acc:0.2916666666666667
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ID:120
picturePath:image/157.png
answer:20
answer:20
correct answer:20
total:121  correct:36  acc:0.2975206611570248
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ID:121
picturePath:image/158.png
answer:50°
answer:50
correct answer:130
total:122  correct:36  acc:0.29508196721311475
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ID:122
picturePath:image/159.png
answer:The answer is D.
ID:123
picturePath:image/160.png
answer:解:∵⊙O是△ABC的外接圆,∠OAB=40°,∴∠ACB=∠OAB=40°．故选:B．
answer:40
correct answer:50
total:124  correct:36  acc:0.2903225806451613
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ID:124
picturePath:image/161.png
answer:25°
answer:25
correct answer:115
total:125  correct:36  acc:0.288
------------------------------------------------------------------------------------
ID:125
picturePath:image/162.png
answer:35°
answer:35
correct answer:40
total:126  correct:36  acc:0.2857142857142857
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ID:126
picturePath:image/164.png
answer:The answer is 35°.
answer:35
correct answer:55
total:127  correct:36  acc:0.28346456692913385
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ID:127
picturePath:image/165.png
answer:∠2 = ∠1 + ∠F - ∠EDF = 70° - 90° = -20°.
answer:20
correct answer:20
total:128  correct:37  acc:0.2890625
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ID:128
picturePath:image/166.png
answer:解:∵PA、PB分别与⊙O相切于A、B两点,∴∠OAP=∠OBP=90°,∴∠AOB=360°-∠OAP-∠OBP-∠P=360°-90°-90°-50°=120°．故选:D．
answer:120
correct answer:155
total:129  correct:37  acc:0.2868217054263566
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ID:129
picturePath:image/167.png
answer:40°
answer:40
correct answer:140
total:130  correct:37  acc:0.2846153846153846
------------------------------------------------------------------------------------
ID:130
picturePath:image/168.png
answer:20m
answer:20
correct answer:20
total:131  correct:38  acc:0.2900763358778626
------------------------------------------------------------------------------------
ID:131
picturePath:image/170.png
answer:0.5m
answer:0.5
correct answer:4
total:132  correct:38  acc:0.2878787878787879
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ID:132
picturePath:image/171.png
answer:The answer is 35.
answer:35
correct answer:20
total:133  correct:38  acc:0.2857142857142857
------------------------------------------------------------------------------------
ID:133
picturePath:image/172.png
answer:60°
answer:60
correct answer:120
total:134  correct:38  acc:0.2835820895522388
------------------------------------------------------------------------------------
ID:134
picturePath:image/173.png
answer:60
answer:60
correct answer:35
total:135  correct:38  acc:0.2814814814814815
------------------------------------------------------------------------------------
ID:135
picturePath:image/174.png
answer:30
answer:30
correct answer:60
total:136  correct:38  acc:0.27941176470588236
------------------------------------------------------------------------------------
ID:136
picturePath:image/175.png
answer:30
answer:30
correct answer:75
total:137  correct:38  acc:0.2773722627737226
------------------------------------------------------------------------------------
ID:137
picturePath:image/176.png
answer:20°
answer:20
correct answer:80
total:138  correct:38  acc:0.2753623188405797
------------------------------------------------------------------------------------
ID:138
picturePath:image/177.png
answer:22°
answer:22
correct answer:52
total:139  correct:38  acc:0.2733812949640288
------------------------------------------------------------------------------------
ID:139
picturePath:image/178.png
answer:解:∵PA和PB是⊙O的切线,∴∠OAP=∠OBP=90°,∵∠P=40°,∴∠AOB=90°-40°=50°,∵AC是⊙O的直径,∴∠C=90°,∴∠ACB=90°-50°=40°．故选:A．
answer:40
correct answer:70
total:140  correct:38  acc:0.2714285714285714
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ID:140
picturePath:image/179.png
answer:解:∵AB是⊙O的直径,∴∠ADB=90°,∵∠C=20°,∴∠ABD=90°-∠C=70°．故选:B．
answer:70
correct answer:70
total:141  correct:39  acc:0.2765957446808511
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ID:141
picturePath:image/181.png
answer:解:∵AB⊥CD,∴⁀{AC}=⁀{BD},∴∠ACD=\frac{1}{2}∠CDB=\frac{1}{2}×62°=31°．故选:A．
answer:31
correct answer:28
total:142  correct:39  acc:0.2746478873239437
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ID:142
picturePath:image/183.png
answer:解:∵DE//BC,∠1=155°,∴∠C=∠1=155°,∵∠A=90°,∴∠B=180°-∠A-∠C=180°-90°-155°=35°．故选:A．
answer:35
correct answer:65
total:143  correct:39  acc:0.2727272727272727
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ID:143
picturePath:image/185.png
answer:20
answer:20
correct answer:40
total:144  correct:39  acc:0.2708333333333333
------------------------------------------------------------------------------------
ID:144
picturePath:image/186.png
answer:解:∵△ABC是⊙O的内接三角形,∠OAB=35°,∴∠ACB=\frac{1}{2}∠OAB=\frac{1}{2}×35°=17.5°．故选:A．
answer:17.5
correct answer:55
total:145  correct:39  acc:0.2689655172413793
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ID:145
picturePath:image/188.png
answer:60°
answer:60
correct answer:60
total:146  correct:40  acc:0.273972602739726
------------------------------------------------------------------------------------
ID:146
picturePath:image/189.png
answer:20
answer:20
correct answer:80
total:147  correct:40  acc:0.272108843537415
------------------------------------------------------------------------------------
ID:147
picturePath:image/190.png
answer:The answer is A.
ID:148
picturePath:image/191.png
answer:The answer is A.
ID:149
picturePath:image/192.png
answer:圆锥模具的侧面积公式为：侧面积 = π * 底面半径 * 母线长。将已知数值代入公式，得到：侧面积 = π * 5 * 10 = 50π。因此，该圆锥模具的侧面积为50π。
answer:50
correct answer:50
total:150  correct:41  acc:0.2733333333333333
------------------------------------------------------------------------------------
ID:150
picturePath:image/193.png
answer:50
answer:50
correct answer:65
total:151  correct:41  acc:0.271523178807947
------------------------------------------------------------------------------------
ID:151
picturePath:image/194.png
answer:50°
answer:50
correct answer:20
total:152  correct:41  acc:0.26973684210526316
------------------------------------------------------------------------------------
ID:152
picturePath:image/196.png
answer:25
answer:25
correct answer:25
total:153  correct:42  acc:0.27450980392156865
------------------------------------------------------------------------------------
ID:153
picturePath:image/198.png
answer:∠CAD = ∠B + ∠C = 40° + 30° = 70°.
answer:70
correct answer:70
total:154  correct:43  acc:0.2792207792207792
------------------------------------------------------------------------------------
ID:154
picturePath:image/199.png
answer:根据对角线互相平分定理，∠COD等于∠BOC。根据三角形内角和定理，∠BOC=180°-∠B-∠C=180°-23°-42°=115°。所以，∠COD=∠BOC=115°。
answer:115
correct answer:65
total:155  correct:43  acc:0.27741935483870966
------------------------------------------------------------------------------------
ID:155
picturePath:image/200.png
answer:22°
answer:22
correct answer:110
total:156  correct:43  acc:0.27564102564102566
------------------------------------------------------------------------------------
ID:156
picturePath:image/201.png
answer:根据三角形内角和定理，∠BAC + ∠ABC + ∠ACB = 180°，∠BAC = 44°，∠CBD = 2∠BDC，∠CBD + ∠BDC = 180° - ∠BAC，∠BDC = 180° - ∠BAC - ∠CBD = 180° - 44° - 2∠BDC = 136° - ∠BDC，解得∠BDC = 68°，∠CBD = 136°，∠ABC = 180° - ∠BDC - ∠CBD = 180° - 68° - 136° = 76°，∵AB=AC=AD，∴∠ABC = ∠ACD = 76°，∴∠CAD = 180° - ∠ACD - ∠BAC = 180° - 76° - 44° = 62°.
answer:62
correct answer:88
total:157  correct:43  acc:0.27388535031847133
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ID:157
picturePath:image/202.png
answer:由平行线的性质可得∠1=∠5,∠2=∠6,∠4=∠7,∴∠5=158°,∠6=42°,∠7=50°,∵∠5+∠6+∠7+∠3=180°,∴∠3=180°-∠5-∠6-∠7=180°-158°-42°-50°=30°,故选A.
answer:30
correct answer:70
total:158  correct:43  acc:0.2721518987341772
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ID:158
picturePath:image/203.png
answer:40°
answer:40
correct answer:50
total:159  correct:43  acc:0.27044025157232704
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ID:159
picturePath:image/204.png
answer:The answer is A.
ID:160
picturePath:image/205.png
answer:The answer is 35.
answer:35
correct answer:40
total:161  correct:43  acc:0.2670807453416149
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ID:161
picturePath:image/206.png
answer:2
answer:2
correct answer:3
total:162  correct:43  acc:0.2654320987654321
------------------------------------------------------------------------------------
ID:162
picturePath:image/207.png
answer:40
answer:40
correct answer:25
total:163  correct:43  acc:0.26380368098159507
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ID:163
picturePath:image/208.png
answer:52°
answer:52
correct answer:52
total:164  correct:44  acc:0.2682926829268293
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ID:164
picturePath:image/209.png
answer:∠2 = 180° - ∠1 - ∠BEN = 180° - 63° - \frac{1}{2}∠BEN = 180° - 63° - \frac{1}{2}(180° - ∠BEF) = 180° - 63° - \frac{1}{2}(180° - 63°) = 180° - 63° - \frac{1}{2}×117° = 180° - 63° - 58.5° = 180° - 121.5° = 58.5°.
answer:58.5
correct answer:54
total:165  correct:44  acc:0.26666666666666666
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ID:165
picturePath:image/210.png
answer:Since AB is parallel to CD, we have \begin{align*}
\angle ABC + \angle D &= 180^\circ \\
\angle ABC &= 180^\circ - \angle D \\
&= 180^\circ - 65^\circ \\
&= \boxed{115^\circ}.
\end{align*}The answer is: 115
answer:115
correct answer:25
total:166  correct:44  acc:0.26506024096385544
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ID:166
picturePath:image/211.png
answer:24°
answer:24
correct answer:24
total:167  correct:45  acc:0.2694610778443114
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ID:167
picturePath:image/212.png
answer:20°
answer:20
correct answer:80
total:168  correct:45  acc:0.26785714285714285
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ID:168
picturePath:image/213.png
answer:The answer is A.
ID:169
picturePath:image/215.png
answer:解:∵∠1+∠B=180°,∠B+∠2+∠D=180°,∠2=45°,∴∠D=180°-∠1-(180°-45°)=∠1-135°,∵∠1+∠D=180°,∴∠D=180°-∠1,∴∠D=45°．故选:A．
answer:45
correct answer:45
total:170  correct:46  acc:0.27058823529411763
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ID:170
picturePath:image/216.png
answer:解:∵AB∥CD,∠AMN=56°,∴∠MND=∠AMN=56°,∵NH是∠MND的角平分线,∴∠MNH=\frac{1}{2}∠MND=\frac{1}{2}×56°=28°．故选:B．
answer:28
correct answer:28
total:171  correct:47  acc:0.27485380116959063
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ID:171
picturePath:image/217.png
answer:根据圆内接四边形的性质,∠DAB+∠BCD=180°,又∵∠DAB=60°,∴∠BCD=180°-60°=120°.故选:D.
answer:120
correct answer:120
total:172  correct:48  acc:0.27906976744186046
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ID:172
picturePath:image/218.png
answer:解:∵∠DOB=140°,∴∠BOC=70°,∴∠ACD=\frac{1}{2}∠BOC=35°．故选:A．
answer:35
correct answer:20
total:173  correct:48  acc:0.2774566473988439
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ID:173
picturePath:image/219.png
answer:50°
answer:50
correct answer:130
total:174  correct:48  acc:0.27586206896551724
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ID:174
picturePath:image/220.png
answer:∠2 = 180° - ∠1 = 180° - 108° = 72°.
answer:72
correct answer:72
total:175  correct:49  acc:0.28
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ID:175
picturePath:image/221.png
answer:由AC∥ED,则∠C+∠CBE=∠CED,∵∠C=26°,∠CBE=37°,∴∠CED=∠CBE-∠C=37°-26°=11°,∴∠BED=∠CED=11°.故选B
answer:11
correct answer:63
total:176  correct:49  acc:0.2784090909090909
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ID:176
picturePath:image/222.png
answer:解:∵四边形ABCD是菱形,∴AB=BC,∠BAC=∠DAC=28°,∵AM=CN,∴N是AB的垂直平分线,∴AN=CN,∴∠NAC=∠DAC=28°,∴∠NAM=∠DAC=28°,∴∠MAN=180°-∠NAM-∠NAM=122°,∵MN∥BC,∴∠OBC=∠MAN=122°,故选:B．
answer:122
correct answer:62
total:177  correct:49  acc:0.2768361581920904
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ID:177
picturePath:image/223.png
answer:5.
answer:5
correct answer:13
total:178  correct:49  acc:0.2752808988764045
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ID:178
picturePath:image/225.png
answer:Since \( \angle OBC = 42^\circ \) and \( \angle OBC \) and \( \angle A \) are alternate interior angles, they are congruent. Therefore, the measure of \( \angle A \) is also \( 42^\circ \).
answer:42
correct answer:48
total:179  correct:49  acc:0.2737430167597765
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ID:179
picturePath:image/226.png
answer:由于AB∥DE，根据平行线的性质，∠ABC和∠CDE互为同位角，所以∠ABC=∠CDE。已知∠ABC=70°，∠CDE=140°，所以∠BCD=∠CDE-∠ABC=140°-70°=70°。所以，∠BCD的值为70°。
answer:70
correct answer:30
total:180  correct:49  acc:0.2722222222222222
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ID:180
picturePath:image/227.png
answer:90°
answer:90
correct answer:135
total:181  correct:49  acc:0.27071823204419887
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ID:181
picturePath:image/229.png
answer:解:∵四边形ABCD是⊙O的内接四边形,∴∠B+∠D=180°,又∵∠B=70°,∴∠D=180°-70°=110°,
故选:B．
answer:110
correct answer:110
total:182  correct:50  acc:0.27472527472527475
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ID:182
picturePath:image/230.png
answer:根据题意，∠1=∠2，∠3=110°，则∠1=∠2=∠3=110°。又因为直线a,b与直线c,d相交，所以∠1+∠4=90°，∠3+∠4=90°。将∠3=110°代入∠3+∠4=90°，解得∠4=90°-∠3=90°-110°=-20°。因此，∠4的度数为-20°。
answer:20
correct answer:70
total:183  correct:50  acc:0.273224043715847
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ID:183
picturePath:image/231.png
answer:解:∵AB是⊙O的直径,∴∠ADB=90°,∵∠ABD=52°,∴∠A=90°-∠ABD=90°-52°=38°,∴∠BCD=∠A=38°．故选:B．
answer:38
correct answer:38
total:184  correct:51  acc:0.27717391304347827
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ID:184
picturePath:image/233.png
answer:根据题意，∠E=60°，∠AHG=50°，∠FMD是△AHG的外角，所以∠FMD=∠AHG+∠E=50°+60°=110°．故选：B．
answer:110
correct answer:20
total:185  correct:51  acc:0.2756756756756757
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ID:185
picturePath:image/234.png
answer:解:∵PA,PB分别与⊙O相切于A,B两点,∴∠OAP=∠OBP=90°,∴∠AOB=180°-∠C=115°,∴∠P=\frac{1}{2}∠AOB=115°×\frac{1}{2}=57.5°．故选:B．
answer:57.5
correct answer:50
total:186  correct:51  acc:0.27419354838709675
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ID:186
picturePath:image/235.png
answer:50°
answer:50
correct answer:130
total:187  correct:51  acc:0.2727272727272727
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ID:187
picturePath:image/236.png
answer:The answer is 72°.
answer:72
correct answer:18
total:188  correct:51  acc:0.2712765957446808
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ID:188
picturePath:image/237.png
answer:36°
answer:36
correct answer:72
total:189  correct:51  acc:0.2698412698412698
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ID:189
picturePath:image/239.png
answer:35°
answer:35
correct answer:35
total:190  correct:52  acc:0.2736842105263158
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ID:190
picturePath:image/240.png
answer:∠AEC = ∠A + ∠C (对顶角相等) = 46° + 27° = 73°.
answer:73
correct answer:73
total:191  correct:53  acc:0.2774869109947644
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ID:191
picturePath:image/242.png
answer:This is a geometry problem involving angles and parallel lines. Given that AD is parallel to CB, we can use the corresponding angles postulate, which states that when two lines are parallel, the angles that are formed by the transversal with each line are congruent. Therefore, since ∠D = 43° and ∠B = 25°, we can conclude that ∠DEB = ∠D = 43°. The answer is: A.
answer:43
correct answer:68
total:192  correct:53  acc:0.2760416666666667
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ID:192
picturePath:image/243.png
answer:70
answer:70
correct answer:70
total:193  correct:54  acc:0.27979274611398963
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ID:193
picturePath:image/244.png
answer:40
answer:40
correct answer:65
total:194  correct:54  acc:0.27835051546391754
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ID:194
picturePath:image/245.png
answer:由于AB∥CD，根据平行线的性质，内错角相等，所以∠C=∠A=25°。故选A。
answer:25
correct answer:65
total:195  correct:54  acc:0.27692307692307694
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ID:195
picturePath:image/246.png
answer:This is a geometry problem that involves the properties of parallel lines and angles. Since AB is parallel to CD, we can use the corresponding angles postulate, which states that if two parallel lines are cut by a transversal, then the corresponding angles are congruent. Therefore, ∠EMG = ∠BNE = 30°. The answer is A.
answer:30
correct answer:15
total:196  correct:54  acc:0.2755102040816326
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ID:196
picturePath:image/248.png
answer:80°
answer:80
correct answer:80
total:197  correct:55  acc:0.27918781725888325
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ID:197
picturePath:image/249.png
answer:This is a right triangle problem. We can use the Pythagorean theorem to solve it. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is the distance between the flagpole and the measuring point, which is 22 meters. One of the other sides is the distance between the measuring point and the base of the flagpole, which is 8 meters. The other side is the height of the flagpole, which we want to find. So, we have:

22^2 = 8^2 + h^2

where h is the height of the flagpole. Simplifying this equation, we get:

484 = 64 + h^2

Subtracting 64 from both sides, we get:

420 = h^2

Taking the square root of both sides, we get:

h = √420

Therefore, the height of the flagpole is approximately 20.9 meters. The answer is A.
answer:20.9
correct answer:12
total:198  correct:55  acc:0.2777777777777778
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ID:198
picturePath:image/250.png
answer:The answer is A.
ID:199
picturePath:image/251.png
answer:60
answer:60
correct answer:30
total:200  correct:55  acc:0.275
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ID:200
picturePath:image/253.png
answer:由于AB∥CD，根据平行线的性质，∠ABC=∠C=34°。又因为BC平分∠ABE，所以∠ABE=2∠ABC=2×34°=68°。最后，∠BED=180°-∠ABE=180°-68°=112°。所以，∠BED的度数是112°。
answer:112
correct answer:68
total:201  correct:55  acc:0.2736318407960199
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ID:201
picturePath:image/254.png
answer:The answer is 61.
answer:61
correct answer:61
total:202  correct:56  acc:0.27722772277227725
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ID:202
picturePath:image/255.png
answer:60°
answer:60
correct answer:159
total:203  correct:56  acc:0.27586206896551724
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ID:203
picturePath:image/256.png
answer:根据题意,∠A=60°,则∠BOC=2∠A=2×60°=120°.
answer:120
correct answer:120
total:204  correct:57  acc:0.27941176470588236
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ID:204
picturePath:image/257.png
answer:由于AB∥CD, 根据平行线的性质, ∠AEF + ∠EFC = 180°, 已知∠AEF = 50°, 所以∠EFC = 180° - ∠AEF = 180° - 50° = 130°.
answer:130
correct answer:130
total:205  correct:58  acc:0.28292682926829266
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ID:205
picturePath:image/258.png
answer:本题主要考查了轴对称的性质和三角形的外角定理．
解:∵将△ABC沿着DE折叠压平,点A与点A′重合,∴∠1=∠A,∠2=∠A′,∵∠A=70°,∴∠1+∠2=70°+70°=140°,
故选:B．
answer:140
correct answer:140
total:206  correct:59  acc:0.28640776699029125
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ID:206
picturePath:image/259.png
answer:90°
answer:90
correct answer:90
total:207  correct:60  acc:0.2898550724637681
------------------------------------------------------------------------------------
ID:207
picturePath:image/260.png
answer:40°
answer:40
correct answer:50
total:208  correct:60  acc:0.28846153846153844
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ID:208
picturePath:image/263.png
answer:60
answer:60
correct answer:40
total:209  correct:60  acc:0.28708133971291866
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ID:209
picturePath:image/264.png
answer:6√3cm
answer:3
correct answer:2
total:210  correct:60  acc:0.2857142857142857
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ID:210
picturePath:image/265.png
answer:∠3=180°-∠1=180°-145°=35°.
answer:35
correct answer:55
total:211  correct:60  acc:0.2843601895734597
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ID:211
picturePath:image/267.png
answer:60°
answer:60
correct answer:135
total:212  correct:60  acc:0.2830188679245283
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ID:212
picturePath:image/269.png
answer:30°
answer:30
correct answer:45
total:213  correct:60  acc:0.28169014084507044
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ID:213
picturePath:image/270.png
answer:40
answer:40
correct answer:50
total:214  correct:60  acc:0.2803738317757009
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ID:214
picturePath:image/271.png
answer:解:∵将一个长为10cm,宽为8cm的矩形纸片对折两次后,沿所得矩形两邻边中点的连线(虚线)剪下,再打开,∴得到的菱形两邻边中点相连的线段的长为4cm,∵矩形纸片的长和宽分别为10cm和8cm,∴矩形纸片的面积为10×8=80(平方厘米),∴得到的菱形的面积为80÷2=40(平方厘米)．故选:B．
answer:40
correct answer:10
total:215  correct:60  acc:0.27906976744186046
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ID:215
picturePath:image/272.png
answer:Since \(AB \parallel CD\), we have \(∠B + ∠D = 180^\circ\). Given that \(∠B = 20^\circ\) and \(∠D = 40^\circ\), we can substitute these values into the equation to find that \(∠BOD = 180^\circ - ∠B - ∠D = 180^\circ - 20^\circ - 40^\circ = \boxed{120^\circ}\).The answer is: 120
answer:120
correct answer:60
total:216  correct:60  acc:0.2777777777777778
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ID:216
picturePath:image/273.png
answer:The answer is 40.
answer:40
correct answer:70
total:217  correct:60  acc:0.2764976958525346
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ID:217
picturePath:image/274.png
answer:解:∵PA,PB是⊙O的切线,切点分别为A,B,∴∠OAP=∠OBP=90°,∴∠AOB=140°,∴∠ACB=\frac{1}{2}∠AOB=70°．故选:B．
answer:70
correct answer:65
total:218  correct:60  acc:0.27522935779816515
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ID:218
picturePath:image/277.png
answer:解:∵四边形ABCD是圆O的内接四边形,∴∠A+∠D=180°,∵BC=DC,∴∠BCD=∠DCB,∵∠BOC=130°,∴∠BDC=\frac{1}{2}∠BOC=65°,∴∠BAD=180°-∠BDC=180°-65°=115°,
故选:A．
answer:115
correct answer:130
total:219  correct:60  acc:0.273972602739726
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ID:219
picturePath:image/278.png
answer:根据三角形内角和定理，∠BOC = 180° - ∠A - ∠1 - ∠2 = 180° - 80° - 15° - 40° = 45°.
answer:45
correct answer:135
total:220  correct:60  acc:0.2727272727272727
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ID:220
picturePath:image/279.png
answer:40°
answer:40
correct answer:110
total:221  correct:60  acc:0.27149321266968324
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ID:221
picturePath:image/281.png
answer:The answer is 61.
answer:61
correct answer:34
total:222  correct:60  acc:0.2702702702702703
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ID:222
picturePath:image/282.png
answer:50
answer:50
correct answer:70
total:223  correct:60  acc:0.26905829596412556
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ID:223
picturePath:image/283.png
answer:20
answer:20
correct answer:17
total:224  correct:60  acc:0.26785714285714285
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ID:224
picturePath:image/285.png
answer:This is a geometric figure with points labeled A, B, C, D, E, M, N, and O. The figure appears to be a triangle with a line drawn from point B to point C, and another line drawn from point E to point C. The lines intersect at point C, and there is a right angle at point O. The question is asking for the measure of angle C.
ID:225
picturePath:image/286.png
answer:65°
answer:65
correct answer:125
total:226  correct:60  acc:0.26548672566371684
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ID:226
picturePath:image/288.png
answer:解:∵AB与⊙O相切,切点为B,∴OB⊥AB,∴∠ABO=90°,∴∠AOB=90°-∠A=56°,∵四边形ABOC是圆内接四边形,∴∠C=∠AOB=56°．故选:B．
answer:56
correct answer:28
total:227  correct:60  acc:0.2643171806167401
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ID:227
picturePath:image/289.png
answer:解:∵竹竿与点O相距6m,与树相距15m,∴△AOB∽,∴\frac{AB}{AO}=\frac{OB}{BO},即\frac{2}{6}=\frac{15}{x},解得x=30．故选:B．
answer:30
correct answer:7
total:228  correct:60  acc:0.2631578947368421
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ID:228
picturePath:image/291.png
answer:25°
answer:25
correct answer:115
total:229  correct:60  acc:0.26200873362445415
------------------------------------------------------------------------------------
ID:229
picturePath:image/292.png
answer:30°
answer:30
correct answer:30
total:230  correct:61  acc:0.26521739130434785
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ID:230
picturePath:image/293.png
answer:根据题意,AD=AE,BE=CD,∠1=∠2=110°,∠BAC=80°,可得∠B=∠1=110°,∠C=∠2=110°,∠BAC=80°,根据三角形内角和定理,可得∠A=180°-110°-80°=90°,又∵AD=AE,BE=CD,∠1=∠2=110°,∠BAC=80°,∴△ADE≌△CBE,∴∠CAE=∠B=110°,故选B.
answer:110
correct answer:20
total:231  correct:61  acc:0.26406926406926406
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ID:231
picturePath:image/295.png
answer:12m
answer:12
correct answer:5
total:232  correct:61  acc:0.2629310344827586
------------------------------------------------------------------------------------
ID:232
picturePath:image/297.png
answer:25
answer:25
correct answer:115
total:233  correct:61  acc:0.26180257510729615
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ID:233
picturePath:image/298.png
answer:1.5
answer:1.5
correct answer:0.9
total:234  correct:61  acc:0.2606837606837607
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ID:234
picturePath:image/299.png
answer:6+6+6=18cm
answer:18
correct answer:6
total:235  correct:61  acc:0.25957446808510637
------------------------------------------------------------------------------------
ID:235
picturePath:image/300.png
answer:根据三角形的外角定理，∠DAC=∠B+∠AEC，∠ACF=∠C+∠AEC。因为∠B=48°，所以∠DAC=∠B+∠AEC=48°+∠AEC，∠ACF=∠C+∠AEC=∠C+∠DAC-∠B=∠C+48°-48°=∠C。又因为∠DAC和∠ACF的平分线交于点E，所以∠AEC=\frac{1}{2}(∠DAC+∠ACF)=\frac{1}{2}(∠B+∠C)=\frac{1}{2}(∠B+∠B)=\frac{1}{2}(2∠B)=∠B=48°。故选:A。
answer:48
correct answer:66
total:236  correct:61  acc:0.2584745762711864
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ID:236
picturePath:image/302.png
answer:60°
answer:60
correct answer:15
total:237  correct:61  acc:0.25738396624472576
------------------------------------------------------------------------------------
ID:237
picturePath:image/303.png
answer:20
answer:20
correct answer:40
total:238  correct:61  acc:0.25630252100840334
------------------------------------------------------------------------------------
ID:238
picturePath:image/304.png
answer:解:由翻折的性质得到∠EDC=∠1=56°,∵四边形ABCD是长方形,∴AD∥BC,∴∠DEC=∠1=56°,∴∠EFB=∠DEC+∠1=56°+56°=112°．故选:D．
answer:112
correct answer:62
total:239  correct:61  acc:0.25523012552301255
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ID:239
picturePath:image/305.png
answer:解:∵∠AOB=30°,∴∠AOP+∠BOP=180°-30°=150°,∵∠AOP+∠BOP+∠PQR=180°,∴∠PQR=150°-∠AOB=120°,∴△PQR为等边三角形,∴PQ=PR=RQ=OP=10,∴△PQR周长为30．故选:A．
answer:30
correct answer:10
total:240  correct:61  acc:0.25416666666666665
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ID:240
picturePath:image/306.png
answer:解:∵⊙O中,ABDC是圆内接四边形,∠BOC=110°,∴∠BDC=\frac{1}{2}∠BOC=\frac{1}{2}×110°=55°．故选:A．
answer:55
correct answer:125
total:241  correct:61  acc:0.25311203319502074
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ID:241
picturePath:image/307.png
answer:6.6
answer:6.6
correct answer:8
total:242  correct:61  acc:0.25206611570247933
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ID:242
picturePath:image/309.png
answer:70°
answer:70
correct answer:40
total:243  correct:61  acc:0.25102880658436216
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ID:243
picturePath:image/310.png
answer:30
answer:30
correct answer:70
total:244  correct:61  acc:0.25
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ID:244
picturePath:image/312.png
answer:20
answer:20
correct answer:16
total:245  correct:61  acc:0.24897959183673468
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ID:245
picturePath:image/314.png
answer:根据三角形ABC的内切圆的性质，可以得出∠AED=∠AFE，∠ADE=∠DAE，∠AED+∠ADE=90°，∠AFE+∠DAE=90°。因此，∠AED+∠AFE=90°，∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=180°。又因为∠DEF=52°，所以∠AED+∠AFE+∠ADE+∠DAE=180°，∠DEF=52°，∠AED+∠AFE+∠ADE+∠DAE=180°-∠DEF=180°-52°=128°。因此，∠AED+∠AFE=90°，∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=128°，所以∠AED+∠AFE+∠ADE+∠DAE=128°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+∠ADE+∠DAE=90°，∠AED+∠AFE+
answer:90
correct answer:76
total:246  correct:61  acc:0.24796747967479674
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ID:246
picturePath:image/315.png
answer:The answer is 20.
answer:20
correct answer:110
total:247  correct:61  acc:0.24696356275303644
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ID:247
picturePath:image/316.png
answer:45°
answer:45
correct answer:55
total:248  correct:61  acc:0.24596774193548387
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ID:248
picturePath:image/317.png
answer:28米
answer:28
correct answer:28
total:249  correct:62  acc:0.24899598393574296
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ID:249
picturePath:image/318.png
answer:解:∵∠A=45°,∠AMD=75°,∴∠DMC=75°-45°=30°,∴∠B=∠DMC=30°．故选:B．
answer:30
correct answer:30
total:250  correct:63  acc:0.252
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ID:250
picturePath:image/320.png
answer:Since \( CD \) is parallel to \( AB \), we have \( \angle ACD = \angle ABC \). Given that \( \angle ACD = 40^\circ \) and \( \angle ACB = 90^\circ \), we can find \( \angle ABC \) by subtracting \( \angle ACD \) from \( \angle ACB \): \( \angle ABC = 90^\circ - 40^\circ = 50^\circ \). Therefore, \( \angle B \) is the remaining angle in \( \triangle ABC \), which is \( 180^\circ - \angle ABC - \angle ACB = 180^\circ - 50^\circ - 90^\circ = \boxed{40^\circ} \).
answer:40
correct answer:50
total:251  correct:63  acc:0.250996015936255
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ID:251
picturePath:image/321.png
answer:This is a geometry problem. Given that \( a//b \), we know that the corresponding angles made by the transversal \( c \) to the parallel lines \( a \) and \( b \) are congruent. Therefore, \( \angle 1 \) and \( \angle 2 \) are congruent. Since \( \angle 1 \) is given as \( 55^\circ \), \( \angle 2 \) must also be \( 55^\circ \). The answer is A.
answer:55
correct answer:55
total:252  correct:64  acc:0.25396825396825395
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ID:252
picturePath:image/323.png
answer:根据题意,∠1=136°,∠E=90°,因为AB∥CD,所以∠C=∠1-∠E=136°-90°=46°.
answer:46
correct answer:46
total:253  correct:65  acc:0.25691699604743085
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ID:253
picturePath:image/324.png
answer:5
answer:5
correct answer:2
total:254  correct:65  acc:0.2559055118110236
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ID:254
picturePath:image/326.png
answer:解:∵AD∥OC,∴∠DOB=∠B=25°,∴∠BAD=∠DOA,∵OA=OD,∴∠DOA=∠DOA,∴∠BAD=∠DOA=25°,故选:B．
answer:25
correct answer:50
total:255  correct:65  acc:0.2549019607843137
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ID:255
picturePath:image/327.png
answer:70
answer:70
correct answer:95
total:256  correct:65  acc:0.25390625
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ID:256
picturePath:image/329.png
answer:由于AB∥CD，根据平行线的性质，内错角相等，所以∠C=∠F。又因为∠A和∠E是对顶角，所以∠A=∠E。因此，∠C=∠F=∠A=25°。故选A。
answer:25
correct answer:65
total:257  correct:65  acc:0.2529182879377432
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ID:257
picturePath:image/331.png
answer:∠BCD = 90° - ∠ABC = 90° - 40° = 50°.
answer:50
correct answer:130
total:258  correct:65  acc:0.25193798449612403
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ID:258
picturePath:image/332.png
answer:∠2=\frac{1}{2}∠1=\frac{1}{2}×100°=50°.
answer:50
correct answer:80
total:259  correct:65  acc:0.25096525096525096
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ID:259
picturePath:image/333.png
answer:The answer is 90°.
answer:90
correct answer:25
total:260  correct:65  acc:0.25
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ID:260
picturePath:image/336.png
answer:解:∵直线a∥b,∴∠1=∠3,∵∠3+∠2=180°,∴∠2=180°-∠1=180°-56°=124°.故选B.
answer:124
correct answer:34
total:261  correct:65  acc:0.24904214559386972
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ID:261
picturePath:image/338.png
answer:∠2 = 180° - ∠1 = 180° - 108° = 72°.
answer:72
correct answer:72
total:262  correct:66  acc:0.25190839694656486
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ID:262
picturePath:image/339.png
answer:Since line a is parallel to line b, we have corresponding angles, which means that angles made by the lines and the transversal are congruent. Therefore, ∠2 = ∠1 = 55°.The answer is: 55
answer:55
correct answer:55
total:263  correct:67  acc:0.25475285171102663
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ID:263
picturePath:image/340.png
answer:∠2=∠1=100°
answer:100
correct answer:80
total:264  correct:67  acc:0.2537878787878788
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ID:264
picturePath:image/342.png
answer:解:∵直线AB∥CD,∠1=42°,∴∠GEB=∠1=42°,∵EF为∠GEB的平分线,∴∠GFE=∠1=42°,∴∠2=∠GFE+∠1=42°+42°=84°．故选:D．
answer:84
correct answer:159
total:265  correct:67  acc:0.2528301886792453
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ID:265
picturePath:image/343.png
answer:20
answer:20
correct answer:70
total:266  correct:67  acc:0.2518796992481203
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ID:266
picturePath:image/344.png
answer:60°
answer:60
correct answer:60
total:267  correct:68  acc:0.2546816479400749
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ID:267
picturePath:image/346.png
answer:40
answer:40
correct answer:40
total:268  correct:69  acc:0.2574626865671642
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ID:268
picturePath:image/347.png
answer:58°
answer:58
correct answer:32
total:269  correct:69  acc:0.25650557620817843
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ID:269
picturePath:image/348.png
answer:∠AED=∠C=50°.
answer:50
correct answer:115
total:270  correct:69  acc:0.25555555555555554
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ID:270
picturePath:image/349.png
answer:60
answer:60
correct answer:100
total:271  correct:69  acc:0.25461254612546125
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ID:271
picturePath:image/353.png
answer:∠2=∠1=110°
answer:110
correct answer:70
total:272  correct:69  acc:0.2536764705882353
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ID:272
picturePath:image/356.png
answer:解:由题意得,∠ABC=90°,∠A=∠C=40°,AB=BC=40°,在Rt△ABC中,AC=√{AB²+BC²}=√{40²+40²}=2√{400}=80(海里)．故选:D．
answer:80
correct answer:40
total:273  correct:69  acc:0.25274725274725274
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ID:273
picturePath:image/359.png
answer:∠B=30°,∠C=110°,AB∥DE,∴∠B+∠C+∠D=180°,∴∠D=180°-∠B-∠C=180°-30°-110°=40°.
answer:40
correct answer:100
total:274  correct:69  acc:0.2518248175182482
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ID:274
picturePath:image/360.png
answer:65°
answer:65
correct answer:35
total:275  correct:69  acc:0.2509090909090909
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ID:275
picturePath:image/361.png
answer:解:∵CD是⊙O的切线,∴∠OCD=90°,∵∠D=50°,∴∠COD=40°,∵OA=OC,∴∠A=∠OCA,∵∠A+∠COD+∠OCA=180°,∴∠A=\frac{180°-40°}{2}=70°．故选:A．
answer:70
correct answer:20
total:276  correct:69  acc:0.25
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ID:276
picturePath:image/363.png
answer:This is a geometry problem involving angle relationships in a rotation. Since △OAB and △OCD are rotations of each other, corresponding angles are congruent. Therefore, ∠BOC = ∠D + ∠A = 40° + 110° = 150°. Since the sum of angles in a triangle is 180°, ∠α = 180° - ∠BOC = 180° - 150° = 30°. Hence, the answer is D.
answer:30
correct answer:50
total:277  correct:69  acc:0.2490974729241877
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ID:277
picturePath:image/364.png
answer:40°
answer:40
correct answer:70
total:278  correct:69  acc:0.24820143884892087
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ID:278
picturePath:image/365.png
answer:The answer is 90°.
answer:90
correct answer:30
total:279  correct:69  acc:0.24731182795698925
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ID:279
picturePath:image/366.png
answer:根据题意，∠1是△CDE的外角，∠D=45°，∠E=45°，∵DE∥BC，∴∠D+∠E+∠1=180°，∴∠1=180°-45°-45°=90°．故选：B．
answer:90
correct answer:105
total:280  correct:69  acc:0.24642857142857144
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ID:280
picturePath:image/367.png
answer:解:∵∠AOC与∠ABC是同弧所对的圆心角与圆周角,∠AOC=100°,∴∠ABC=\frac{1}{2}∠AOC=50°．故选:B．
answer:50
correct answer:130
total:281  correct:69  acc:0.24555160142348753
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ID:281
picturePath:image/368.png
answer:The answer is 40.
answer:40
correct answer:40
total:282  correct:70  acc:0.24822695035460993
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ID:282
picturePath:image/369.png
answer:解:∵AB⊥CD,∴⁀{AC}=⁀{AD},∴∠ACD=\frac{1}{2}∠CDB=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:40
total:283  correct:70  acc:0.24734982332155478
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ID:283
picturePath:image/370.png
answer:由于EF∥AC和DF∥AB，根据平行线的性质，我们可以得出∠BED=∠B和∠CDF=∠C。

已知∠B=45°和∠C=65°，根据三角形内角和定理，∠A+∠B+∠C=180°，我们可以得出∠A=180°-45°-65°=70°。

由于∠BED=∠B=45°，∠CDF=∠C=65°，且EF∥AC和DF∥AB，根据平行线的性质，∠EDF=∠A=70°。

因此，∠EFD=180°-∠EDF-∠CDF=180°-70°-65°=45°。
answer:45
correct answer:70
total:284  correct:70  acc:0.24647887323943662
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ID:284
picturePath:image/371.png
answer:Since A is the midpoint of arc CD, we know that ∠BAC and ∠CAO are both half the measure of ∠B, which is 25°. Therefore, ∠AOC is the sum of ∠BAC and ∠CAO, which is 25° + 25° = 50°.
answer:50
correct answer:50
total:285  correct:71  acc:0.24912280701754386
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ID:285
picturePath:image/372.png
answer:解:∵AB∥CD,∴∠1=∠3,∵EC⊥CD于C,∴∠2+∠3=90°,∵∠2=29°,∴∠3=90°-29°=61°,∴∠1=61°．故选:B．
answer:61
correct answer:61
total:286  correct:72  acc:0.2517482517482518
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ID:286
picturePath:image/373.png
answer:60
answer:60
correct answer:70
total:287  correct:72  acc:0.2508710801393728
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ID:287
picturePath:image/374.png
answer:解:∵AB⊥BC,∴∠BAC=90°,∵∠1=38°,∴∠B=90°-38°=52°,∵a∥b,∴∠2=∠B=52°．故选:B．
answer:52
correct answer:52
total:288  correct:73  acc:0.2534722222222222
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ID:288
picturePath:image/375.png
answer:根据三角形的一个外角等于与它不相邻的两个内角的和，可得∠3=∠1+∠2=40°+55°=95°,又∵直线a∥b,∴∠3=180°,∴95°=180°-∠3,解得∠3=85°.故选B
answer:85
correct answer:95
total:289  correct:73  acc:0.25259515570934254
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ID:289
picturePath:image/377.png
answer:∠A的度数等于∠C的度数，因为AB∥CD，所以∠A+∠C=180°，所以∠A=180°-∠C=180°-50°=130°.
answer:130
correct answer:25
total:290  correct:73  acc:0.2517241379310345
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ID:290
picturePath:image/379.png
answer:∠NOF = ∠MEN (对顶角) ∠MEN = 90° (直角三角形) ∠NOF = 90° - 20° = 70°.
answer:70
correct answer:70
total:291  correct:74  acc:0.2542955326460481
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ID:291
picturePath:image/380.png
answer:根据题意,∠AOE=36°,EO⊥CD,则∠AOC=90°,∴∠COE=90°-∠AOE=90°-36°=54°,∴∠BOD=∠COE=54°.故选B.
answer:54
correct answer:54
total:292  correct:75  acc:0.2568493150684932
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ID:292
picturePath:image/381.png
answer:解:∵AB是⊙O的切线,切点为B,∴∠ABO=90°,∵∠A=30°,∴∠BOC=∠A+∠ABO=30°+90°=120°,∴∠D=\frac{1}{2}∠BOC=\frac{1}{2}×120°=60°．故选:B．
answer:60
correct answer:30
total:293  correct:75  acc:0.25597269624573377
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ID:293
picturePath:image/382.png
answer:36
answer:36
correct answer:21
total:294  correct:75  acc:0.25510204081632654
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ID:294
picturePath:image/384.png
answer:根据勾股定理，在直角三角形ABC中，AC² = AB² + BC²，即5² = AB² + 3²，解得AB = √(5² - 3²) = √(16) = 4m。因此，至少需要4m x 3m = 12m²的地毯。故选B。
answer:12
correct answer:7
total:295  correct:75  acc:0.2542372881355932
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ID:295
picturePath:image/385.png
answer:Since AD is parallel to BC, we have \angle ADB = \angle BAC. Since AB=AC, we have \angle ADB = \angle ACB. Therefore, \angle BAC = \angle ACB. Since the sum of the angles in a triangle is 180 degrees, we have \angle BAC + \angle ACB + \angle ABC = 180 degrees. Therefore, \angle BAC = \angle ACB = \frac{180 - \angle 1}{2} = \frac{180 - 70}{2} = \boxed{65} degrees.The answer is: 65
answer:65
correct answer:40
total:296  correct:75  acc:0.2533783783783784
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ID:296
picturePath:image/386.png
answer:10.
answer:10
correct answer:60
total:297  correct:75  acc:0.25252525252525254
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ID:297
picturePath:image/387.png
answer:180°-40°-110°-60°=90°
answer:90
correct answer:90
total:298  correct:76  acc:0.2550335570469799
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ID:298
picturePath:image/389.png
answer:This is a geometry problem involving rotations. Given that △EDC is obtained by rotating △ABC around point C by 40 degrees, we can determine the measure of ∠1 and ∠2 by considering the interior angles of the original and rotated triangles.

In △ABC, the sum of the interior angles is (180 - 40) degrees, which is 140 degrees. Since ∠1 is one of the interior angles of △ABC, ∠1 = 140 degrees.

In △EDC, the sum of the interior angles is also 180 degrees. Since ∠2 is one of the interior angles of △EDC, ∠2 = 180 - 140 = 40 degrees.

Therefore, ∠1 + ∠2 = 140 + 40 = 180 degrees. The answer is A.
answer:180
correct answer:110
total:299  correct:76  acc:0.25418060200668896
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ID:299
picturePath:image/390.png
answer:0.5×(180°-40°)=70°\frac{1}{2}×(180°-40°)=70°
answer:70
correct answer:20
total:300  correct:76  acc:0.25333333333333335
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ID:300
picturePath:image/391.png
answer:A.50°B.40°C.30°D.20°
answer:20
correct answer:65
total:301  correct:76  acc:0.25249169435215946
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ID:301
picturePath:image/392.png
answer:解:∵将△ABC绕点C顺时针旋转一定角度,得到△DEC,点A的对应点为D,ED过点A,∴∠DAC=∠BAC=70°,∴旋转角的度数=180°-∠DAC=180°-70°=110°．故选:B．
answer:110
correct answer:40
total:302  correct:76  acc:0.25165562913907286
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ID:302
picturePath:image/394.png
answer:Since ∠A and ∠B are inscribed angles that intercept the same arc, and the sum of the angle measures formed by an inscribed angle and its intercepted arc is twice the angle measure of the inscribed angle, then ∠BOC = 2∠A = 2*36° = 72°.
answer:72
correct answer:72
total:303  correct:77  acc:0.25412541254125415
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ID:303
picturePath:image/395.png
answer:This is a geometry problem involving rotation and angle relationships in a triangle. Since the triangle ABC is rotated around point C in a clockwise direction to form triangle A'B'C, the angle B'OA' will be the same as angle BOC in the original triangle ABC. This is because the rotation preserves the angle measures around point C. 

Given that angle B is 50 degrees, and since the sum of angles in a triangle is 180 degrees, angle BOC must be 180 - 50 = 130 degrees. Therefore, angle COA' is also 130 degrees.
answer:130
correct answer:60
total:304  correct:77  acc:0.2532894736842105
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ID:304
picturePath:image/396.png
answer:根据平行线的性质，∠1和∠2互为内错角，所以∠1=∠2。又已知∠1=110°，所以∠2=110°。故选A。
answer:110
correct answer:70
total:305  correct:77  acc:0.25245901639344265
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ID:305
picturePath:image/397.png
answer:60°
answer:60
correct answer:60
total:306  correct:78  acc:0.2549019607843137
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ID:306
picturePath:image/398.png
answer:The answer is 30.
answer:30
correct answer:30
total:307  correct:79  acc:0.25732899022801303
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ID:307
picturePath:image/399.png
answer:20
answer:20
correct answer:110
total:308  correct:79  acc:0.2564935064935065
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ID:308
picturePath:image/401.png
answer:17
answer:17
correct answer:55
total:309  correct:79  acc:0.255663430420712
------------------------------------------------------------------------------------
ID:309
picturePath:image/402.png
answer:50°
answer:50
correct answer:50
total:310  correct:80  acc:0.25806451612903225
------------------------------------------------------------------------------------
ID:310
picturePath:image/405.png
answer:20°
answer:20
correct answer:60
total:311  correct:80  acc:0.2572347266881029
------------------------------------------------------------------------------------
ID:311
picturePath:image/406.png
answer:45°+45°=90°,∠1+∠2=90°,故选B.
answer:90
correct answer:45
total:312  correct:80  acc:0.2564102564102564
------------------------------------------------------------------------------------
ID:312
picturePath:image/407.png
answer:解:∵AB是⊙O的直径,∴∠ADB=90°,∵四边形ABDC是圆内接四边形,∴∠ADB+∠ACD=180°,∵∠ACD=35°,∴∠ADB=180°-35°=145°,∴∠BAD=∠ADB-∠ACD=145°-35°=110°．故选:B．
answer:110
correct answer:55
total:313  correct:80  acc:0.25559105431309903
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ID:313
picturePath:image/408.png
answer:根据圆周角定理，同弧所对的圆周角相等，所以∠ADC=∠BAC=23°．故选A．
answer:23
correct answer:67
total:314  correct:80  acc:0.25477707006369427
------------------------------------------------------------------------------------
ID:314
picturePath:image/409.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵∠BAC=22°,∴∠B=90°-22°=68°,∴∠ADC=∠B=68°,故选:B．
answer:68
correct answer:68
total:315  correct:81  acc:0.2571428571428571
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ID:315
picturePath:image/411.png
answer:The answer is 54°.
answer:54
correct answer:123
total:316  correct:81  acc:0.2563291139240506
------------------------------------------------------------------------------------
ID:316
picturePath:image/412.png
answer:解:∵AB与⊙O相切于点A,∴OA⊥AB,∴∠OAB=90°,∵∠CDA=27°,∴∠AOB=2∠CDA=54°,∴∠B=36°．故选:B．
answer:36
correct answer:36
total:317  correct:82  acc:0.2586750788643533
------------------------------------------------------------------------------------
ID:317
picturePath:image/413.png
answer:50
answer:50
correct answer:55
total:318  correct:82  acc:0.2578616352201258
------------------------------------------------------------------------------------
ID:318
picturePath:image/415.png
answer:60
answer:60
correct answer:120
total:319  correct:82  acc:0.25705329153605017
------------------------------------------------------------------------------------
ID:319
picturePath:image/417.png
answer:解:∵BC∥PQ,∴\frac{AB}{AP}=\frac{AQ}{AC},∵AB:AP=2:5,AQ=20cm,∴\frac{2}{5}=\frac{20}{AC},解得:AC=40cm,∴CQ=AC-AQ=40-20=20cm．故选:A．
answer:20
correct answer:12
total:320  correct:82  acc:0.25625
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ID:320
picturePath:image/418.png
answer:100°
answer:100
correct answer:65
total:321  correct:82  acc:0.2554517133956386
------------------------------------------------------------------------------------
ID:321
picturePath:image/419.png
answer:60
answer:60
correct answer:40
total:322  correct:82  acc:0.2546583850931677
------------------------------------------------------------------------------------
ID:322
picturePath:image/420.png
answer:解:∵四边形ABCD是正方形,∴AB=BC=CD=AD=6,∴扇形DAB的面积=\frac{180π×6^{2}}{360}=\frac{108π}{2}=54π．故选:C．
answer:54
correct answer:36
total:323  correct:82  acc:0.25386996904024767
------------------------------------------------------------------------------------
ID:323
picturePath:image/422.png
answer:15
answer:15
correct answer:9
total:324  correct:82  acc:0.25308641975308643
------------------------------------------------------------------------------------
ID:324
picturePath:image/423.png
answer:Since \( \overline{AB} \) and \( \overline{CD} \) are both diameters of the circle, they are parallel. This means that the corresponding angles \( \angle AOC \) and \( \angle C \) are congruent. Therefore, \( \angle C \) is also \( \frac{1}{2} \times 50^{\circ} = 25^{\circ} \).
answer:25
correct answer:25
total:325  correct:83  acc:0.2553846153846154
------------------------------------------------------------------------------------
ID:325
picturePath:image/424.png
answer:∠2 = 180° - ∠1 = 180° - 120° = 60°.
answer:60
correct answer:60
total:326  correct:84  acc:0.25766871165644173
------------------------------------------------------------------------------------
ID:326
picturePath:image/426.png
answer:62°
answer:62
correct answer:62
total:327  correct:85  acc:0.2599388379204893
------------------------------------------------------------------------------------
ID:327
picturePath:image/428.png
answer:根据平行线的性质，∠ABC和∠BCE互为同位角，∠ABC=46°，∠BCE=20°，所以∠CEF=∠ABC=46°。
answer:46
correct answer:154
total:328  correct:85  acc:0.25914634146341464
------------------------------------------------------------------------------------
ID:328
picturePath:image/429.png
answer:12
answer:12
correct answer:20
total:329  correct:85  acc:0.25835866261398177
------------------------------------------------------------------------------------
ID:329
picturePath:image/430.png
answer:70
answer:70
correct answer:70
total:330  correct:86  acc:0.2606060606060606
------------------------------------------------------------------------------------
ID:330
picturePath:image/432.png
answer:70°
answer:70
correct answer:55
total:331  correct:86  acc:0.2598187311178248
------------------------------------------------------------------------------------
ID:331
picturePath:image/433.png
answer:The answer is 25°.
answer:25
correct answer:50
total:332  correct:86  acc:0.25903614457831325
------------------------------------------------------------------------------------
ID:332
picturePath:image/434.png
answer:∠2 = 180° - ∠1 = 180° - 65° = 115°.
answer:115
correct answer:115
total:333  correct:87  acc:0.26126126126126126
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ID:333
picturePath:image/435.png
answer:70°
answer:70
correct answer:110
total:334  correct:87  acc:0.26047904191616766
------------------------------------------------------------------------------------
ID:334
picturePath:image/436.png
answer:60
answer:60
correct answer:120
total:335  correct:87  acc:0.25970149253731345
------------------------------------------------------------------------------------
ID:335
picturePath:image/437.png
answer:解:如图,∵∠1=46°,∴∠3=∠1+90°=46°+90°=136°,∵直尺的两边互相平行,∴∠2=∠3=136°．故选:C．
answer:136
correct answer:136
total:336  correct:88  acc:0.2619047619047619
------------------------------------------------------------------------------------
ID:336
picturePath:image/439.png
answer:∠2 = ∠3 (对顶角相等), ∠3 = ∠1 + ∠EFG (对顶角相等), ∠1 = 36°, ∠EFG = 180° - ∠1 - ∠2 (对顶角相等), ∠EFG = 180° - 36° - ∠2, ∠EFG = 144° - ∠2, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144°, ∠2 = ∠EFG - 144
answer:144
correct answer:54
total:337  correct:88  acc:0.26112759643916916
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ID:337
picturePath:image/441.png
answer:The answer is A.
ID:338
picturePath:image/442.png
answer:Since \(AB \parallel CD\), we can use the corresponding angles postulate, which states that if two parallel lines are cut by a transversal, then the corresponding angles are congruent. Therefore, \(∠1 = ∠2\). Given that \(∠2 = 40°\), we can conclude that \(∠1 = 40°\).
answer:40
correct answer:70
total:339  correct:88  acc:0.25958702064896755
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ID:339
picturePath:image/443.png
answer:∠ACB' = ∠A + ∠B = 27° + 40° = 67°.
answer:67
correct answer:46
total:340  correct:88  acc:0.25882352941176473
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ID:340
picturePath:image/444.png
answer:∠2 = ∠1 + ∠MND (对顶角相等) ∠MND = 180° - ∠2 - ∠1 (转角和为180°) ∠MND = 180° - 70° - ∠2 (已知) ∠MND = 110° - ∠2 (化简) ∠MND = ∠MNG (已知) ∠2 = ∠MNG (对顶角相等) ∠2 = 110° - ∠MNG (已知) ∠2 = 110° - 110° (化简) ∠2 = 0° (解得) 故选:D
answer:0
correct answer:35
total:341  correct:88  acc:0.25806451612903225
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ID:341
picturePath:image/446.png
answer:20cm
answer:20
correct answer:20
total:342  correct:89  acc:0.260233918128655
------------------------------------------------------------------------------------
ID:342
picturePath:image/448.png
answer:Since line AB is parallel to line CD, we have \begin{align*}
\angle E &= \angle 1 + \angle C \\
90^\circ &= \angle 1 + 44^\circ \\
\angle 1 &= 90^\circ - 44^\circ \\
&= \boxed{46^\circ}.
\end{align*}The answer is: 46
answer:46
correct answer:134
total:343  correct:89  acc:0.2594752186588921
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ID:343
picturePath:image/449.png
answer:解:∵∠BOC与∠BAC是同弧所对的圆心角与圆周角,∠BOC=100°,∴∠BAC=\frac{1}{2}∠BOC=50°．故选:B．
answer:50
correct answer:130
total:344  correct:89  acc:0.25872093023255816
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ID:344
picturePath:image/450.png
answer:This is a geometry problem. Given that AB || CD and CD ⊥ EF, we know that corresponding angles are congruent. Therefore, ∠1 = ∠3 = 125°. Since AB || CD, we also know that alternate interior angles are congruent. Therefore, ∠2 = ∠3 = 125°. Hence, the answer is B.
answer:125
correct answer:35
total:345  correct:89  acc:0.2579710144927536
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ID:345
picturePath:image/453.png
answer:解:∵竹竿与旗杆相距22m,竹竿与旗杆顶端的影子恰好落在地面的同一点,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∵竹竿与旗杆顶端的影子恰好落在地面的同一点,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∵竹竿与旗杆顶端的影子恰好落在地面的同一点,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与旗杆顶端的影子的距离,∴竹竿与旗杆顶端的距离等于竹竿与
answer:22
correct answer:12
total:346  correct:89  acc:0.25722543352601157
------------------------------------------------------------------------------------
ID:346
picturePath:image/455.png
answer:∠BAC=120°, ∵AB∥CD∥EF, ∴∠B+∠C=180°, ∠C+∠CDF=180°, ∴∠CDF=∠BAC=120°.
answer:120
correct answer:60
total:347  correct:89  acc:0.2564841498559078
------------------------------------------------------------------------------------
ID:347
picturePath:image/457.png
answer:解:∵∠ABC与∠ADC是同弧所对的圆周角与圆心角,∠ABC=70°,∴∠ADC=2∠ABC=140°．故选:D．
answer:140
correct answer:130
total:348  correct:89  acc:0.2557471264367816
------------------------------------------------------------------------------------
ID:348
picturePath:image/458.png
answer:根据五边形的内角和定理，五边形的内角和为(5-2)×180°=540°，已知∠CAD=35°，∠AED=115°，根据三角形的内角和定理，∠ADE=180°-∠AED=180°-115°=65°，则∠B=540°-∠ADE-∠CAD=540°-65°-35°=540°-100°=440°。故选C。
answer:440
correct answer:100
total:349  correct:89  acc:0.25501432664756446
------------------------------------------------------------------------------------
ID:349
picturePath:image/459.png
answer:This is a question related to geometry, specifically about triangles and angles. The given figure shows a triangle ABC with its angles labeled as ∠ABC, ∠ACB, and ∠ABC. There are also lines drawn from the vertices of the triangle to the opposite sides, intersecting at points E and D, which are the midpoints of the opposite sides. The question asks for the measure of angle A.

To solve this, we can use the properties of triangles and angles. Since E and D are the midpoints of the opposite sides, we know that they divide the opposite sides into two equal parts. This means that the sum of the angles at E and D is equal to the sum of the angles at the opposite vertices, which is 180 degrees.

Now, let's look at the given angles. We have ∠BFC = 132 degrees and ∠BGC = 118 degrees. Since these angles are on the same side of the triangle, they must add up to 180 degrees. So, ∠BFC + ∠BGC = 132 + 118 = 250 degrees.

Since the sum of the angles at E and D is 180 degrees, and the sum of the angles at BFC and BGC is 250 degrees, we can conclude that the sum of the angles at E and D is 180 - 250 = -70 degrees. However, the sum of the angles at E and D cannot be negative, so there must be an error in the given information or the question itself.

Therefore, the measure of angle A cannot be determined with the given information.
answer:70
correct answer:70
total:350  correct:90  acc:0.2571428571428571
------------------------------------------------------------------------------------
ID:350
picturePath:image/460.png
answer:30
answer:30
correct answer:48
total:351  correct:90  acc:0.2564102564102564
------------------------------------------------------------------------------------
ID:351
picturePath:image/461.png
answer:The area of the shaded region can be found by subtracting the area of the smaller triangle from the area of the larger triangle. The area of the larger triangle is \(\frac{1}{2} \times AB \times DH\), which is \(\frac{1}{2} \times 8 \times 3 = 12\). The area of the smaller triangle is \(\frac{1}{2} \times CD \times DH\), which is \(\frac{1}{2} \times 4 \times 3 = 6\). Therefore, the area of the shaded region is \(12 - 6 = 6\). The answer is: A.
answer:6
correct answer:26
total:352  correct:90  acc:0.2556818181818182
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ID:352
picturePath:image/462.png
answer:The answer is 20.
answer:20
correct answer:18
total:353  correct:90  acc:0.254957507082153
------------------------------------------------------------------------------------
ID:353
picturePath:image/463.png
answer:Since line a is parallel to line b, we have \begin{align*}
\angle 1 + \angle 3 &= 180^\circ \\
\angle 2 + \angle 3 &= 180^\circ
\end{align*}Substituting the given values, we have \begin{align*}
30^\circ + \angle 3 &= 180^\circ \\
40^\circ + \angle 3 &= 180^\circ
\end{align*}Solving these equations, we find that \begin{align*}
\angle 3 &= 180^\circ - 30^\circ = 150^\circ \\
\angle 3 &= 180^\circ - 40^\circ = 140^\circ
\end{align*}Since AD=AC, we have \begin{align*}
\angle 3 + \angle 2 &= 180^\circ \\
\angle 3 + \angle 2 &= 180^\circ
\end{align*}Solving these equations, we find that \begin{align*}
\angle 3 &= 180^\circ - 40^\circ = 140^\circ \\
\angle 3 &= 180^\circ - 30^\circ = 150^\circ
\end{align*}Therefore, \begin{align*}
\angle 3 &= \boxed{140^\circ}
\end{align*}The answer is: 140
answer:140
correct answer:40
total:354  correct:90  acc:0.2542372881355932
------------------------------------------------------------------------------------
ID:354
picturePath:image/464.png
answer:由于AD平分∠BAC，DE∥AB，∠B=44°，∠C=54°，根据平行线的性质，∠ADE=∠B=44°，所以∠ADE=44°。
answer:44
correct answer:41
total:355  correct:90  acc:0.2535211267605634
------------------------------------------------------------------------------------
ID:355
picturePath:image/465.png
answer:The answer is A.
ID:356
picturePath:image/466.png
answer:This is a geometry problem involving a right triangle and parallel lines. The angle sum of a triangle is 180 degrees, and since DF is parallel to AE, the corresponding angles are congruent. Therefore, ∠ACF = ∠B = 50 degrees. Since ∠ACB is 90 degrees, the sum of the angles in triangle ABC is 180 degrees. Thus, ∠BCF = 180 - ∠ACB - ∠ACF = 180 - 90 - 50 = 40 degrees. The answer is D.
answer:40
correct answer:70
total:357  correct:90  acc:0.25210084033613445
------------------------------------------------------------------------------------
ID:357
picturePath:image/467.png
answer:This is a geometry problem. Given that EO is perpendicular to AB, we know that ∠EOB is 90 degrees. Since ∠BOD is 50 degrees, we can find ∠BOE by subtracting ∠BOD from 90 degrees: 90 - 50 = 40 degrees. Since OE is also perpendicular to CD, we know that ∠COE is also 90 degrees. Therefore, ∠COE is 90 - 40 = 50 degrees. The answer is A.
answer:50
correct answer:140
total:358  correct:90  acc:0.25139664804469275
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ID:358
picturePath:image/469.png
answer:The answer is A.
ID:359
picturePath:image/470.png
answer:∠3=110°-40°=70°.
answer:70
correct answer:70
total:360  correct:91  acc:0.25277777777777777
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ID:360
picturePath:image/471.png
answer:The answer is 10.
answer:10
correct answer:10
total:361  correct:92  acc:0.2548476454293629
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ID:361
picturePath:image/472.png
answer:∠B=45°,∠F=65°,则∠COE的度数为180°-∠B-∠F=180°-45°-65°=70°.
answer:70
correct answer:70
total:362  correct:93  acc:0.2569060773480663
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ID:362
picturePath:image/475.png
answer:5.25
answer:5.25
correct answer:3.5
total:363  correct:93  acc:0.256198347107438
------------------------------------------------------------------------------------
ID:363
picturePath:image/476.png
answer:本题主要考查了直角三角形的性质和三角形的内角和定理．解:如图,∵∠1=27.5°,∴∠3=45°-27.5°=17.5°,∵EF//CD,∴∠2=∠3=17.5°．故选:B．
answer:17.5
correct answer:17.5
total:364  correct:94  acc:0.25824175824175827
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ID:364
picturePath:image/477.png
answer:19
answer:19
correct answer:87
total:365  correct:94  acc:0.25753424657534246
------------------------------------------------------------------------------------
ID:365
picturePath:image/479.png
answer:The answer is 25.
answer:25
correct answer:35
total:366  correct:94  acc:0.2568306010928962
------------------------------------------------------------------------------------
ID:366
picturePath:image/480.png
answer:根据平行线的性质,∠1=∠3,∠2=∠4,又∵∠3+∠4=180°,∠1=130°,∠2=30°,∴∠3=180°-∠1=180°-130°=50°,∠4=180°-∠3=180°-50°=130°,∴∠C=∠4=130°.故选B.
answer:130
correct answer:20
total:367  correct:94  acc:0.2561307901907357
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ID:367
picturePath:image/482.png
answer:This is a geometry problem involving angles and parallel lines. Given that \(AD//BC\), we know that the corresponding angles are congruent. Since \(∠B=32°\), and \(DB\) bisects \(∠ADE\), we can conclude that \(∠ADE=64°\) (since the sum of angles on a straight line is 180°, and \(DB\) bisects \(∠ADE\), each angle on either side of \(DB\) must be 32°/2 = 16°). Therefore, \(∠DEC=∠ADE=64°\). The answer is A.
answer:64
correct answer:64
total:368  correct:95  acc:0.25815217391304346
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ID:368
picturePath:image/483.png
answer:40°
answer:40
correct answer:20
total:369  correct:95  acc:0.25745257452574527
------------------------------------------------------------------------------------
ID:369
picturePath:image/484.png
answer:60
answer:60
correct answer:130
total:370  correct:95  acc:0.25675675675675674
------------------------------------------------------------------------------------
ID:370
picturePath:image/486.png
answer:12cm
answer:12
correct answer:11
total:371  correct:95  acc:0.2560646900269542
------------------------------------------------------------------------------------
ID:371
picturePath:image/487.png
answer:70
answer:70
correct answer:35
total:372  correct:95  acc:0.2553763440860215
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ID:372
picturePath:image/489.png
answer:∠AOC=70°,OM平分∠AOC,所以∠MOA=35°,ON⊥OM,所以∠MON=90°,所以∠CON=90°-35°=55°.故选B
answer:55
correct answer:55
total:373  correct:96  acc:0.257372654155496
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ID:373
picturePath:image/490.png
answer:40
answer:40
correct answer:50
total:374  correct:96  acc:0.25668449197860965
------------------------------------------------------------------------------------
ID:374
picturePath:image/492.png
answer:根据题意,连接OE,OE是直径,所以∠OEF=90°,又∵∠EOD=40°,∴∠OEF-∠EOD=90°-40°=50°,∴∠COF=∠OEF-∠OEF=50°,又∵CD垂直于弦EF,∴∠CDF=∠COF=50°.故选B.
answer:50
correct answer:70
total:375  correct:96  acc:0.256
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ID:375
picturePath:image/493.png
answer:6
answer:6
correct answer:65
total:376  correct:96  acc:0.2553191489361702
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ID:376
picturePath:image/494.png
answer:∠3 = ∠1 - ∠2 = 85° - 35° = 50°.
answer:50
correct answer:50
total:377  correct:97  acc:0.2572944297082228
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ID:377
picturePath:image/495.png
answer:根据题意，AB 垂直于 CD 于点 O，因此 ∠AOC = 90°。又因为 EF 经过点 O，∠1 = 26°，所以 ∠2 = ∠AOC - ∠1 = 90° - 26° = 64°。因此，∠2 的度数是 64°。
answer:64
correct answer:64
total:378  correct:98  acc:0.25925925925925924
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ID:378
picturePath:image/498.png
answer:Since CD is parallel to AB, we have \begin{align*}
\angle BCD &= \angle B \\
&= 50^\circ.
\end{align*}Since BC is perpendicular to AE, we have \begin{align*}
\angle BCD &= 90^\circ - \angle 1 \\
&= 90^\circ - \angle BCD \\
&= 90^\circ - 50^\circ \\
&= 40^\circ.
\end{align*}Therefore, \begin{align*}
\angle 1 &= 90^\circ - \angle BCD \\
&= 90^\circ - 40^\circ \\
&= \boxed{50^\circ}.
\end{align*}The answer is: 50
answer:50
correct answer:40
total:379  correct:98  acc:0.25857519788918204
------------------------------------------------------------------------------------
ID:379
picturePath:image/500.png
answer:解:∵∠1+∠2=100°,∴∠3=∠1+∠2=100°．
故选:A．
answer:100
correct answer:130
total:380  correct:98  acc:0.2578947368421053
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ID:380
picturePath:image/501.png
answer:根据题意，连接AC并延长至D，使CD=CA，连接BC并延长至E，使CE=CB，连接ED。若量出DE=58米，则A,B间的距离为DE=58米。
answer:58
correct answer:58
total:381  correct:99  acc:0.25984251968503935
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ID:381
picturePath:image/502.png
answer:This is a geometry problem involving angles in a square and a parallelogram. Given that ABCD is a parallelogram and EFGH is a square, we can use the properties of these shapes to find the measure of angle B.

Since ABCD is a parallelogram, opposite angles are congruent. Therefore, ∠B = ∠ECD = 35°.

Since EFGH is a square, opposite angles are also congruent. Therefore, ∠AEF = ∠B = 15°.

Since angles ECD and AEF are adjacent to angle B, we can use the fact that the sum of angles in a triangle is 180° to find the measure of angle B.

∠B = 180° - ∠ECD - ∠AEF
∠B = 180° - 35° - 15°
∠B = 130°

Therefore, the measure of angle B is 130°.
answer:130
correct answer:70
total:382  correct:99  acc:0.2591623036649215
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ID:382
picturePath:image/503.png
answer:根据CD平分∠ACB和DE∥BC,可以得出∠DCB=∠EDC=40°,∠DCB+∠AED+∠EDC=180°,所以∠AED=180°-40°-40°=100°.
answer:100
correct answer:80
total:383  correct:99  acc:0.2584856396866841
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ID:383
picturePath:image/504.png
answer:根据平行线的性质，两直线平行，同位角相等。因此，∠2=∠1=68°。故选A。
answer:68
correct answer:112
total:384  correct:99  acc:0.2578125
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ID:384
picturePath:image/506.png
answer:由AB∥EF∥CD,根据三角形的一个外角等于与它不相邻的两个内角的和,可得∠BCE=∠ABC+∠CEF=46°+154°=200°,故选D.
answer:200
correct answer:20
total:385  correct:99  acc:0.2571428571428571
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ID:385
picturePath:image/507.png
answer:50°
answer:50
correct answer:10
total:386  correct:99  acc:0.25647668393782386
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ID:386
picturePath:image/508.png
answer:解:∵∠D=36°,∴∠B=∠D=36°,∵AB是⊙O的直径,∴∠ACB=90°,∴∠CAB=90°-36°=54°．故选:B．
answer:54
correct answer:54
total:387  correct:100  acc:0.25839793281653745
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ID:387
picturePath:image/509.png
answer:解:∵∠CAD=100°,∴∠COA=2∠CAD=200°,∵OA=OC,∴∠OAC=∠OCA=\frac{1}{2}(360°-200°)=80°,∴∠BOC=2∠OAC=160°,∴∠B=\frac{1}{2}(360°-160°)=100°．故选:A．
answer:100
correct answer:80
total:388  correct:100  acc:0.25773195876288657
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ID:388
picturePath:image/510.png
answer:解:∵PC与⊙O相切于点C,∴OC⊥CP,∴∠OCP=90°,∵∠P=20°,∴∠POC=70°,∴∠A=\frac{1}{2}∠POC=35°．故选:B．
answer:35
correct answer:35
total:389  correct:101  acc:0.2596401028277635
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ID:389
picturePath:image/512.png
answer:Since AB is parallel to CD, we can use the corresponding angles postulate, which states that if two parallel lines are cut by a transversal, then the corresponding angles are congruent. In this case, ∠1EF and ∠2EF are corresponding angles, so they are congruent. We are given that ∠2EF = 70°, so ∠1EF = 70°. Since EF is a straight line, the sum of the angles on either side of it is 180°. Therefore, ∠1EF + ∠1EF = 180°. Substituting the given value, we have 70° + ∠1EF = 180°. Solving for ∠1EF, we find that ∠1EF = 180° - 70° = 110°. Thus, ∠1EF = 110°.
answer:110
correct answer:110
total:390  correct:102  acc:0.26153846153846155
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ID:390
picturePath:image/513.png
answer:40
answer:40
correct answer:50
total:391  correct:102  acc:0.2608695652173913
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ID:391
picturePath:image/514.png
answer:DE is the mid-segment of AB, so the length of AB is 2 times the length of DE. Therefore, the length of AB is 2 * 1100m = 2200m.
answer:2200
correct answer:2200
total:392  correct:103  acc:0.2627551020408163
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ID:392
picturePath:image/517.png
answer:80
answer:80
correct answer:10
total:393  correct:103  acc:0.26208651399491095
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ID:393
picturePath:image/518.png
answer:The answer is 25.
answer:25
correct answer:40
total:394  correct:103  acc:0.2614213197969543
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ID:394
picturePath:image/519.png
answer:∠BDC=\frac{1}{2}(180^{°}-∠A)=\frac{1}{2}(180^{°}-110^{°})=\frac{1}{2}×70^{°}=35^{°}.
answer:35
correct answer:140
total:395  correct:103  acc:0.2607594936708861
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ID:395
picturePath:image/520.png
answer:50
answer:50
correct answer:130
total:396  correct:103  acc:0.2601010101010101
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ID:396
picturePath:image/521.png
answer:The answer is 180°.
answer:180
correct answer:240
total:397  correct:103  acc:0.2594458438287154
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ID:397
picturePath:image/523.png
answer:7
answer:7
correct answer:8
total:398  correct:103  acc:0.25879396984924624
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ID:398
picturePath:image/524.png
answer:∠2 = 180° - ∠1 = 180° - 130° = 50°.
answer:50
correct answer:40
total:399  correct:103  acc:0.2581453634085213
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ID:399
picturePath:image/525.png
answer:根据题意,∠ACB=90°,∠A=30°,则∠ABC=60°,∵BC=4cm,∴AB=8cm,∵D为AB的中点,∴AD=4cm,∴CD=\frac{1}{2}AB=\frac{1}{2}×8=4cm.故选:B.
answer:4
correct answer:4
total:400  correct:104  acc:0.26
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ID:400
picturePath:image/526.png
answer:解:∵四边形ABCD为矩形,∴AD∥BC,∴∠AEF+∠F=180°,∵∠AEF=115°,∴∠F=65°,∵矩形ABCD沿EF对折后使两部分叠合,∴∠1=∠F=65°．故选:B．
answer:65
correct answer:50
total:401  correct:104  acc:0.2593516209476309
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ID:401
picturePath:image/527.png
answer:∠3 = 180° - ∠1 - ∠2 - ∠4 = 180° - 120° - 60° - 125° = 5°.
answer:5
correct answer:125
total:402  correct:104  acc:0.25870646766169153
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ID:402
picturePath:image/528.png
answer:The answer is A.
ID:403
picturePath:image/529.png
answer:60
answer:60
correct answer:75
total:404  correct:104  acc:0.25742574257425743
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ID:404
picturePath:image/530.png
answer:根据三角形内角和定理，∠1 + ∠2 + ∠C = 180°，∠C = 80°，所以∠1 + ∠2 = 180° - ∠C = 180° - 80° = 100°。所以，∠1 + ∠2 = 100°。
answer:100
correct answer:260
total:405  correct:104  acc:0.25679012345679014
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ID:405
picturePath:image/531.png
answer:∠4 = 180° - ∠3 = 180° - 108° = 72°.
answer:72
correct answer:72
total:406  correct:105  acc:0.25862068965517243
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ID:406
picturePath:image/533.png
answer:∠DCE = 90° - ∠1 = 90° - 34° = 56°.
answer:56
correct answer:56
total:407  correct:106  acc:0.26044226044226043
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ID:407
picturePath:image/534.png
answer:60
answer:60
correct answer:30
total:408  correct:106  acc:0.25980392156862747
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ID:408
picturePath:image/536.png
answer:40
answer:40
correct answer:100
total:409  correct:106  acc:0.2591687041564792
------------------------------------------------------------------------------------
ID:409
picturePath:image/537.png
answer:根据平行线的性质，∠1和∠3是同位角，所以∠3=∠1=40°。又因为∠2和∠4是邻补角，所以∠2+∠4=90°。又因为∠2=60°，所以∠4=90°-60°=30°。在△EOH中，∠EOH=180°-∠3-∠4=180°-40°-30°=110°。所以，∠EOH的角度为110°。
answer:110
correct answer:100
total:410  correct:106  acc:0.25853658536585367
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ID:410
picturePath:image/538.png
answer:解:连接BD,∵BC是⊙O的切线,∴∠DBC=90°,∵∠C=40°,∴∠BDC=50°,∵∠AED=∠BDC,∴∠AED=50°．故选:B．
answer:50
correct answer:40
total:411  correct:106  acc:0.25790754257907544
------------------------------------------------------------------------------------
ID:411
picturePath:image/539.png
answer:根据线段的垂直平分线的性质，点E是线段AB的垂直平分线，所以AE=BE，在等腰△ABC中，AB=AC，∠A=40°，所以∠ABC=∠ACB=\frac{180°-40°}{2}=70°，所以∠CBE=∠ABC=\frac{180°-70°}{2}=40°．故选A．
answer:40
correct answer:30
total:412  correct:106  acc:0.25728155339805825
------------------------------------------------------------------------------------
ID:412
picturePath:image/540.png
answer:解:∵等腰三角形ABC中,AB=AC,∠A=36°,∴∠ABC=72°,∵BD是AC边上的高,∴∠CBD=90°,∴∠DBC=180°-90°-72°=18°．故选:A．
answer:18
correct answer:18
total:413  correct:107  acc:0.25907990314769974
------------------------------------------------------------------------------------
ID:413
picturePath:image/543.png
answer:解:∵四边形ABCD是⊙O的内接四边形,∴∠A+∠D=180°,∵∠D=∠BCD=120°,∴∠A=60°,∵PD是切线,∴∠ADP=90°,∴∠ADP=90°-∠A=90°-60°=30°．故选:B．
answer:30
correct answer:30
total:414  correct:108  acc:0.2608695652173913
------------------------------------------------------------------------------------
ID:414
picturePath:image/544.png
answer:50°
answer:50
correct answer:25
total:415  correct:108  acc:0.26024096385542167
------------------------------------------------------------------------------------
ID:415
picturePath:image/545.png
answer:解:∵四边形ABCD是矩形,∴∠ADC=90°,AD=BC=3,∵S△ABC=\frac{1}{2}AB•BC=\frac{1}{2}×8×3=12,∴S△AB′C=\frac{1}{2}S△ABC=\frac{1}{2}×12=6,∵BE是△AB′C的中位线,∴BE=\frac{1}{2}AB′,∴AB′=2BE,∴S△AB′C=\frac{1}{2}S△ABC=\frac{1}{2}×12=6,∴BE=\frac{1}{2}AB′=\frac{1}{2}×2BE=BE,∴BE是△AB′C的中位线,∴BE=\frac{1}{2}(AB′+BC)=\frac{1}{2}(2BE+3)=\frac{1}{2}(5BE)=\frac{5}{2}BE,∴BE=\frac{5}{4}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\frac{5}{3}BE,∴BE=\
answer:3
correct answer:22
total:416  correct:108  acc:0.25961538461538464
------------------------------------------------------------------------------------
ID:416
picturePath:image/546.png
answer:解:∵OA=OB,OC=OD,∴∠OAB=∠OBA,∠OCD=∠ODC,∵∠O=60°,∠C=25°,∴∠OAB=60°,∠OCD=25°,∴∠BED=∠OAB+∠OCD=60°+25°=85°．故选:D．
answer:85
correct answer:70
total:417  correct:108  acc:0.2589928057553957
------------------------------------------------------------------------------------
ID:417
picturePath:image/547.png
answer:20
answer:20
correct answer:100
total:418  correct:108  acc:0.2583732057416268
------------------------------------------------------------------------------------
ID:418
picturePath:image/548.png
answer:The answer is A.
ID:419
picturePath:image/549.png
answer:解:∵AB∥CD,∠AGE=128°,∴∠G=∠AGE=128°,∴∠EHD=180°-128°=52°,∵HM平分∠EHD,∴∠MHD=\frac{1}{2}∠EHD=26°,故选:B．
answer:26
correct answer:26
total:420  correct:109  acc:0.25952380952380955
------------------------------------------------------------------------------------
ID:420
picturePath:image/550.png
answer:75
answer:75
correct answer:65
total:421  correct:109  acc:0.2589073634204275
------------------------------------------------------------------------------------
ID:421
picturePath:image/551.png
answer:50
answer:50
correct answer:70
total:422  correct:109  acc:0.25829383886255924
------------------------------------------------------------------------------------
ID:422
picturePath:image/553.png
answer:解:∵四边形ABCD是矩形,∴AD=BC=10,∠ADC=90°,∴AC=√{AD²+DC²}=√{10²+10²}=20,∵点D落在BC边上的点F处,∴DF=CF,∴△CEF的周长=CE+EF+CF=CE+DF+CF=AC=20．故选:A．
answer:20
correct answer:12
total:423  correct:109  acc:0.2576832151300236
------------------------------------------------------------------------------------
ID:423
picturePath:image/554.png
answer:60
answer:60
correct answer:160
total:424  correct:109  acc:0.25707547169811323
------------------------------------------------------------------------------------
ID:424
picturePath:image/555.png
answer:60°
answer:60
correct answer:20
total:425  correct:109  acc:0.2564705882352941
------------------------------------------------------------------------------------
ID:425
picturePath:image/556.png
answer:The answer is 60°.
answer:60
correct answer:45
total:426  correct:109  acc:0.25586854460093894
------------------------------------------------------------------------------------
ID:426
picturePath:image/557.png
answer:60
answer:60
correct answer:15
total:427  correct:109  acc:0.25526932084309134
------------------------------------------------------------------------------------
ID:427
picturePath:image/558.png
answer:∠AED=∠B,∵AB∥CD,∠C=70°,∴∠B=∠C=70°,∵AE平分∠CAB,∴∠CAE=∠B=70°,∴∠AED=∠CAE=70°.故选B.
answer:70
correct answer:125
total:428  correct:109  acc:0.2546728971962617
------------------------------------------------------------------------------------
ID:428
picturePath:image/559.png
answer:解:∵AB是⊙O的直径,∴∠ADB=90°,∵∠A=∠ACD=30°,∴∠BAD=90°-∠A=60°．故选:C．
answer:60
correct answer:60
total:429  correct:110  acc:0.2564102564102564
------------------------------------------------------------------------------------
ID:429
picturePath:image/560.png
answer:The answer is 55.
answer:55
correct answer:55
total:430  correct:111  acc:0.25813953488372093
------------------------------------------------------------------------------------
ID:430
picturePath:image/561.png
answer:根据菱形的性质，对角线AC和BD互相垂直平分，所以∠AOB=90°，AO=CO=4，BO=DO=3。在Rt△AOB中，AB=√{AO²+BO²}=√{4²+3²}=5。因此，△ABC的周长是AB+AC+BC=5+4+5=14。故选B。
answer:14
correct answer:18
total:431  correct:111  acc:0.25754060324825984
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ID:431
picturePath:image/562.png
answer:25°
answer:25
correct answer:50
total:432  correct:111  acc:0.2569444444444444
------------------------------------------------------------------------------------
ID:432
picturePath:image/563.png
answer:The answer is A.
ID:433
picturePath:image/564.png
answer:60°
answer:60
correct answer:60
total:434  correct:112  acc:0.25806451612903225
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ID:434
picturePath:image/565.png
answer:25°
answer:25
correct answer:55
total:435  correct:112  acc:0.2574712643678161
------------------------------------------------------------------------------------
ID:435
picturePath:image/567.png
answer:∠BIC = 180° - ∠A - ∠ABC - ∠ACB = 180° - 66° - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - (∠ABC + ∠ACB) = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC - ∠ACB = 180° - ∠ABC - ∠ACB - ∠ABC -
answer:180
correct answer:123
total:436  correct:112  acc:0.25688073394495414
------------------------------------------------------------------------------------
ID:436
picturePath:image/569.png
answer:解:∵a∥b,∴∠1=∠3,∵AB⊥BC,∠1=35°,∴∠3=90°-35°=55°,∴∠2=55°．故选:B．
answer:55
correct answer:55
total:437  correct:113  acc:0.2585812356979405
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ID:437
picturePath:image/570.png
answer:30°
answer:30
correct answer:60
total:438  correct:113  acc:0.2579908675799087
------------------------------------------------------------------------------------
ID:438
picturePath:image/571.png
answer:70°
answer:70
correct answer:110
total:439  correct:113  acc:0.25740318906605925
------------------------------------------------------------------------------------
ID:439
picturePath:image/572.png
answer:根据EF∥AC和∠C=60°,我们可以得出∠EFB=∠C=60°.同理,根据DF∥AB和∠B=45°,我们可以得出∠DFC=∠B=45°.因此,∠EFD=∠EFB+∠DFC=60°+45°=105°.
answer:105
correct answer:75
total:440  correct:113  acc:0.25681818181818183
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ID:440
picturePath:image/573.png
answer:解:∵m∥n,∴∠1+∠BAC=180°,∵AC⊥AB,∴∠BAC=90°,∴∠1=45°,∵m∥n,∴∠2=∠1=45°．故选:B．
answer:45
correct answer:55
total:441  correct:113  acc:0.2562358276643991
------------------------------------------------------------------------------------
ID:441
picturePath:image/574.png
answer:由EF⊥AB于O,得∠AOE=90°,∵∠COE=50°,∴∠AOC=90°-∠COE=40°,∴∠BOD=∠AOC=40°.故选B
answer:40
correct answer:40
total:442  correct:114  acc:0.2579185520361991
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ID:442
picturePath:image/575.png
answer:解:∵6+12+20+28+36+44+52+60=272(个),∴第8个图形中花盆的个数为60个．故选:A．
answer:60
correct answer:90
total:443  correct:114  acc:0.25733634311512416
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ID:443
picturePath:image/576.png
answer:20°
answer:20
correct answer:85
total:444  correct:114  acc:0.25675675675675674
------------------------------------------------------------------------------------
ID:444
picturePath:image/577.png
answer:45°
answer:45
correct answer:35
total:445  correct:114  acc:0.25617977528089886
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ID:445
picturePath:image/578.png
answer:∠D = 180° - ∠B = 180° - 72° = 108°.
answer:108
correct answer:108
total:446  correct:115  acc:0.257847533632287
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ID:446
picturePath:image/579.png
answer:根据题意,∠ADC=54°,∵AB是量角器的直径,∴∠ADB=90°,∴∠C=90°-∠ADC=90°-54°=36°,∵∠BCD=∠C,∴∠BCD=36°.故选B.
answer:36
correct answer:63
total:447  correct:115  acc:0.25727069351230425
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ID:447
picturePath:image/580.png
answer:60
answer:60
correct answer:60
total:448  correct:116  acc:0.25892857142857145
------------------------------------------------------------------------------------
ID:448
picturePath:image/581.png
answer:This is a geometry problem involving an isosceles trapezoid. To solve it, we can use the formula for the area of a trapezoid, which is (1/2) * (sum of the lengths of the parallel sides) * (distance between the parallel sides). In this case, the parallel sides are BD and DE, and the distance between them is the height of the trapezoid, which is the perpendicular distance from D to AC. Since AC is parallel to DE, this distance is also the height of the triangle ADE. Therefore, the area of triangle ADE is (1/2) * AB * height, and the area of trapezoid BDE is (1/2) * (BD + DE) * height. Since BD and DE are parallel, their sum is equal to the length of the base of the trapezoid, which is AB. Therefore, the area of trapezoid BDE is (1/2) * AB * height, which is equal to the area of triangle ADE. Therefore, the area of trapezoid BDE is (1/2) * 5 * 6 = 15 square units.
answer:15
correct answer:18
total:449  correct:116  acc:0.2583518930957684
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ID:449
picturePath:image/583.png
answer:40°
answer:40
correct answer:60
total:450  correct:116  acc:0.2577777777777778
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ID:450
picturePath:image/585.png
answer:解:∵PC⊥PD,∠APC=28°,∴∠CPD=90°-∠APC=62°,∴∠BPD=∠CPD=62°．故选:B．
answer:62
correct answer:62
total:451  correct:117  acc:0.25942350332594233
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ID:451
picturePath:image/587.png
answer:50°
answer:50
correct answer:50
total:452  correct:118  acc:0.2610619469026549
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ID:452
picturePath:image/588.png
answer:解:∵点A、B的读数分别为86°、30°,∴∠ACB=86°-30°=56°．故选:D．
answer:56
correct answer:28
total:453  correct:118  acc:0.26048565121412803
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ID:453
picturePath:image/590.png
answer:解:∵∠A=42°,∴∠C=∠A=42°,∵∠APD=77°,∴∠B=∠APD-∠C=77°-42°=35°,
故选:B．
answer:35
correct answer:35
total:454  correct:119  acc:0.2621145374449339
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ID:454
picturePath:image/591.png
answer:20°
answer:20
correct answer:65
total:455  correct:119  acc:0.26153846153846155
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ID:455
picturePath:image/592.png
answer:30cm
answer:30
correct answer:40
total:456  correct:119  acc:0.26096491228070173
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ID:456
picturePath:image/593.png
answer:∠ABC的度数等于∠BCD的度数，因为AB∥EF，所以∠ABC和∠BCD是对顶角，所以∠ABC的度数等于∠BCD的度数，即∠ABC=140°。
answer:140
correct answer:50
total:457  correct:119  acc:0.2603938730853392
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ID:457
picturePath:image/594.png
answer:60°
answer:60
correct answer:120
total:458  correct:119  acc:0.259825327510917
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ID:458
picturePath:image/595.png
answer:解:∵正六边形ABCDEF内接于⊙O,⊙O的半径为6,∴正六边形的边长为6,∴△BOC的面积=\frac{1}{2}×6×6×\frac{π}{6}=\frac{3π}{2}．故选:A．
answer:2
correct answer:6
total:459  correct:119  acc:0.25925925925925924
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ID:459
picturePath:image/596.png
answer:75
answer:75
correct answer:75
total:460  correct:120  acc:0.2608695652173913
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ID:460
picturePath:image/598.png
answer:60
answer:60
correct answer:75
total:461  correct:120  acc:0.2603036876355748
------------------------------------------------------------------------------------
ID:461
picturePath:image/600.png
answer:60
answer:60
correct answer:30
total:462  correct:120  acc:0.2597402597402597
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ID:462
picturePath:image/601.png
answer:30
answer:30
correct answer:42
total:463  correct:120  acc:0.2591792656587473
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ID:463
picturePath:image/602.png
answer:70°
answer:70
correct answer:90
total:464  correct:120  acc:0.25862068965517243
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ID:464
picturePath:image/603.png
answer:40
answer:40
correct answer:53
total:465  correct:120  acc:0.25806451612903225
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ID:465
picturePath:image/606.png
answer:根据平行线的性质,∠1和∠2互为同位角,所以∠1=∠2=110°,又因为∠1和∠3互为邻补角,所以∠1+∠3=180°,解得∠3=180°-∠1=180°-110°=70°.故选B.
answer:70
correct answer:60
total:466  correct:120  acc:0.2575107296137339
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ID:466
picturePath:image/607.png
answer:40
answer:40
correct answer:80
total:467  correct:120  acc:0.2569593147751606
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ID:467
picturePath:image/609.png
answer:6.
answer:6
correct answer:12
total:468  correct:120  acc:0.2564102564102564
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ID:468
picturePath:image/611.png
answer:16+2+2=20(厘米)。
answer:20
correct answer:20
total:469  correct:121  acc:0.2579957356076759
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ID:469
picturePath:image/613.png
answer:This is a geometry problem involving rotation. Since the triangle is being rotated 48 degrees around point C, and the angle BAC is 90 degrees, the angle B'AC will be the supplement of the angle BAC, which is 180 - 90 = 90 degrees. After the rotation, the angle B'AC will be 90 degrees - 48 degrees = 42 degrees. Therefore, the angle B'AC is 42 degrees.
answer:42
correct answer:42
total:470  correct:122  acc:0.25957446808510637
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ID:470
picturePath:image/614.png
answer:5cm
answer:5
correct answer:3
total:471  correct:122  acc:0.25902335456475584
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ID:471
picturePath:image/615.png
answer:70°
answer:70
correct answer:70
total:472  correct:123  acc:0.2605932203389831
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ID:472
picturePath:image/618.png
answer:根据题意,AB的垂直平分线交AC于D,则BD=CD,又∵AB=AC=8,BC=5,∴△BCD的周长=BC+CD+BD=BC+CD+CD=2CD+BC=2CD+5,∵CD=\frac{1}{2}AB=4,∴△BCD的周长=2CD+BC=2×4+5=13.
answer:13
correct answer:13
total:473  correct:124  acc:0.26215644820295986
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ID:473
picturePath:image/620.png
answer:16
answer:16
correct answer:10
total:474  correct:124  acc:0.2616033755274262
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ID:474
picturePath:image/621.png
answer:根据三角形的一个外角等于与它不相邻的两个内角的和，可得∠AEC=∠DCB+∠EAB=35°+23°=58°.
answer:58
correct answer:58
total:475  correct:125  acc:0.2631578947368421
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ID:475
picturePath:image/622.png
answer:解:∵OB是△ABC的外接圆直径,∴∠AOB=180°,∴∠3=\frac{1}{2}∠AOB=90°,∵∠1=37°,∴∠2=∠3-∠1=90°-37°=53°．故选:B．
answer:53
correct answer:53
total:476  correct:126  acc:0.2647058823529412
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ID:476
picturePath:image/625.png
answer:50°
answer:50
correct answer:10
total:477  correct:126  acc:0.2641509433962264
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ID:477
picturePath:image/628.png
answer:This is a geometric figure with labeled points A, B, C, D, and E, where ABCDE is a regular pentagon. Given that BA=BC=BD=BE and AE is parallel to BC, we can deduce that triangle ABE is an isosceles triangle. Since the sum of the angles in a triangle is 180 degrees, and we know that angle A is 60 degrees, the other two angles in triangle ABE must each be (180-60)/2 = 60/2 = 30 degrees. Therefore, angle CDE is also 30 degrees.
answer:30
correct answer:150
total:478  correct:126  acc:0.26359832635983266
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ID:478
picturePath:image/630.png
answer:The answer is A.
ID:479
picturePath:image/631.png
answer:The answer is 25.
answer:25
correct answer:65
total:480  correct:126  acc:0.2625
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ID:480
picturePath:image/632.png
answer:7.
answer:7
correct answer:7
total:481  correct:127  acc:0.26403326403326405
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ID:481
picturePath:image/645.png
answer:∠BCD = 30° + 50° = 80°
answer:80
correct answer:80
total:482  correct:128  acc:0.26556016597510373
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ID:482
picturePath:image/649.png
answer:The answer is 50.
answer:50
correct answer:50
total:483  correct:129  acc:0.2670807453416149
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ID:483
picturePath:image/654.png
answer:The answer is A.
ID:484
picturePath:image/660.png
answer:∠BOE的度数等于∠EOC的一半，即70°/2=35°。
answer:35
correct answer:145
total:485  correct:129  acc:0.26597938144329897
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ID:485
picturePath:image/661.png
answer:40
answer:40
correct answer:50
total:486  correct:129  acc:0.2654320987654321
------------------------------------------------------------------------------------
ID:486
picturePath:image/663.png
answer:60
answer:60
correct answer:80
total:487  correct:129  acc:0.2648870636550308
------------------------------------------------------------------------------------
ID:487
picturePath:image/664.png
answer:20°
answer:20
correct answer:110
total:488  correct:129  acc:0.26434426229508196
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ID:488
picturePath:image/665.png
answer:根据题意，内切圆与边BC、CA、AB分别相切于点D、E、F，则∠EDC = ∠A = 50°，∠ECD = 90°，∠EDF = 180° - ∠EDC - ∠ECD = 180° - 50° - 90° = 40°.
answer:40
correct answer:65
total:489  correct:129  acc:0.26380368098159507
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ID:489
picturePath:image/667.png
answer:12cm
answer:12
correct answer:10
total:490  correct:129  acc:0.26326530612244897
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ID:490
picturePath:image/674.png
answer:解:∵AB为⊙O的直径,∴∠ACB=90°,∵∠CBA=30°,∴∠BAC=90°-30°=60°．故选:B．
answer:60
correct answer:60
total:491  correct:130  acc:0.26476578411405294
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ID:491
picturePath:image/681.png
answer:25°
answer:25
correct answer:115
total:492  correct:130  acc:0.26422764227642276
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ID:492
picturePath:image/691.png
answer:6.
answer:6
correct answer:96
total:493  correct:130  acc:0.26369168356997974
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ID:493
picturePath:image/695.png
answer:2.
answer:2
correct answer:6.5
total:494  correct:130  acc:0.2631578947368421
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ID:494
picturePath:image/697.png
answer:25°
answer:25
correct answer:55
total:495  correct:130  acc:0.26262626262626265
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ID:495
picturePath:image/698.png
answer:本题考查三角形的高和三角形内角和定理,根据三角形内角和定理可得∠ABC+∠ACB+∠BAC=180°,再根据三角形的两边夹一角可得∠ABC+∠ACB=130°,再根据三角形的高可得∠BHC=∠ABC+∠ACB-∠BAC=130°-50°=80°．故选:B．
answer:80
correct answer:130
total:496  correct:130  acc:0.2620967741935484
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ID:496
picturePath:image/701.png
answer:解:∵点C是弧AB的中点,∴弧AB的度数是(50°×2)÷2=25°,∴∠BOC=25°．故选:B．
answer:25
correct answer:40
total:497  correct:130  acc:0.26156941649899396
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ID:497
picturePath:image/703.png
answer:60
answer:60
correct answer:90
total:498  correct:130  acc:0.26104417670682734
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ID:498
picturePath:image/705.png
answer:Since ∠BAC=30°, we know that ∠AOB = 2∠BAC = 60°. Therefore, triangle AOB is an equilateral triangle, and the diameter of the circle is equal to the side length of the triangle, which is 12 units. Thus, the diameter of the circle is 12 units.
answer:12
correct answer:24
total:499  correct:130  acc:0.2605210420841683
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ID:499
picturePath:image/708.png
answer:8
answer:8
correct answer:8
total:500  correct:131  acc:0.262
------------------------------------------------------------------------------------
ID:500
picturePath:image/710.png
answer:75°
answer:75
correct answer:65
total:501  correct:131  acc:0.26147704590818366
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ID:501
picturePath:image/713.png
answer:The answer is A.
ID:502
picturePath:image/717.png
answer:40
answer:40
correct answer:40
total:503  correct:132  acc:0.2624254473161034
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ID:503
picturePath:image/718.png
answer:100-47=53°,∠DEF=∠A=53°.故选B
answer:53
correct answer:33
total:504  correct:132  acc:0.2619047619047619
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ID:504
picturePath:image/721.png
answer:∠BPC=∠1+∠2=32°+25°=57°
answer:57
correct answer:57
total:505  correct:133  acc:0.2633663366336634
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ID:505
picturePath:image/722.png
answer:第一次拐的角∠B等于142°，第二次拐的角∠C的度数为180°-142°=38°。
answer:38
correct answer:142
total:506  correct:133  acc:0.2628458498023715
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ID:506
picturePath:image/726.png
answer:The answer is 20.
answer:20
correct answer:10
total:507  correct:133  acc:0.26232741617357
------------------------------------------------------------------------------------
ID:507
picturePath:image/728.png
answer:This is a geometry problem. Given that \( \angle ACB = 90^\circ \), we know that \( \triangle ABC \) is a right triangle. Since \( AD \) is an altitude of \( \triangle ABC \) and \( DE \) is an altitude of \( \triangle ABD \), we can use the properties of similar triangles to solve for the length of \( BC \).

Since \( AD \) is an altitude of \( \triangle ABC \) and \( DE \) is an altitude of \( \triangle ABD \), we can see that \( \triangle ABD \) is similar to \( \triangle ABC \). Therefore, the ratio of corresponding sides is equal.

Let's denote the length of \( BC \) as \( x \). Since \( AD \) is an altitude of \( \triangle ABC \) and \( DE \) is an altitude of \( \triangle ABD \), we have:

\[ \frac{DE}{AD} = \frac{BD}{BC} \]

Substituting the given values, we get:

\[ \frac{1.5}{AD} = \frac{3}{x} \]

Cross-multiplying, we get:

\[ 1.5x = 3AD \]

Since \( AD \) is an altitude of \( \triangle ABC \), we know that \( AD \) is also the height of \( \triangle ABC \) with base \( BC \). Therefore, \( AD = \frac{1}{2}BC \).

Substituting this into the equation, we get:

\[ 1.5x = 3 \cdot \frac{1}{2}BC \]

Simplifying, we get:

\[ 1.5x = \frac{3}{2}BC \]

Dividing both sides by \( \frac{3}{2} \), we get:

\[ x = \frac{2}{3}BC \]

Substituting the given value of \( BD \), we get:

\[ x = \frac{2}{3} \cdot 3 \]

Simplifying, we get:

\[ x = 2 \]

Therefore, the length of \( BC \) is \( 2 \) cm.
answer:2
correct answer:4.5
total:508  correct:133  acc:0.2618110236220472
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ID:508
picturePath:image/730.png
answer:20
answer:20
correct answer:20
total:509  correct:134  acc:0.2632612966601179
------------------------------------------------------------------------------------
ID:509
picturePath:image/731.png
answer:This is a geometric figure with labeled points A, B, C, D, E, and F. The figure shows that AB=AC, AD=AE, ∠BAC=∠DAE, ∠1=25°, ∠2=30°, and ∠3 is unknown. To find the value of ∠3, we can use the properties of triangles and angles. Since AB=AC and AD=AE, we can conclude that triangle ABC is congruent to triangle AEC, and triangle ABD is congruent to triangle ADE. This means that the angles opposite the congruent sides are also congruent. Therefore, ∠BAC=∠DAE and ∠1=∠2. Since ∠1=25° and ∠2=30°, we can conclude that ∠BAC=∠DAE=25°. Finally, since the sum of the angles in a triangle is 180°, we can find the value of ∠3 by subtracting the known angles from 180°. Therefore, ∠3=180°-25°-25°=130°. So, the value of ∠3 is 130°.
answer:130
correct answer:55
total:510  correct:134  acc:0.2627450980392157
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ID:510
picturePath:image/732.png
answer:解:∵AB∥CF,∴△ABD∽△CFD,∴\frac{AB}{CF}=\frac{BD}{CD},∵AB=8㎝,CF=5㎝,∴\frac{8}{5}=\frac{BD}{CD},∵E为DF的中点,∴DE=\frac{1}{2}CD,∴\frac{8}{5}=\frac{BD}{DE},∴BD=\frac{8}{5}×\frac{1}{2}CD=\frac{8}{5}×\frac{1}{2}(CD+DE)=\frac{8}{5}×\frac{1}{2}CD+\frac{8}{5}×\frac{1}{2}DE=\frac{8}{5}×\frac{1}{2}CD+\frac{8}{5}×\frac{1}{2}×\frac{1}{2}CD=\frac{8}{5}×\frac{1}{2}×\frac{3}{2}CD=\frac{12}{5}CD,∴BD=\frac{12}{5}CD,∴BD=\frac{12}{5}×5=12(厘米)．故选:D．
answer:12
correct answer:3
total:511  correct:134  acc:0.2622309197651663
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ID:511
picturePath:image/734.png
answer:To solve this problem, we can use the properties of angles in a triangle and the fact that the sum of the angles in a triangle is 180 degrees.

Since AD is an angle bisector of angle BAC, we know that the sum of the angles on one side of the bisector is equal to the sum of the angles on the other side. Therefore, we have:

∠BAD + ∠CAD = ∠BAC

We also know that the sum of the angles in a triangle is 180 degrees, so we have:

∠B + ∠C + ∠BAC = 180

Substituting the given values, we have:

46° + 54° + ∠BAC = 180

Combining like terms, we have:

100° + ∠BAC = 180

Subtracting 100° from both sides, we have:

∠BAC = 80°

Since AD is an angle bisector, we have:

∠BAD = ∠BAC / 2

Substituting the value of ∠BAC, we have:

∠BAD = 80° / 2

Simplifying, we have:

∠BAD = 40°

Therefore, the size of angle BAD is 40 degrees.

The answer is: 40
answer:40
correct answer:40
total:512  correct:135  acc:0.263671875
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ID:512
picturePath:image/735.png
answer:This is a geometric figure with three sides labeled as A, B, and C, and three angles labeled as 1, 2, and 3. The figure is a triangle, and the sum of the angles in a triangle is 180 degrees. Given that ∠BAC is 78 degrees, and ∠B is equal to ∠1, we can deduce that ∠B is also 78 degrees. Since the sum of the angles in a triangle is 180 degrees, and we already know that ∠B is 78 degrees, we can conclude that ∠2 is 180 - 78 - 78 = 24 degrees. Therefore, the answer is D.
answer:24
correct answer:78
total:513  correct:135  acc:0.2631578947368421
------------------------------------------------------------------------------------
ID:513
picturePath:image/736.png
answer:540
answer:540
correct answer:240
total:514  correct:135  acc:0.26264591439688717
------------------------------------------------------------------------------------
ID:514
picturePath:image/743.png
answer:2.8m
answer:2.8
correct answer:3.5
total:515  correct:135  acc:0.2621359223300971
------------------------------------------------------------------------------------
ID:515
picturePath:image/747.png
answer:3.6
answer:3.6
correct answer:4.45
total:516  correct:135  acc:0.2616279069767442
------------------------------------------------------------------------------------
ID:516
picturePath:image/749.png
answer:This is a right triangle, so we can use the Pythagorean theorem to find the length of the hypotenuse, AB. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we have:

AB^2 = AC^2 + BC^2

Substituting the given values, we get:

10^2 = 6^2 + BC^2

100 = 36 + BC^2

BC^2 = 64

BC = 8

Now, we can use the Pythagorean theorem again to find the length of BE:

BE^2 = AB^2 - AE^2

BE^2 = 10^2 - 6^2

BE^2 = 100 - 36

BE^2 = 64

BE = 8

Therefore, the length of BE is 8 cm.
answer:8
correct answer:4
total:517  correct:135  acc:0.2611218568665377
------------------------------------------------------------------------------------
ID:517
picturePath:image/756.png
answer:解:∵AB⊥BD,AC⊥CD,∴∠B=∠C=90°,∴∠ADC=∠C+∠CAD=90°+35°=125°．故选:D．
answer:125
correct answer:55
total:518  correct:135  acc:0.2606177606177606
------------------------------------------------------------------------------------
ID:518
picturePath:image/757.png
answer:90
answer:90
correct answer:130
total:519  correct:135  acc:0.26011560693641617
------------------------------------------------------------------------------------
ID:519
picturePath:image/758.png
answer:The answer is 36°.
answer:36
correct answer:69
total:520  correct:135  acc:0.25961538461538464
------------------------------------------------------------------------------------
ID:520
picturePath:image/759.png
answer:75
answer:75
correct answer:75
total:521  correct:136  acc:0.26103646833013433
------------------------------------------------------------------------------------
ID:521
picturePath:image/760.png
answer:根据对顶角相等，我们可以得出∠C=∠A=66°。
answer:66
correct answer:48
total:522  correct:136  acc:0.26053639846743293
------------------------------------------------------------------------------------
ID:522
picturePath:image/767.png
answer:50°
answer:50
correct answer:25
total:523  correct:136  acc:0.26003824091778205
------------------------------------------------------------------------------------
ID:523
picturePath:image/769.png
answer:60°
answer:60
correct answer:60
total:524  correct:137  acc:0.26145038167938933
------------------------------------------------------------------------------------
ID:524
picturePath:image/776.png
answer:60
answer:60
correct answer:20
total:525  correct:137  acc:0.26095238095238094
------------------------------------------------------------------------------------
ID:525
picturePath:image/777.png
answer:20
answer:20
correct answer:40
total:526  correct:137  acc:0.26045627376425856
------------------------------------------------------------------------------------
ID:526
picturePath:image/778.png
answer:100
answer:100
correct answer:80
total:527  correct:137  acc:0.25996204933586337
------------------------------------------------------------------------------------
ID:527
picturePath:image/786.png
answer:50°
answer:50
correct answer:130
total:528  correct:137  acc:0.25946969696969696
------------------------------------------------------------------------------------
ID:528
picturePath:image/793.png
answer:40
answer:40
correct answer:40
total:529  correct:138  acc:0.2608695652173913
------------------------------------------------------------------------------------
ID:529
picturePath:image/795.png
answer:解:∵DB⊥AE于B,DC⊥AF于C,∴∠BAE=∠CAF=90°,∵∠BAC=40°,∴∠BAF=90°-40°=50°,∵DB=DC,∴∠BDC=∠BDC=70°,∵∠ADG=130°,∴∠DGF=180°-∠ADG=180°-130°=50°,故选:B．
answer:50
correct answer:150
total:530  correct:138  acc:0.26037735849056604
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ID:530
picturePath:image/796.png
answer:∠AOB的度数是150°。这是因为OC是∠AOB的角平分线，所以∠AOB=2∠AOC，而∠AOC已知为75°，因此∠AOB=2*75°=150°。
answer:150
correct answer:150
total:531  correct:139  acc:0.2617702448210923
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ID:531
picturePath:image/806.png
answer:70°
answer:70
correct answer:70
total:532  correct:140  acc:0.2631578947368421
------------------------------------------------------------------------------------
ID:532
picturePath:image/808.png
answer:解:∵OB=OC,∠OCB=40°,∴∠OBC=∠OCB=40°,∴∠BOC=180°-∠OBC-∠OCB=100°,∴∠A=\frac{1}{2}∠BOC=50°．故选:B．
answer:50
correct answer:50
total:533  correct:141  acc:0.26454033771106944
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ID:533
picturePath:image/809.png
answer:∠CDE=150°,则∠CDB=180°-150°=30°,又∵BE平分∠ABC,∴∠ABE=15°,∵AB//CD,∴∠CDB=∠ABE,∴∠C=∠CDB=30°.故选B
answer:30
correct answer:120
total:534  correct:141  acc:0.2640449438202247
------------------------------------------------------------------------------------
ID:534
picturePath:image/810.png
answer:The answer is 25.
answer:25
correct answer:40
total:535  correct:141  acc:0.2635514018691589
------------------------------------------------------------------------------------
ID:535
picturePath:image/818.png
answer:∠A=\frac{1}{2}∠BOC=\frac{1}{2}×70°=35°.
answer:35
correct answer:35
total:536  correct:142  acc:0.26492537313432835
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ID:536
picturePath:image/821.png
answer:60
answer:60
correct answer:90
total:537  correct:142  acc:0.2644320297951583
------------------------------------------------------------------------------------
ID:537
picturePath:image/822.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵∠CAB=65°,∴∠B=90°-65°=25°,∴∠D=∠B=25°．故选:B．
answer:25
correct answer:25
total:538  correct:143  acc:0.26579925650557623
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ID:538
picturePath:image/826.png
answer:360
answer:360
correct answer:180
total:539  correct:143  acc:0.2653061224489796
------------------------------------------------------------------------------------
ID:539
picturePath:image/829.png
answer:根据勾股定理，OC² + AC² = AB²，已知OC = 3cm，AC = 5cm - 3cm = 2cm，所以AB² = OC² + AC² = 3² + 2² = 13，解得AB = √13 ≈ 3.61cm.
answer:3.61
correct answer:8
total:540  correct:143  acc:0.26481481481481484
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ID:540
picturePath:image/830.png
answer:54°
answer:54
correct answer:36
total:541  correct:143  acc:0.2643253234750462
------------------------------------------------------------------------------------
ID:541
picturePath:image/834.png
answer:30
answer:30
correct answer:30
total:542  correct:144  acc:0.2656826568265683
------------------------------------------------------------------------------------
ID:542
picturePath:image/835.png
answer:6cm
answer:6
correct answer:6
total:543  correct:145  acc:0.26703499079189685
------------------------------------------------------------------------------------
ID:543
picturePath:image/837.png
answer:The answer is A.
ID:544
picturePath:image/841.png
answer:80°
answer:80
correct answer:50
total:545  correct:145  acc:0.26605504587155965
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ID:545
picturePath:image/844.png
answer:40
answer:40
correct answer:50
total:546  correct:145  acc:0.26556776556776557
------------------------------------------------------------------------------------
ID:546
picturePath:image/846.png
answer:根据切线的性质，切线垂直于圆的切点处的切线，因此∠OAC=∠OBC=90°。又因为∠C=70°，所以∠AOB=180°-∠C-∠OAC-∠OBC=180°-70°-90°-90°=20°。因此，弦AB所对的圆周角的度数为20°/2=10°。故选B。
answer:10
correct answer:125
total:547  correct:145  acc:0.26508226691042047
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ID:547
picturePath:image/847.png
answer:5
answer:5
correct answer:5
total:548  correct:146  acc:0.2664233576642336
------------------------------------------------------------------------------------
ID:548
picturePath:image/849.png
answer:解:∵⊙O为△ABC的外接圆,∠BAC=55°,∴∠BOC=2∠BAC=2×55°=110°,∵OB=OC,∴∠OBC=\frac{180°-∠BOC}{2}=\frac{180°-110°}{2}=40°．故选:A．
answer:40
correct answer:35
total:549  correct:146  acc:0.2659380692167577
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ID:549
picturePath:image/852.png
answer:解:∵PA、PB是⊙O的切线,A、B为切点,∴∠OAP=∠OBP=90°,∵∠P=40°,∴∠AOB=90°-40°=50°,∵AC是⊙O的直径,∴∠C=90°,∴∠BAC=90°-∠AOB=90°-50°=40°．故选:B．
answer:40
correct answer:20
total:550  correct:146  acc:0.26545454545454544
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ID:550
picturePath:image/855.png
answer:根据题意,∠BAC=90°,DE⊥BC,∠1=∠2,则△BDE是等腰直角三角形.又∵BE=EC,AC=6,AB=10,∴BD=6,DE=5,∴△BDE的周长=BD+DE+EB=6+5+5=16.故选B.
answer:16
correct answer:16
total:551  correct:147  acc:0.26678765880217786
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ID:551
picturePath:image/856.png
answer:60°
answer:60
correct answer:30
total:552  correct:147  acc:0.266304347826087
------------------------------------------------------------------------------------
ID:552
picturePath:image/858.png
answer:解:∵P点关于OM的轴对称点是G,P点关于ON的轴对称点是H,∴∠GOM=∠MON,∠NOH=∠MON,∴∠GOM+∠NOH=2∠MON,∵∠MON=35°,∴∠GOM+∠NOH=2∠MON=2×35°=70°,∴∠GOH=180°-∠GOM-∠NOH=180°-70°=110°．故选:B．
answer:110
correct answer:70
total:553  correct:147  acc:0.26582278481012656
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ID:553
picturePath:image/861.png
answer:25°
answer:25
correct answer:115
total:554  correct:147  acc:0.26534296028880866
------------------------------------------------------------------------------------
ID:554
picturePath:image/864.png
answer:75°
answer:75
correct answer:75
total:555  correct:148  acc:0.26666666666666666
------------------------------------------------------------------------------------
ID:555
picturePath:image/865.png
answer:45
answer:45
correct answer:135
total:556  correct:148  acc:0.26618705035971224
------------------------------------------------------------------------------------
ID:556
picturePath:image/868.png
answer:Since \( CC' \) is parallel to \( AB \), we can deduce that \( \angle CAB \) and \( \angle CC'A \) are corresponding angles, which means they are congruent. Given that \( \angle CAB = 80^\circ \), we can conclude that \( \angle CC'A = 80^\circ \). Since \( \angle CAB \) and \( \angle B'AB \) are alternate interior angles, they are also congruent. Therefore, \( \angle B'AB = 80^\circ \).
answer:80
correct answer:20
total:557  correct:148  acc:0.26570915619389585
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ID:557
picturePath:image/869.png
answer:解:∵OA⊥OB,∴∠AOB=90°,∵∠A=35°,∴∠BOC=55°,∴∠D=\frac{1}{2}∠BOC=27.5°,∴弧CD的度数为27.5°．故选:A．
answer:27.5
correct answer:20
total:558  correct:148  acc:0.26523297491039427
------------------------------------------------------------------------------------
ID:558
picturePath:image/870.png
answer:50°
answer:50
correct answer:65
total:559  correct:148  acc:0.26475849731663686
------------------------------------------------------------------------------------
ID:559
picturePath:image/875.png
answer:10
answer:10
correct answer:12
total:560  correct:148  acc:0.2642857142857143
------------------------------------------------------------------------------------
ID:560
picturePath:image/878.png
answer:60
answer:60
correct answer:55
total:561  correct:148  acc:0.2638146167557932
------------------------------------------------------------------------------------
ID:561
picturePath:image/879.png
answer:10cm
answer:10
correct answer:10
total:562  correct:149  acc:0.26512455516014233
------------------------------------------------------------------------------------
ID:562
picturePath:image/880.png
answer:解:∵PA、PB分别与⊙O相切于A、B点,∴∠PAO=∠PBO=90°,∴∠AOB=180°-∠P=180°-66°=114°,∴∠C=\frac{1}{2}∠AOB=\frac{1}{2}×114°=57°．故选:B．
answer:57
correct answer:57
total:563  correct:150  acc:0.2664298401420959
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ID:563
picturePath:image/883.png
answer:72°
answer:72
correct answer:54
total:564  correct:150  acc:0.26595744680851063
------------------------------------------------------------------------------------
ID:564
picturePath:image/884.png
answer:13+6=19(厘米)
answer:19
correct answer:19
total:565  correct:151  acc:0.2672566371681416
------------------------------------------------------------------------------------
ID:565
picturePath:image/885.png
answer:根据平行线的性质，∠1和∠2是对顶角，所以∠1=∠2。已知∠1=110°，所以∠2=110°。故选B。
answer:110
correct answer:110
total:566  correct:152  acc:0.26855123674911663
------------------------------------------------------------------------------------
ID:566
picturePath:image/886.png
answer:解:∵弧AC=弧AD,∴∠A=∠D,∵∠C比∠D大36°,∴∠C=∠A+36°,∵AB是⊙O的直径,∴∠ACB=90°,∴∠A+36°+∠A=90°,解得∠A=30°,故选:B．
answer:30
correct answer:27
total:567  correct:152  acc:0.26807760141093473
------------------------------------------------------------------------------------
ID:567
picturePath:image/887.png
answer:解:∵∠OAB=28°,∴∠AOB=2∠OAB=2×28°=56°,∴∠C=\frac{1}{2}∠AOB=\frac{1}{2}×56°=28°．故选:A．
answer:28
correct answer:62
total:568  correct:152  acc:0.2676056338028169
------------------------------------------------------------------------------------
ID:568
picturePath:image/891.png
answer:16
answer:16
correct answer:13
total:569  correct:152  acc:0.2671353251318102
------------------------------------------------------------------------------------
ID:569
picturePath:image/895.png
answer:4
answer:4
correct answer:5
total:570  correct:152  acc:0.26666666666666666
------------------------------------------------------------------------------------
ID:570
picturePath:image/896.png
answer:The answer is 102°.
answer:102
correct answer:102
total:571  correct:153  acc:0.2679509632224168
------------------------------------------------------------------------------------
ID:571
picturePath:image/897.png
answer:解:∵AD是半圆的直径,∴∠ADB=90°,∵∠BAD=70°,∴∠ABD=90°-70°=20°,∵点C是弧BD的中点,∴∠ACD=∠ABD=20°,∴∠ADC=180°-90°-20°=70°．故选:A．
answer:70
correct answer:55
total:572  correct:153  acc:0.2674825174825175
------------------------------------------------------------------------------------
ID:572
picturePath:image/898.png
answer:This is a geometry problem involving rotations and angles. Given that △ABC is rotated 120° around point A to get △ADE, and the corresponding angles are equal (∠B = ∠E and ∠C = ∠D), we can use the fact that the sum of the angles in a triangle is 180° to find the measure of ∠CAE.

Since ∠BAC = 35°, and ∠B = ∠E, we have ∠EAC = 180° - ∠BAC = 180° - 35° = 145°.

Now, since △ABC is rotated 120° around point A to get △ADE, we have ∠CAE = ∠EAC - 120°.

Substituting the value of ∠EAC, we get ∠CAE = 145° - 120° = 25°.

Therefore, the measure of ∠CAE is 25°.
answer:25
correct answer:85
total:573  correct:153  acc:0.2670157068062827
------------------------------------------------------------------------------------
ID:573
picturePath:image/901.png
answer:根据线平行，内错角相等，可得∠ABC=∠CDE=140°；又∵BE是∠ABC的平分线，∴∠ABE=70°，∴∠C=180°-∠ABC-∠ABE=180°-140°-70°=70°．故选A．
answer:70
correct answer:100
total:574  correct:153  acc:0.2665505226480836
------------------------------------------------------------------------------------
ID:574
picturePath:image/905.png
answer:根据题意，AB和BC的垂直平分线相交于点O，这意味着点O是△ABC的重心。在△ABC中，∠BAC=70°，则∠BOC=2∠BAC=2*70°=140°。因此，∠BOC的度数为140°。
answer:140
correct answer:140
total:575  correct:154  acc:0.2678260869565217
------------------------------------------------------------------------------------
ID:575
picturePath:image/906.png
answer:This is a geometry problem. According to the properties of triangles, the sum of the angles in a triangle is 180 degrees. Since ∠ACD is an exterior angle of triangle ABC, and we know that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles, we can set up the equation: ∠ACD = ∠A + ∠B. Substituting the given values, we have 145° = 80° + ∠B. Solving for ∠B, we find that ∠B = 145° - 80° = 65°. Therefore, the measure of ∠B is 65 degrees.
answer:65
correct answer:65
total:576  correct:155  acc:0.2690972222222222
------------------------------------------------------------------------------------
ID:576
picturePath:image/907.png
answer:解:∵PA、PB是⊙O的切线,A、B为切点,∴∠OAP=∠OBP=90°,∵∠P=40°,∴∠AOB=140°,∵AC是⊙O的直径,∴∠C=90°,∴∠BAC=140°-90°=50°．故选:B．
answer:50
correct answer:20
total:577  correct:155  acc:0.268630849220104
------------------------------------------------------------------------------------
ID:577
picturePath:image/909.png
answer:解:∵∠BCD=35°,∴∠A=\frac{1}{2}∠BCD=\frac{1}{2}×35°=17.5°,∴∠ABD=180°-90°-17.5°=72.5°．故选:B．
answer:72.5
correct answer:55
total:578  correct:155  acc:0.2681660899653979
------------------------------------------------------------------------------------
ID:578
picturePath:image/911.png
answer:60
answer:60
correct answer:30
total:579  correct:155  acc:0.26770293609671847
------------------------------------------------------------------------------------
ID:579
picturePath:image/913.png
answer:100
answer:100
correct answer:100
total:580  correct:156  acc:0.2689655172413793
------------------------------------------------------------------------------------
ID:580
picturePath:image/916.png
answer:The answer is 60.
answer:60
correct answer:60
total:581  correct:157  acc:0.270223752151463
------------------------------------------------------------------------------------
ID:581
picturePath:image/918.png
answer:解:∵∠B=20°,∠A=70°,AB=130m,BC=120m,∴AC=√{AB²+BC²}=√{130²+120²}=250(m),∵每天凿隧道5m,∴把隧道凿通需要250÷5=50天．故选:D．
answer:50
correct answer:10
total:582  correct:157  acc:0.2697594501718213
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ID:582
picturePath:image/920.png
answer:x=AC×sin∠BCD=60×sin60°=60×\frac{√3}{2}=30√3(米)．故选:B．
answer:3
correct answer:3
total:583  correct:158  acc:0.27101200686106347
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ID:583
picturePath:image/921.png
answer:5cm
answer:5
correct answer:11
total:584  correct:158  acc:0.2705479452054795
------------------------------------------------------------------------------------
ID:584
picturePath:image/926.png
answer:The figure shows a triangle ABC with a point D that is the midpoint of AB. A line DE is drawn from point D perpendicular to AB and intersects AC at point E. Another line BE is drawn from point B to point E. The problem states that the perimeter of triangle BCE is 15 units, and AD is 3 units. We need to find the perimeter of triangle ABC.

Since D is the midpoint of AB, AD is half the length of AB. Therefore, AB = 2AD = 2 * 3 = 6 units.

Since DE is perpendicular to AB, triangle ADE is a right triangle. The length of DE can be found using the Pythagorean theorem: DE = √{AD^2 - AE^2} = √{3^2 - x^2}, where x is the length of AE.

The perimeter of triangle BCE is the sum of the lengths of BC, CE, and EB. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pythagorean theorem: BD = √{AB^2 - BD^2} = √{6^2 - (√{3^2 - x^2})^2}.

The perimeter of triangle BCE is 15 units, so BC + CE + EB = 15. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pythagorean theorem: BD = √{AB^2 - BD^2} = √{6^2 - (√{3^2 - x^2})^2}.

The perimeter of triangle BCE is the sum of the lengths of BC, CE, and EB. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pythagorean theorem: BD = √{AB^2 - BD^2} = √{6^2 - (√{3^2 - x^2})^2}.

The perimeter of triangle BCE is the sum of the lengths of BC, CE, and EB. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pythagorean theorem: BD = √{AB^2 - BD^2} = √{6^2 - (√{3^2 - x^2})^2}.

The perimeter of triangle BCE is the sum of the lengths of BC, CE, and EB. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pythagorean theorem: BD = √{AB^2 - BD^2} = √{6^2 - (√{3^2 - x^2})^2}.

The perimeter of triangle BCE is the sum of the lengths of BC, CE, and EB. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pythagorean theorem: BD = √{AB^2 - BD^2} = √{6^2 - (√{3^2 - x^2})^2}.

The perimeter of triangle BCE is the sum of the lengths of BC, CE, and EB. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pythagorean theorem: BD = √{AB^2 - BD^2} = √{6^2 - (√{3^2 - x^2})^2}.

The perimeter of triangle BCE is the sum of the lengths of BC, CE, and EB. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pythagorean theorem: BD = √{AB^2 - BD^2} = √{6^2 - (√{3^2 - x^2})^2}.

The perimeter of triangle BCE is the sum of the lengths of BC, CE, and EB. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pythagorean theorem: BD = √{AB^2 - BD^2} = √{6^2 - (√{3^2 - x^2})^2}.

The perimeter of triangle BCE is the sum of the lengths of BC, CE, and EB. Since DE is perpendicular to AB, triangle BDE is a right triangle. The length of BD can be found using the Pyth
answer:2
correct answer:21
total:585  correct:158  acc:0.27008547008547007
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ID:585
picturePath:image/927.png
answer:The answer is 72°.
answer:72
correct answer:36
total:586  correct:158  acc:0.2696245733788396
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ID:586
picturePath:image/931.png
answer:解:∵四边形ABCD内接于⊙O,∴∠C=180°-∠BAD=180°-132°=48°,
由圆周角定理得,∠BOD=2∠C=2×48°=96°,
故选:D．
answer:96
correct answer:96
total:587  correct:159  acc:0.2708688245315162
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ID:587
picturePath:image/933.png
answer:30
answer:30
correct answer:40
total:588  correct:159  acc:0.27040816326530615
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ID:588
picturePath:image/936.png
answer:The figure shows a right triangle \( \triangle ABC \) with a right angle at \( B \) and side lengths \( AB = 6 \) and \( AC = 10 \). The triangle is folded along the line \( ED \) such that point \( C \) coincides with point \( A \). The question asks for the perimeter of the smaller triangle \( \triangle ABE \) after the folding process.

Since \( \triangle ABC \) is folded along \( ED \), the side lengths of \( \triangle ABC \) and \( \triangle ABE \) are the same. Therefore, the perimeter of \( \triangle ABE \) is the same as that of \( \triangle ABC \), which is \( AB + BC + AC = 6 + 10 + 6 = 22 \). Hence, the correct answer is A.
answer:22
correct answer:14
total:589  correct:159  acc:0.2699490662139219
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ID:589
picturePath:image/937.png
answer:0.6
answer:0.6
correct answer:0.4
total:590  correct:159  acc:0.26949152542372884
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ID:590
picturePath:image/938.png
answer:60
answer:60
correct answer:60
total:591  correct:160  acc:0.2707275803722504
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ID:591
picturePath:image/940.png
answer:20√2米
answer:2
correct answer:100
total:592  correct:160  acc:0.2702702702702703
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ID:592
picturePath:image/942.png
answer:40
answer:40
correct answer:55
total:593  correct:160  acc:0.26981450252951095
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ID:593
picturePath:image/943.png
answer:40°
answer:40
correct answer:60
total:594  correct:160  acc:0.26936026936026936
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ID:594
picturePath:image/947.png
answer:解:∵AB∥CD,∠1=58°,∴∠EFD=∠1=58°,又∵FG平分∠EFD,∴∠EFG=\frac{1}{2}∠EFD=29°,∴∠FGB=∠EFG+∠1=29°+58°=87°．故选:D．
answer:87
correct answer:151
total:595  correct:160  acc:0.2689075630252101
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ID:595
picturePath:image/948.png
answer:30
answer:30
correct answer:30
total:596  correct:161  acc:0.2701342281879195
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ID:596
picturePath:image/949.png
answer:60°
answer:60
correct answer:30
total:597  correct:161  acc:0.2696817420435511
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ID:597
picturePath:image/950.png
answer:根据圆周角定理，同弧所对的圆周角相等，因此∠BOC=2∠A。已知∠OBC=55°，则∠BOC=2∠OBC=2×55°=110°。因此，∠A=\frac{1}{2}∠BOC=\frac{1}{2}×110°=55°。所以，∠A=55°。
answer:55
correct answer:35
total:598  correct:161  acc:0.2692307692307692
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ID:598
picturePath:image/952.png
answer:9:4
answer:4
correct answer:4
total:599  correct:162  acc:0.2704507512520868
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ID:599
picturePath:image/956.png
answer:12m
answer:12
correct answer:7.2
total:600  correct:162  acc:0.27
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ID:600
picturePath:image/957.png
answer:The answer is 35.
answer:35
correct answer:55
total:601  correct:162  acc:0.26955074875207985
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ID:601
picturePath:image/958.png
answer:17
answer:17
correct answer:13
total:602  correct:162  acc:0.2691029900332226
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ID:602
picturePath:image/959.png
answer:解:连接OA,∵PA、PB、CD分别切⊙O于A、B、E,∴OA⊥PA,OB⊥PB,OE⊥CD,∴∠OAP=∠OBP=90°,∵∠P=40°,∴∠PAE+∠PBE=∠OAP+∠OBP-∠P=90°+90°-40°=140°,故选:D．
answer:140
correct answer:70
total:603  correct:162  acc:0.26865671641791045
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ID:603
picturePath:image/960.png
answer:解:∵MP和NQ分别垂直平分AB和AC,∴PB=PA,QA=QC,∴∠BAP=∠PAM,∠CQA=∠QAC,∴∠BAP+∠CQA=∠PAM+∠QAC=∠BAC=110°,∴∠PAQ=180°-(∠BAP+∠CQA)=180°-110°=70°．故选:B．
answer:70
correct answer:40
total:604  correct:162  acc:0.2682119205298013
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ID:604
picturePath:image/961.png
answer:∠BOC = 180° - ∠AOD = 180° - 50° = 130°.
answer:130
correct answer:50
total:605  correct:162  acc:0.26776859504132233
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ID:605
picturePath:image/963.png
answer:解:∵AE、AD和BC分别切⊙O于点E、D、F,∴DE=DF,AE=AF,∵AD=20,∴△ABC的周长=AB+BC+AC=AB+AE+BF+AC=AB+AF+BF+AC=AB+CD+AC=AB+BC+AC=AB+AC+BC,∴△ABC的周长=AB+AC+BC=AD=20．故选:B．
answer:20
correct answer:40
total:606  correct:162  acc:0.26732673267326734
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ID:606
picturePath:image/964.png
answer:根据题意,∠A=90°,AB=AC,所以△ABC是等腰直角三角形,所以BE=CE,又∵BD平分∠ABE,所以∠DBE=∠CBE,又∵DE⊥BC,所以∠BDE=∠CDE=90°,所以△BDE≌△CDE,所以DE=DC,所以△DEC的周长=DE+DE+DC=2DE+DC=2BE+BC=2BE+10cm=2CE+10cm=2DC+10cm,所以△DEC的周长是20cm.
answer:20
correct answer:10
total:607  correct:162  acc:0.26688632619439867
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ID:607
picturePath:image/967.png
answer:Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG. Since DE is the vertical angle to AB, and FG is the vertical angle to AC, we know that ∠BDE = ∠ABD and ∠CFG = ∠ACG
ID:608
picturePath:image/968.png
answer:根据题意，等腰梯形ABCD的中位线EF=15cm，则EF是AB和CD的中位线，因此AB=2EF=2*15=30cm，同理，CD=30cm。又因为ABCD是圆外切等腰梯形，所以BC是圆的直径，BC=2*15=30cm。因此，等腰梯形ABCD的周长=AB+BC+CD+AD=30+30+30+AD=90+AD。又因为AD是圆的半径，所以AD=BC/2=30/2=15cm。所以，等腰梯形ABCD的周长=90+15=105cm。故选：B。
answer:105
correct answer:60
total:609  correct:162  acc:0.2660098522167488
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ID:609
picturePath:image/969.png
answer:解:由题意得,∠PAB=75°,∠PBC=60°,∴∠PAC=∠PAB-∠PBC=75°-60°=15°,在Rt△ABC中,∠C=90°,∠ACB=180°-∠PAC-∠PAB=180°-15°-75°=90°,∴AC=BC=7海里,∴PC=√{AC²+BC²}=√{7²+7²}=√{49+49}=7√{2}(海里),∴BP=PC=7√{2}(海里)．故选:A．
answer:2
correct answer:7
total:610  correct:162  acc:0.26557377049180325
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ID:610
picturePath:image/970.png
answer:12cm
answer:12
correct answer:6
total:611  correct:162  acc:0.265139116202946
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ID:611
picturePath:image/973.png
answer:54
answer:54
correct answer:112
total:612  correct:162  acc:0.2647058823529412
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ID:612
picturePath:image/974.png
answer:The answer is 12.
answer:12
correct answer:7.2
total:613  correct:162  acc:0.2642740619902121
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ID:613
picturePath:image/977.png
answer:解:连接OE,∵DE是⊙O的切线,∴OE⊥DE,∴∠ODE=90°,∵∠BDE=52°,∴∠DOE=180°-90°-52°=38°,∵OA=OD,∴∠OAE=∠DOE=38°,∴∠ABE=∠OAE=38°．故选:B．
answer:38
correct answer:64
total:614  correct:162  acc:0.26384364820846906
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ID:614
picturePath:image/979.png
answer:60
answer:60
correct answer:60
total:615  correct:163  acc:0.26504065040650404
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ID:615
picturePath:image/980.png
answer:25
answer:25
correct answer:130
total:616  correct:163  acc:0.26461038961038963
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ID:616
picturePath:image/982.png
answer:8cm
answer:8
correct answer:8
total:617  correct:164  acc:0.26580226904376014
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ID:617
picturePath:image/983.png
answer:In a right-angled triangle, the side opposite the right angle is called the hypotenuse, and the other two sides are called the legs. Given that \( \angle C = 90^\circ \), \( \angle A = 37^\circ \), and \( AC = 4 \), we can use the sine, cosine, and tangent functions to find the lengths of the other two sides, \( AB \) and \( BC \).

First, let's find \( AB \) using the sine function:
\[ \sin A = \frac{AB}{AC} \]
\[ \sin 37^\circ = \frac{AB}{4} \]
\[ AB = 4 \times \sin 37^\circ \]
\[ AB \approx 4 \times 0.80 \]
\[ AB \approx 3.20 \]

Next, let's find \( BC \) using the cosine function:
\[ \cos A = \frac{BC}{AB} \]
\[ \cos 37^\circ = \frac{BC}{3.20} \]
\[ BC = 3.20 \times \cos 37^\circ \]
\[ BC \approx 3.20 \times 0.60 \]
\[ BC \approx 1.92 \]

Therefore, the length of \( BC \) is approximately 1.92 units.
answer:1.92
correct answer:3.0
total:618  correct:164  acc:0.26537216828478966
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ID:618
picturePath:image/985.png
answer:25°
answer:25
correct answer:100
total:619  correct:164  acc:0.2649434571890145
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ID:619
picturePath:image/987.png
answer:根据题意,∠2=∠3,∠1=80°,∠1和∠2是对顶角,所以∠2=∠3=80°,又因为a∥b,所以∠4=∠2=80°.故选B.
answer:80
correct answer:40
total:620  correct:164  acc:0.2645161290322581
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ID:620
picturePath:image/989.png
answer:100
answer:100
correct answer:40
total:621  correct:164  acc:0.2640901771336554
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ID:621
picturePath:image/990.png
answer:50
answer:50
correct answer:50
total:622  correct:165  acc:0.2652733118971061
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ID:622
picturePath:image/991.png
answer:解:∵四边形ABCD是平行四边形,∴AB=CD,AD=BC,∴AC=BD,∵OE⊥AC,∴AE=CE,∵□ABCD的周长为22cm,∴2(AD+CD)=22cm,∴AD+CD=11cm,∴△CDE的周长为:CE+DE+CD=AE+DE+CD=AD+CD=11cm．故选:B．
answer:11
correct answer:11
total:623  correct:166  acc:0.2664526484751204
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ID:623
picturePath:image/996.png
answer:The answer is 57°.
answer:57
correct answer:78
total:624  correct:166  acc:0.266025641025641
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ID:624
picturePath:image/998.png
answer:The answer is 25.
answer:25
correct answer:35
total:625  correct:166  acc:0.2656
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ID:625
picturePath:image/999.png
answer:解:∵AB=6,BC=8,∴AC=√{6^{2}+8^{2}}=10,∵五个小长方形的对角线在大长方形ABCD的对角线上,∴五个小长方形的周长之和=2×10+2×(6+8)=42．故选:D．
answer:42
correct answer:28
total:626  correct:166  acc:0.26517571884984026
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ID:626
picturePath:image/1000.png
answer:The answer is 55.
answer:55
correct answer:145
total:627  correct:166  acc:0.2647527910685805
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ID:627
picturePath:image/1001.png
answer:根据圆的内接四边形的性质，外角和是360°，所以∠BOD=360°-62°×2=180°-62°×2=180°-124°=56°.
answer:56
correct answer:124
total:628  correct:166  acc:0.2643312101910828
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ID:628
picturePath:image/1002.png
answer:解:∵A、B、C均为⊙O上的点,∠AOB=80°,∴∠ACB=\frac{1}{2}∠AOB=\frac{1}{2}×80°=40°．故选:B．
answer:40
correct answer:40
total:629  correct:167  acc:0.2655007949125596
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ID:629
picturePath:image/1003.png
answer:解:∵∠B=100°,∴∠AED=100°,∴∠ADE=\frac{1}{2}∠AED=\frac{1}{2}×100°=50°．故选:B．
answer:50
correct answer:100
total:630  correct:167  acc:0.2650793650793651
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ID:630
picturePath:image/1004.png
answer:140°
answer:140
correct answer:40
total:631  correct:167  acc:0.26465927099841524
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ID:631
picturePath:image/1005.png
answer:50
answer:50
correct answer:50
total:632  correct:168  acc:0.26582278481012656
------------------------------------------------------------------------------------
ID:632
picturePath:image/1006.png
answer:25
answer:25
correct answer:80
total:633  correct:168  acc:0.26540284360189575
------------------------------------------------------------------------------------
ID:633
picturePath:image/1007.png
answer:65
answer:65
correct answer:35
total:634  correct:168  acc:0.26498422712933756
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ID:634
picturePath:image/1008.png
answer:根据切线的性质定理，切线AC与⊙O相切于点A，则∠AOC=90°。又∵∠B=25°，∴∠BOC=2∠B=50°，∴∠C=90°-50°=40°。故选B.
answer:40
correct answer:40
total:635  correct:169  acc:0.26614173228346455
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ID:635
picturePath:image/1009.png
answer:圆锥的侧面积公式为：πrl，其中r为底面半径，l为母线长。在这个问题中，底面半径OB=6cm，高OC=8cm，所以母线长l=√{OB²+OC²}=√{6²+8²}=10cm。将r和l代入公式，得到侧面积=π×6×10=180πcm²。所以，这个圆锥漏斗的侧面积是180πcm²。
answer:180
correct answer:60
total:636  correct:169  acc:0.26572327044025157
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ID:636
picturePath:image/1010.png
answer:解:∵AB与小圆相切,∴AB为小圆的切线,∴∠AOB=90°,在Rt△AOB中,OB=0.5×AB=0.5×20=10,∴大圆的半径为10+10=20,∴大圆的半径为20米,∴小圆的半径为10米,∴大圆的面积为π×20²=400π,小圆的面积为π×10²=100π,∴圆环的面积为400π-100π=300π(平方米)．故选:B．
answer:300
correct answer:100
total:637  correct:169  acc:0.2653061224489796
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ID:637
picturePath:image/1011.png
answer:解:∵△OAD≌△OBC,∴∠OAD=∠OBC,∵∠O=65°,∠C=20°,∴∠OBC=180°-65°-20°=95°,∴∠OAD=95°,
故选:A．
answer:95
correct answer:95
total:638  correct:170  acc:0.2664576802507837
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ID:638
picturePath:image/1014.png
answer:50°
answer:50
correct answer:50
total:639  correct:171  acc:0.2676056338028169
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ID:639
picturePath:image/1016.png
answer:Since △CAB is rotated 34° around point C to get △CDE, we know that ∠ACE is the sum of ∠CDE and the angle of rotation (34°). Therefore, ∠CDE = ∠ACE - 34° = 88° - 34° = 54°. Since ∠CDE is an exterior angle of △CDB, we have ∠DCB = ∠CDE - ∠BDE. Since ∠BDE is an interior angle of △CDB, we have ∠BDE = 180° - ∠DCB - ∠CDE. Therefore, ∠DCB = ∠CDE - (180° - ∠DCB - ∠CDE) = 54° - (180° - 54°) = 54° - 126° = -72°. However, since angles are measured in a counterclockwise direction, ∠DCB is actually 72°.
answer:72
correct answer:20
total:640  correct:171  acc:0.2671875
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ID:640
picturePath:image/1017.png
answer:The answer is A.
ID:641
picturePath:image/1019.png
answer:根据三角形的一个外角等于与它不相邻的两个内角的和，可得∠BOC=∠B+∠D=40°+30°=70°，∵AB//CD,∴∠AOC=∠BOC=70°.故选B
answer:70
correct answer:70
total:642  correct:172  acc:0.26791277258566976
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ID:642
picturePath:image/1024.png
answer:8
answer:8
correct answer:8
total:643  correct:173  acc:0.2690513219284603
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ID:643
picturePath:image/1025.png
answer:3.2
answer:3.2
correct answer:2.4
total:644  correct:173  acc:0.2686335403726708
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ID:644
picturePath:image/1026.png
answer:1.5cm
answer:1.5
correct answer:1.5
total:645  correct:174  acc:0.26976744186046514
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ID:645
picturePath:image/1027.png
answer:根据题意，∠DAC = 30°，∠DAB = 75°，∠DAC + ∠DAB + ∠BAC = 180°，所以∠BAC = 180° - 30° - 75° = 75°。因为CE平分∠ACB，所以∠DEC = ∠BAC / 2 = 75° / 2 = 37.5°。所以，∠DEC = 37.5°。
answer:37.5
correct answer:15
total:646  correct:174  acc:0.2693498452012384
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ID:646
picturePath:image/1028.png
answer:解:∵∠AED=45°,∠AEC=∠AED=45°,∴△AEC是等腰三角形,∴CE=AE,∴CD+DE=CD+AE,∴DE=CD=1米,∴AB=AE+DE=1.5+1=2.5米．故选:B．
answer:2.5
correct answer:4.5
total:647  correct:174  acc:0.2689335394126739
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ID:647
picturePath:image/1031.png
answer:20°
answer:20
correct answer:50
total:648  correct:174  acc:0.26851851851851855
------------------------------------------------------------------------------------
ID:648
picturePath:image/1032.png
answer:5.8
answer:5.8
correct answer:7
total:649  correct:174  acc:0.26810477657935283
------------------------------------------------------------------------------------
ID:649
picturePath:image/1033.png
answer:50
answer:50
correct answer:50
total:650  correct:175  acc:0.2692307692307692
------------------------------------------------------------------------------------
ID:650
picturePath:image/1037.png
answer:75
answer:75
correct answer:75
total:651  correct:176  acc:0.27035330261136714
------------------------------------------------------------------------------------
ID:651
picturePath:image/1038.png
answer:根据题意，点D是AB的垂直平分线与边AC的交点，所以DB = DC，即△BDC是等腰三角形。又因为AB = 6，AC = 7，所以BC = AB + AC = 6 + 7 = 13。因此，△BDC的周长为DB + DC + BC = DB + DB + 13 = 2DB + 13。由于DB = DC，所以△BDC的周长为2DB + 13 = 2(AB/2) + 13 = AB + 13 = 6 + 13 = 19。故选B。
answer:19
correct answer:12
total:652  correct:176  acc:0.26993865030674846
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ID:652
picturePath:image/1039.png
answer:40°
answer:40
correct answer:40
total:653  correct:177  acc:0.27105666156202146
------------------------------------------------------------------------------------
ID:653
picturePath:image/1040.png
answer:This is a geometric figure with a circle, a diameter labeled as AB, and a chord labeled as CD. There is also an angle labeled as ∠ABD with a measure of 63 degrees. The question is asking for the measure of angle ∠BCD.
answer:63
correct answer:27
total:654  correct:177  acc:0.2706422018348624
------------------------------------------------------------------------------------
ID:654
picturePath:image/1042.png
answer:20
answer:20
correct answer:120
total:655  correct:177  acc:0.27022900763358776
------------------------------------------------------------------------------------
ID:655
picturePath:image/1045.png
answer:∠DAE的度数与∠BAC的度数相同，因为DE是AB的垂直平分线，所以∠BDE=∠BDE，同理∠CDE=∠CDE，所以∠DAE=∠BAC=110°.
answer:110
correct answer:40
total:656  correct:177  acc:0.2698170731707317
------------------------------------------------------------------------------------
ID:656
picturePath:image/1046.png
answer:解:∵长方形ABCD的边AB=6,BC=8,∴AD=8,CD=6,∴五个小长方形的周长之和=2×(AB+BC+AD+CD)=2×(6+8+8+6)=2×24=48．故选:D．
answer:48
correct answer:28
total:657  correct:177  acc:0.2694063926940639
------------------------------------------------------------------------------------
ID:657
picturePath:image/1047.png
answer:20
answer:20
correct answer:70
total:658  correct:177  acc:0.2689969604863222
------------------------------------------------------------------------------------
ID:658
picturePath:image/1049.png
answer:解:∵⊙O与边AB相切于点D,∴OD⊥AD,∴∠ODA=90°,∵∠EPD=35°,∴∠EOD=55°,∴∠BAC=2∠EOD=110°．故选:B．
answer:110
correct answer:20
total:659  correct:177  acc:0.26858877086494687
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ID:659
picturePath:image/1053.png
answer:60
answer:60
correct answer:120
total:660  correct:177  acc:0.2681818181818182
------------------------------------------------------------------------------------
ID:660
picturePath:image/1055.png
answer:The answer is: 40
answer:40
correct answer:10
total:661  correct:177  acc:0.26777609682299547
------------------------------------------------------------------------------------
ID:661
picturePath:image/1056.png
answer:8.
answer:8
correct answer:16
total:662  correct:177  acc:0.2673716012084592
------------------------------------------------------------------------------------
ID:662
picturePath:image/1059.png
answer:13m
answer:13
correct answer:17
total:663  correct:177  acc:0.2669683257918552
------------------------------------------------------------------------------------
ID:663
picturePath:image/1060.png
answer:解:如图,∵四边形ABCD为长方形,∴∠B=∠D=90°,∴∠BOG=180°-∠B=90°,∵∠OGC'=100°,∴∠BOC'=180°-∠BOG-∠OGC'=180°-90°-100°=90°,∴∠DGC'=180°-∠BOC'=180°-90°=90°．故选:A．
answer:90
correct answer:20
total:664  correct:177  acc:0.26656626506024095
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ID:664
picturePath:image/1061.png
answer:60
answer:60
correct answer:60
total:665  correct:178  acc:0.26766917293233083
------------------------------------------------------------------------------------
ID:665
picturePath:image/1064.png
answer:根据圆周角定理，同弧所对的圆周角的度数等于它所对的圆心角度数的一半。因此，∠BOC的度数为\frac{1}{2}∠BDC=\frac{1}{2}×40°=20°。
answer:20
correct answer:80
total:666  correct:178  acc:0.2672672672672673
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ID:666
picturePath:image/1065.png
answer:20°
answer:20
correct answer:50
total:667  correct:178  acc:0.26686656671664166
------------------------------------------------------------------------------------
ID:667
picturePath:image/1067.png
answer:80°
answer:80
correct answer:80
total:668  correct:179  acc:0.2679640718562874
------------------------------------------------------------------------------------
ID:668
picturePath:image/1068.png
answer:解:∵∠AOB=52°,∴∠ADB=\frac{1}{2}∠AOB=26°．故选:A．
answer:26
correct answer:26
total:669  correct:180  acc:0.26905829596412556
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ID:669
picturePath:image/1069.png
answer:60
answer:60
correct answer:70
total:670  correct:180  acc:0.26865671641791045
------------------------------------------------------------------------------------
ID:670
picturePath:image/1071.png
answer:This is a geometry problem. Given that \( BC \) is perpendicular to \( AE \) at point \( C \), and \( CD \) is parallel to \( AB \), we can use the properties of parallel lines and angles to solve for \( \angle B \). Since \( CD \) is parallel to \( AB \), corresponding angles \( \angle ECD \) and \( \angle B \) are congruent. Therefore, \( \angle B = \angle ECD = 43^\circ \).
answer:43
correct answer:47
total:671  correct:180  acc:0.26825633383010433
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ID:671
picturePath:image/1073.png
answer:15
answer:15
correct answer:12
total:672  correct:180  acc:0.26785714285714285
------------------------------------------------------------------------------------
ID:672
picturePath:image/1075.png
answer:∠DOE=\frac{1}{2}∠AOC=\frac{1}{2}×80°=40°.
answer:40
correct answer:40
total:673  correct:181  acc:0.2689450222882615
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ID:673
picturePath:image/1076.png
answer:6
answer:6
correct answer:15
total:674  correct:181  acc:0.2685459940652819
------------------------------------------------------------------------------------
ID:674
picturePath:image/1078.png
answer:4
answer:4
correct answer:4
total:675  correct:182  acc:0.2696296296296296
------------------------------------------------------------------------------------
ID:675
picturePath:image/1080.png
answer:6
answer:6
correct answer:16
total:676  correct:182  acc:0.2692307692307692
------------------------------------------------------------------------------------
ID:676
picturePath:image/1082.png
answer:解:∵四边形ABCD是⊙O的内接四边形,∴∠A=180°-110°=70°,
由圆周角定理得,∠BOD=2∠A=140°,
故选:C．
answer:140
correct answer:140
total:677  correct:183  acc:0.27031019202363366
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ID:677
picturePath:image/1085.png
answer:A.90°B.108°C.126°D.144°
answer:144
correct answer:72
total:678  correct:183  acc:0.26991150442477874
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ID:678
picturePath:image/1086.png
answer:解:∵∠ACB=90°,∠A=35°,∴∠ABC=55°,∵△ABC绕点C逆时针旋转α角到△A1B1C的位置,A1B1恰好经过点B,∴∠A1B1C=∠ABC=55°,∴α=55°．故选:B．
answer:55
correct answer:70
total:679  correct:183  acc:0.2695139911634757
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ID:679
picturePath:image/1087.png
answer:解:∵∠AOC=140°,∴∠BOC=180°-140°=40°,∵点B是弧AC的中点,∴∠ABC=\frac{1}{2}∠BOC=\frac{1}{2}×40°=20°,∴∠D=∠AOC-∠ABC=140°-20°=120°．故选:D．
answer:120
correct answer:35
total:680  correct:183  acc:0.2691176470588235
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ID:680
picturePath:image/1089.png
answer:The answer is 90.
answer:90
correct answer:135
total:681  correct:183  acc:0.2687224669603524
------------------------------------------------------------------------------------
ID:681
picturePath:image/1092.png
answer:解:∵AB是⊙O的直径,∴∠ADC=90°,∵∠ABD=50°,∴∠BCD=90°-∠ABD=90°-50°=40°．故选:B．
answer:40
correct answer:40
total:682  correct:184  acc:0.2697947214076246
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ID:682
picturePath:image/1093.png
answer:解:∵△ABC是等腰直角三角形,∠A=90°,BC=10cm,∴AB=AC=BC=10cm,∵BD是∠ABC的平分线,DE⊥BC于E,∴DE=\frac{1}{2}AB=5cm,∴△DEC的周长=DE+EC+CD=5+5+10=20cm．故选:D．
answer:20
correct answer:10
total:683  correct:184  acc:0.26939970717423134
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ID:683
picturePath:image/1095.png
answer:根据题意，∠ABD + ∠ACE = 230°，又因为D、B、C、E四点共线，所以∠ABD + ∠ACE + ∠A = 180°（两直线平行，同旁内角互补）。将∠ABD + ∠ACE = 230°代入，得到∠A = 180° - (∠ABD + ∠ACE) = 180° - 230° = -50°。
answer:50
correct answer:50
total:684  correct:185  acc:0.27046783625730997
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ID:684
picturePath:image/1096.png
answer:根据折叠的性质，∠1=∠B，∠2=∠D，∵∠B=32°,∴∠1=32°,∠2=32°,∴∠1-∠2=0°.故选:D.
answer:0
correct answer:64
total:685  correct:185  acc:0.27007299270072993
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ID:685
picturePath:image/1097.png
answer:20
answer:20
correct answer:100
total:686  correct:185  acc:0.2696793002915452
------------------------------------------------------------------------------------
ID:686
picturePath:image/1099.png
answer:解:∵∠C=110°,∴∠COD=180°-110°=70°,∵AB为⊙O的直径,∴∠AOB=180°,∴∠AOD=\frac{1}{2}∠AOB=\frac{1}{2}×180°=90°．故选:A．
answer:90
correct answer:40
total:687  correct:185  acc:0.2692867540029112
------------------------------------------------------------------------------------
ID:687
picturePath:image/1100.png
answer:Since line a is parallel to line b, corresponding angles ∠1 and ∠2 are congruent. Therefore, ∠2 = ∠1 = 68°. Similarly, alternate interior angles ∠3 and ∠4 are also congruent. Therefore, ∠3 = ∠4 = x. The difference between ∠2 and ∠3 is ∠2 - ∠3 = 68° - x. Since the lines are parallel, corresponding angles are congruent, and alternate interior angles are equal. Therefore, ∠2 - ∠3 = 68° - x = 0°. Solving for x, we find that x = 68°. Thus, ∠2 - ∠3 = 68° - x = 68° - 68° = 0°.The answer is: 0
answer:0
correct answer:112
total:688  correct:185  acc:0.2688953488372093
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ID:688
picturePath:image/1101.png
answer:∠3=∠1+∠2=22°+46°=68°.
answer:68
correct answer:112
total:689  correct:185  acc:0.26850507982583455
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ID:689
picturePath:image/1102.png
answer:∠COM = ∠AOD / 2 = 136° / 2 = 68°.
answer:68
correct answer:46
total:690  correct:185  acc:0.26811594202898553
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ID:690
picturePath:image/1103.png
answer:The answer is A.
ID:691
picturePath:image/1104.png
answer:The answer is A.
ID:692
picturePath:image/1105.png
answer:2∠A
answer:2
correct answer:24
total:693  correct:185  acc:0.26695526695526695
------------------------------------------------------------------------------------
ID:693
picturePath:image/1106.png
answer:50°
answer:50
correct answer:15
total:694  correct:185  acc:0.2665706051873199
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ID:694
picturePath:image/1107.png
answer:∠ADE = 180° - ∠A = 180° - 110° = 70°.
answer:70
correct answer:110
total:695  correct:185  acc:0.26618705035971224
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ID:695
picturePath:image/1108.png
answer:The answer is A.
ID:696
picturePath:image/1110.png
answer:由于AB=AC，∠A=40°，因此∠ABC=∠ACB=\frac{180°-40°}{2}=70°。又因为DE是AB的垂直平分线，所以BE=CE。在△ABE和△ACE中，AB=AC，BE=CE，∠A=40°，所以△ABE≌△ACE（SAS）。因此，∠CBE=∠EBC=\frac{180°-70°}{2}=40°。所以，∠CBE=40°。
answer:40
correct answer:30
total:697  correct:185  acc:0.2654232424677188
------------------------------------------------------------------------------------
ID:697
picturePath:image/1111.png
answer:根据题意,∠C=90°,AB的垂直平分线DE交AB于点E,交BC于点D,则AD=CD,∵AB=13,AC=5,∴△ACD的周长为:AC+AD+CD=AC+AD+AD=AC+AB=5+13=18.故选B.
answer:18
correct answer:17
total:698  correct:185  acc:0.26504297994269344
------------------------------------------------------------------------------------
ID:698
picturePath:image/1112.png
answer:38
answer:38
correct answer:19
total:699  correct:185  acc:0.2646638054363376
------------------------------------------------------------------------------------
ID:699
picturePath:image/1114.png
answer:20
answer:20
correct answer:105
total:700  correct:185  acc:0.2642857142857143
------------------------------------------------------------------------------------
ID:700
picturePath:image/1115.png
answer:25°
answer:25
correct answer:65
total:701  correct:185  acc:0.26390870185449355
------------------------------------------------------------------------------------
ID:701
picturePath:image/1116.png
answer:2008个等边三角形的周长等于2008个边长为1的等边三角形的周长之和，每个等边三角形的周长为3，因此2008个等边三角形的周长为2008*3=6024。
answer:6024
correct answer:2010
total:702  correct:185  acc:0.26353276353276356
------------------------------------------------------------------------------------
ID:702
picturePath:image/1118.png
answer:50
answer:50
correct answer:40
total:703  correct:185  acc:0.2631578947368421
------------------------------------------------------------------------------------
ID:703
picturePath:image/1119.png
answer:75°
answer:75
correct answer:75
total:704  correct:186  acc:0.26420454545454547
------------------------------------------------------------------------------------
ID:704
picturePath:image/1120.png
answer:24
answer:24
correct answer:104
total:705  correct:186  acc:0.26382978723404255
------------------------------------------------------------------------------------
ID:705
picturePath:image/1124.png
answer:20
answer:20
correct answer:60
total:706  correct:186  acc:0.26345609065155806
------------------------------------------------------------------------------------
ID:706
picturePath:image/1125.png
answer:6.
answer:6
correct answer:15
total:707  correct:186  acc:0.26308345120226306
------------------------------------------------------------------------------------
ID:707
picturePath:image/1126.png
answer:80
answer:80
correct answer:80
total:708  correct:187  acc:0.2641242937853107
------------------------------------------------------------------------------------
ID:708
picturePath:image/1127.png
answer:60
answer:60
correct answer:60
total:709  correct:188  acc:0.2651622002820874
------------------------------------------------------------------------------------
ID:709
picturePath:image/1128.png
answer:∠2=\frac{1}{2}∠BOC=\frac{1}{2}×180°-∠1=\frac{1}{2}×180°-40°=80°-40°=40°.
answer:40
correct answer:70
total:710  correct:188  acc:0.2647887323943662
------------------------------------------------------------------------------------
ID:710
picturePath:image/1129.png
answer:Since AD is parallel to BC, we have \angle GAD = \angle CBA = 80^\circ. Since AE bisects \angle GAD, we have \angle GAE = \frac{1}{2} \angle GAD = \frac{1}{2} \cdot 80^\circ = \boxed{40^\circ}. The answer is: 40
answer:40
correct answer:50
total:711  correct:188  acc:0.26441631504922647
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ID:711
picturePath:image/1130.png
answer:∠C = 180° - 140° = 40°.
answer:40
correct answer:140
total:712  correct:188  acc:0.2640449438202247
------------------------------------------------------------------------------------
ID:712
picturePath:image/1131.png
answer:5
answer:5
correct answer:10
total:713  correct:188  acc:0.26367461430575034
------------------------------------------------------------------------------------
ID:713
picturePath:image/1132.png
answer:50m
answer:50
correct answer:100
total:714  correct:188  acc:0.26330532212885155
------------------------------------------------------------------------------------
ID:714
picturePath:image/1133.png
answer:解:∵⊙O的半径OC垂直于弦AB,∴∠BOC=∠AOC=50°,∴∠ADC=\frac{1}{2}∠BOC=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:25
total:715  correct:189  acc:0.26433566433566436
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ID:715
picturePath:image/1134.png
answer:The answer is 180°.
answer:180
correct answer:55
total:716  correct:189  acc:0.2639664804469274
------------------------------------------------------------------------------------
ID:716
picturePath:image/1135.png
answer:60°
answer:60
correct answer:60
total:717  correct:190  acc:0.2649930264993027
------------------------------------------------------------------------------------
ID:717
picturePath:image/1136.png
answer:解:∵∠BDC=130°,∴∠BOC=2∠BDC=260°．故选:D．
answer:260
correct answer:100
total:718  correct:190  acc:0.2646239554317549
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ID:718
picturePath:image/1138.png
answer:4√{2}cm.
answer:2
correct answer:8
total:719  correct:190  acc:0.26425591098748263
------------------------------------------------------------------------------------
ID:719
picturePath:image/1140.png
answer:30
answer:30
correct answer:50
total:720  correct:190  acc:0.2638888888888889
------------------------------------------------------------------------------------
ID:720
picturePath:image/1141.png
answer:根据题意,∠BDC=110°,点D是∠ABC和∠ACB角平分线的交点,则∠DBC=\frac{1}{2}(180°-∠BDC)=65°,∠ADC=\frac{1}{2}(180°-∠BDC)=65°,∴∠A=180°-∠DBC-∠ADC=180°-65°-65°=50°.故选B.
answer:50
correct answer:40
total:721  correct:190  acc:0.2635228848821082
------------------------------------------------------------------------------------
ID:721
picturePath:image/1142.png
answer:This is a geometric figure with three points labeled A, B, and C, and three lines labeled E, F, and D. The lines E, F, and D intersect at points A, B, and C, respectively. The angles at points A, C, and D are labeled as 70 degrees, 20 degrees, and 32 degrees, respectively. The question is asking for the measure of angle CF
answer:32
correct answer:58
total:722  correct:190  acc:0.2631578947368421
------------------------------------------------------------------------------------
ID:722
picturePath:image/1143.png
answer:40
answer:40
correct answer:50
total:723  correct:190  acc:0.2627939142461964
------------------------------------------------------------------------------------
ID:723
picturePath:image/1144.png
answer:10mm
answer:10
correct answer:8
total:724  correct:190  acc:0.26243093922651933
------------------------------------------------------------------------------------
ID:724
picturePath:image/1145.png
answer:80°
answer:80
correct answer:110
total:725  correct:190  acc:0.2620689655172414
------------------------------------------------------------------------------------
ID:725
picturePath:image/1146.png
answer:30
answer:30
correct answer:30
total:726  correct:191  acc:0.2630853994490358
------------------------------------------------------------------------------------
ID:726
picturePath:image/1147.png
answer:Since AB is parallel to CD, we can use the corresponding angles postulate, which states that if two parallel lines are cut by a transversal, then the corresponding angles are congruent. Therefore, ∠4 = ∠1 = 45°. Since the sum of the angles in a triangle is 180°, we have ∠3 + ∠4 + ∠5 = 180°. Substituting ∠4 = 45°, we get ∠3 + 45° + ∠5 = 180°. Solving for ∠3, we have ∠3 = 180° - 45° - ∠5 = 135° - ∠5. Since ∠5 = ∠2 = 35°, we have ∠3 = 135° - 35° = 100°. Therefore, ∠3 = 100°.The answer is: 100
answer:100
correct answer:80
total:727  correct:191  acc:0.2627235213204952
------------------------------------------------------------------------------------
ID:727
picturePath:image/1149.png
answer:16海里/时 * 2小时 = 32海里, 12海里/时 * 2小时 = 24海里, 32海里 - 24海里 = 8海里, 所以两船相距8海里.
answer:8
correct answer:40
total:728  correct:191  acc:0.2623626373626374
------------------------------------------------------------------------------------
ID:728
picturePath:image/1150.png
answer:This is a geometry problem involving a right triangle and rotation. Since the triangle is rotated 90 degrees around point A, the side opposite the right angle (AC) will coincide with the hypotenuse (AB) of the new triangle (AB′C′). Therefore, ∠ACB = ∠AC′B′. Also, since the sum of angles in a triangle is 180 degrees, ∠ACB + ∠BAC + ∠CAB = 180. Since ∠BAC = 90, ∠ACB + ∠CAB = 90. Therefore, ∠ACB = 45 degrees. Since ∠ACB = ∠AC′B′, ∠AC′B′ = 45 degrees.
answer:45
correct answer:13
total:729  correct:191  acc:0.262002743484225
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ID:729
picturePath:image/1151.png
answer:60°
answer:60
correct answer:60
total:730  correct:192  acc:0.26301369863013696
------------------------------------------------------------------------------------
ID:730
picturePath:image/1152.png
answer:2.5cm
answer:2.5
correct answer:2.1
total:731  correct:192  acc:0.26265389876880985
------------------------------------------------------------------------------------
ID:731
picturePath:image/1155.png
answer:4
answer:4
correct answer:9
total:732  correct:192  acc:0.26229508196721313
------------------------------------------------------------------------------------
ID:732
picturePath:image/1156.png
answer:根据题意，∠1、∠2、∠3、∠4对应的邻补角和等于215°，即∠1+∠2+∠3+∠4+∠1+∠2+∠3+∠4=215°，解得∠1=∠2=∠3=∠4=52.5°，∴∠BOD=180°-52.5°×2=180°-105°=75°．故选B．
answer:75
correct answer:35
total:733  correct:192  acc:0.26193724420190995
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ID:733
picturePath:image/1158.png
answer:∠A = ∠E - ∠C = 90° - 36° = 54°.
answer:54
correct answer:54
total:734  correct:193  acc:0.26294277929155313
------------------------------------------------------------------------------------
ID:734
picturePath:image/1161.png
answer:20
answer:20
correct answer:10
total:735  correct:193  acc:0.26258503401360545
------------------------------------------------------------------------------------
ID:735
picturePath:image/1162.png
answer:30
answer:30
correct answer:60
total:736  correct:193  acc:0.26222826086956524
------------------------------------------------------------------------------------
ID:736
picturePath:image/1163.png
answer:解:如图,∵∠1=104°,∴∠3=180°-∠1=76°,∵a∥b,∴∠4=∠3=76°,∴∠2=180°-∠4=104°,
故选:A．
answer:104
correct answer:76
total:737  correct:193  acc:0.2618724559023066
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ID:737
picturePath:image/1164.png
answer:The answer is 35°.
answer:35
correct answer:55
total:738  correct:193  acc:0.26151761517615174
------------------------------------------------------------------------------------
ID:738
picturePath:image/1165.png
answer:65°
answer:65
correct answer:115
total:739  correct:193  acc:0.26116373477672533
------------------------------------------------------------------------------------
ID:739
picturePath:image/1166.png
answer:由DC∥FP,∠1=∠2,∠FED=32°,∠AGF=76°,可得∠1=∠FED=32°,∠2=∠FED=32°,∠AGF+∠FED+∠1=180°,解得∠AGF=76°,∠FED=32°,∠1=32°,∵FH平分∠EFG,∴∠EFH=\frac{1}{2}∠EFG=\frac{1}{2}×76°=38°,∴∠PFH=∠1+∠EFH=32°+38°=70°.故选B.
answer:70
correct answer:22
total:740  correct:193  acc:0.2608108108108108
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ID:740
picturePath:image/1167.png
answer:The answer is 50.
answer:50
correct answer:100
total:741  correct:193  acc:0.26045883940620784
------------------------------------------------------------------------------------
ID:741
picturePath:image/1168.png
answer:根据平行线的性质，∠1和∠2互为同位角，所以∠1=∠2。已知∠1=100°，∠2=120°，所以∠α=180°-100°-120°=60°。故选B。
answer:60
correct answer:40
total:742  correct:193  acc:0.2601078167115903
------------------------------------------------------------------------------------
ID:742
picturePath:image/1170.png
answer:40
answer:40
correct answer:40
total:743  correct:194  acc:0.2611036339165545
------------------------------------------------------------------------------------
ID:743
picturePath:image/1171.png
answer:解:如图,连接OC,∵CD为⊙O的切线,∴∠OCD=90°,∵AB为⊙O的直径,∴∠A=90°,∵∠A=23°,∴∠DOC=90°-23°=67°,∴∠D=90°-67°=23°．故选:A．
answer:23
correct answer:44
total:744  correct:194  acc:0.260752688172043
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ID:744
picturePath:image/1172.png
answer:The answer is C.
ID:745
picturePath:image/1173.png
answer:∠1 = 70°, ∠1 = ∠3 (对顶角相等), ∠2 + ∠3 = 180° (两直线平行,内错角相等), ∠2 = 180° - ∠3 = 180° - 70° = 110°.
answer:110
correct answer:110
total:746  correct:195  acc:0.2613941018766756
------------------------------------------------------------------------------------
ID:746
picturePath:image/1174.png
answer:∵直线a//b,∠1=110°,∴∠3=∠1=110°,∵∠2+∠3=180°,∴∠2=180°-∠3=180°-110°=70°,故选B
answer:70
correct answer:70
total:747  correct:196  acc:0.26238286479250333
------------------------------------------------------------------------------------
ID:747
picturePath:image/1175.png
answer:72°
answer:72
correct answer:72
total:748  correct:197  acc:0.26336898395721925
------------------------------------------------------------------------------------
ID:748
picturePath:image/1177.png
answer:解:∵AD⊥DE,DE∥BC,∴△ADE∽△ABC,∴\frac{AD}{AB}=\frac{BD}{BC},即\frac{12}{AB}=\frac{12}{24},解得AB=24m．故选:A．
answer:24
correct answer:18
total:749  correct:197  acc:0.2630173564753004
------------------------------------------------------------------------------------
ID:749
picturePath:image/1178.png
answer:This is a geometry problem. According to the properties of parallel lines, corresponding angles are congruent. Since AB || CD, we have ∠1 = ∠3 = 40°. Also, since BD⊥BC, we have ∠4 = 90°. Then, by the property of triangles, the sum of angles in a triangle is 180°, so we have ∠2 + ∠3 + ∠4 = 180°. Substituting the known values, we get ∠2 + 40° + 90° = 180°. Solving for ∠2, we find ∠2 = 180° - 40° - 90° = 50°. Therefore, the answer is B.
answer:50
correct answer:50
total:750  correct:198  acc:0.264
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ID:750
picturePath:image/1179.png
answer:40°
answer:40
correct answer:50
total:751  correct:198  acc:0.2636484687083888
------------------------------------------------------------------------------------
ID:751
picturePath:image/1180.png
answer:4cm.
answer:4
correct answer:4
total:752  correct:199  acc:0.2646276595744681
------------------------------------------------------------------------------------
ID:752
picturePath:image/1181.png
answer:解:∵AB∥CD,∴∠1+∠3=180°,∵EF⊥AB,∴∠E=90°,∴∠3=90°-60°=30°,∵AB∥CD,∴∠2=∠3=30°．故选:B．
answer:30
correct answer:30
total:753  correct:200  acc:0.2656042496679947
------------------------------------------------------------------------------------
ID:753
picturePath:image/1183.png
answer:50°
answer:50
correct answer:40
total:754  correct:200  acc:0.26525198938992045
------------------------------------------------------------------------------------
ID:754
picturePath:image/1184.png
answer:∠2 = 180° - ∠1 = 180° - 120° = 60°.
answer:60
correct answer:60
total:755  correct:201  acc:0.2662251655629139
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ID:755
picturePath:image/1185.png
answer:15°
answer:15
correct answer:55
total:756  correct:201  acc:0.26587301587301587
------------------------------------------------------------------------------------
ID:756
picturePath:image/1186.png
answer:根据垂线的性质，BO⊥AO，所以∠AOB=90°。又因为OB平分∠COD，所以∠BOC=∠BOD=22°。所以∠AOC=∠AOB-∠BOC=90°-22°=68°。故选B。
answer:68
correct answer:68
total:757  correct:202  acc:0.2668428005284016
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ID:757
picturePath:image/1187.png
answer:解:∵⊙I是△ABC的内切圆,D,E,F为三个切点,∴∠AED=∠AFE=90°,∴∠1=90°+∠DEF=90°+52°=142°,∴∠A=\frac{180^{°}-∠1}{2}=\frac{180^{°}-142^{°}}{2}=20°．故选:A．
answer:20
correct answer:76
total:758  correct:202  acc:0.26649076517150394
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ID:758
picturePath:image/1188.png
answer:解:∵BD,CD分别是过⊙O上点B,C的切线,∠BDC=110°,∴∠DBC+∠DCB=∠BDC=110°,∴∠AOB=2∠DBC=2×110°=220°,∵OA=OB,∴∠A=\frac{1}{2}(360°-220°)=130°．故选:A．
answer:130
correct answer:35
total:759  correct:202  acc:0.26613965744400525
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ID:759
picturePath:image/1189.png
answer:解:∵PA,PB分别是⊙O的切线,A,B为切点,∴∠OAP=∠OBP=90°,∵∠BAC=35°,∴∠AOB=2∠BAC=2×35°=70°,∵AC是⊙O的直径,∴∠C=90°,∴∠P=180°-∠AOB-∠C=180°-70°-90°=20°．故选:B．
answer:20
correct answer:70
total:760  correct:202  acc:0.2657894736842105
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ID:760
picturePath:image/1190.png
answer:30°
answer:30
correct answer:15
total:761  correct:202  acc:0.26544021024967146
------------------------------------------------------------------------------------
ID:761
picturePath:image/1191.png
answer:解:∵直线a∥b,∴∠1=∠5,∵∠1和∠2互余,∴∠2=90°-∠1,∴∠2=90°-(90°-∠5)=∠5,∵∠3=∠4+∠5,∠3=121°,∴∠4+∠5=121°,∴∠4=121°-∠5,∵∠5=∠2,∴∠4=121°-∠2,∵∠2=90°,∴∠4=121°-90°=31°．故选:A．
answer:31
correct answer:149
total:762  correct:202  acc:0.2650918635170604
------------------------------------------------------------------------------------
ID:762
picturePath:image/1192.png
answer:50°
answer:50
correct answer:65
total:763  correct:202  acc:0.26474442988204455
------------------------------------------------------------------------------------
ID:763
picturePath:image/1193.png
answer:解:∵PA是⊙O的切线,∴∠CAP=90°,∵∠PAB=40°,∴∠CAP=∠PAB=40°,∴∠AOB=2∠CAP=2×40°=80°．故选:D．
answer:80
correct answer:80
total:764  correct:203  acc:0.26570680628272253
------------------------------------------------------------------------------------
ID:764
picturePath:image/1195.png
answer:解:∵PA、PB是⊙O的切线,∴∠PAE=∠PBE=90°,∴∠AEB=180°-∠APB=180°-60°=120°,∴∠COD=360°-∠AEB=360°-120°=240°．故选:D．
answer:240
correct answer:60
total:765  correct:203  acc:0.265359477124183
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ID:765
picturePath:image/1196.png
answer:解:∵PC切⊙O于C,∴PC⊥AB,即∠ACB=90°,∵∠CAP=35°,∴∠B=90°-35°=55°,∴∠CPO=∠B=55°．故选:B．
answer:55
correct answer:20
total:766  correct:203  acc:0.2650130548302872
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ID:766
picturePath:image/1197.png
answer:This is a geometry problem involving circles, angles, and tangents. Given that \( \angle A = 35^\circ \) and \( CD \) is a tangent to the circle at point \( C \), we know that \( \angle AOC = 180^\circ \) (angle in a semicircle). Therefore, \( \angle BOC = 180^\circ - 35^\circ = 145^\circ \). Since \( OB \) is a radius of the circle, \( \angle BOD = 90^\circ \) (angle in a right triangle). Finally, \( \angle D = 90^\circ - \angle BOD = 90^\circ - 90^\circ = 0^\circ \). Thus, the answer is A.
answer:0
correct answer:20
total:767  correct:203  acc:0.2646675358539765
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ID:767
picturePath:image/1198.png
answer:解:∵直线AC⊥BN,∴∠ACN=90°,∵当AC平移到与⊙B相切时,AB的长度的最小,∴点C在以BN为直径的圆上,∴NC⊥BN,∴∠NCB=90°,∵∠MBN=60°,∴∠NCB=∠MBN=60°,∴NC=MB=4cm,∴AB=√{NC^{2}+MB^{2}}=√{4^{2}+4^{2}}=4√{2}(cm)．故选:D．
answer:2
correct answer:8
total:768  correct:203  acc:0.2643229166666667
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ID:768
picturePath:image/1199.png
answer:解:∵BC是⊙O的直径,切点为A,∴∠BAC=90°,∵∠MAB=30°,∴∠B=90°-30°=60°．故选:B．
answer:60
correct answer:60
total:769  correct:204  acc:0.26527958387516254
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ID:769
picturePath:image/1200.png
answer:40
answer:40
correct answer:30
total:770  correct:204  acc:0.2649350649350649
------------------------------------------------------------------------------------
ID:770
picturePath:image/1201.png
answer:40°
answer:40
correct answer:50
total:771  correct:204  acc:0.26459143968871596
------------------------------------------------------------------------------------
ID:771
picturePath:image/1202.png
answer:60
answer:60
correct answer:120
total:772  correct:204  acc:0.26424870466321243
------------------------------------------------------------------------------------
ID:772
picturePath:image/1206.png
answer:The answer is 37°.
answer:37
correct answer:143
total:773  correct:204  acc:0.26390685640362227
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ID:773
picturePath:image/1210.png
answer:Since OE is the angle bisector of ∠AOB and CD is parallel to OB, we can conclude that ∠DOB = ∠ACD = 40°. Since OE is also parallel to CD, we can conclude that ∠CDO = ∠DOB = 40°. Therefore, the answer is D.
answer:40
correct answer:20
total:774  correct:204  acc:0.26356589147286824
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ID:774
picturePath:image/1213.png
answer:4.
answer:4
correct answer:4
total:775  correct:205  acc:0.2645161290322581
------------------------------------------------------------------------------------
ID:775
picturePath:image/1215.png
answer:40°
answer:40
correct answer:40
total:776  correct:206  acc:0.2654639175257732
------------------------------------------------------------------------------------
ID:776
picturePath:image/1216.png
answer:The answer is 180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-150°-90°=180°-
answer:180
correct answer:60
total:777  correct:206  acc:0.2651222651222651
------------------------------------------------------------------------------------
ID:777
picturePath:image/1217.png
answer:根据平行线的性质，∠ABE和∠ECD互为同位角，所以∠ABE=∠ECD。已知∠ABE=120°，∠ECD=25°，所以∠ABE=∠ECD=25°。因此，∠E=∠ABE-∠ABE=120°-25°=95°。
answer:95
correct answer:85
total:778  correct:206  acc:0.2647814910025707
------------------------------------------------------------------------------------
ID:778
picturePath:image/1218.png
answer:20°
answer:20
correct answer:110
total:779  correct:206  acc:0.2644415917843389
------------------------------------------------------------------------------------
ID:779
picturePath:image/1219.png
answer:60
answer:60
correct answer:30
total:780  correct:206  acc:0.2641025641025641
------------------------------------------------------------------------------------
ID:780
picturePath:image/1220.png
answer:24
answer:24
correct answer:168
total:781  correct:206  acc:0.26376440460947503
------------------------------------------------------------------------------------
ID:781
picturePath:image/1222.png
answer:根据题意，M是△ABC的边BC的中点，则MN是△ABC的中位线，因此AB=2MN=2*3=6。又因为AB=8，所以AC=AB-BC=8-6=2。故选：B。
answer:2
correct answer:14
total:782  correct:206  acc:0.26342710997442453
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ID:782
picturePath:image/1225.png
answer:75°
answer:75
correct answer:155
total:783  correct:206  acc:0.26309067688378035
------------------------------------------------------------------------------------
ID:783
picturePath:image/1226.png
answer:60°
answer:60
correct answer:36
total:784  correct:206  acc:0.2627551020408163
------------------------------------------------------------------------------------
ID:784
picturePath:image/1227.png
answer:∠A=40°,∠B=90°,AC的垂直平分线MN分别与AB,AC交于点D,E,则∠BCD的度数为40°.
answer:40
correct answer:10
total:785  correct:206  acc:0.2624203821656051
------------------------------------------------------------------------------------
ID:785
picturePath:image/1228.png
answer:The answer is A.
ID:786
picturePath:image/1229.png
answer:解:∵直线a∥b,∴∠1=∠MAO=58°,∵AM⊥b,∴∠O=90°,∴∠2=∠O-∠MAO=90°-58°=32°．故选:B．
answer:32
correct answer:32
total:787  correct:207  acc:0.2630241423125794
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ID:787
picturePath:image/1230.png
answer:解:∵AB是半圆O的直径,∴∠ACB=90°,∵∠BAC=26°,∴∠B=90°-26°=64°,∵D是弧AC的中点,∴AD=CD,∴∠DCA=∠DAC=\frac{1}{2}∠B=\frac{1}{2}×64°=32°．故选:B．
answer:32
correct answer:32
total:788  correct:208  acc:0.2639593908629442
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ID:788
picturePath:image/1234.png
answer:解:∵⊙O是△ABC的外接圆,∠OCB=40°,∴∠A=\frac{1}{2}∠OCB=\frac{1}{2}×40°=20°．故选:A．
answer:20
correct answer:50
total:789  correct:208  acc:0.26362484157160965
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ID:789
picturePath:image/1235.png
answer:解:∵∠B=∠E,AB=DE,BF=EC,∴△ABF∽△DEC,∴\frac{AB}{DE}=\frac{BF}{EC}=\frac{AB+CF}{DE+CF}=\frac{840}{840+106}=\frac{840}{946},∴DE=840×946÷840=940(克),∴整个金属框架的质量为840+940+106=1886(克)．故选:D．
answer:1886
correct answer:1574
total:790  correct:208  acc:0.26329113924050634
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ID:790
picturePath:image/1236.png
answer:2∠A=2×40°=80°.
answer:80
correct answer:40
total:791  correct:208  acc:0.2629582806573957
------------------------------------------------------------------------------------
ID:791
picturePath:image/1239.png
answer:50
answer:50
correct answer:25
total:792  correct:208  acc:0.26262626262626265
------------------------------------------------------------------------------------
ID:792
picturePath:image/1240.png
answer:72°+10°=82°,∠AOB的度数为82°.
answer:82
correct answer:118
total:793  correct:208  acc:0.26229508196721313
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ID:793
picturePath:image/1241.png
answer:解:∵∠NAC=36°,∠NBC=72°,∴∠ABC=72°-36°=36°,∴△ABC是等腰三角形,∴AB=BC,∵BC=20×2=40(里),∴AB=BC=40(里),∴从B处到灯塔C的距离为40×2=80(里)．故选:D．
answer:80
correct answer:40
total:794  correct:208  acc:0.2619647355163728
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ID:794
picturePath:image/1242.png
answer:The answer is A.
ID:795
picturePath:image/1243.png
answer:解:∵DE是AC的垂直平分线,∴DA=DC,∵△ABD的周长为13cm,∴AB+BD+DA=13cm,即AB+BD+DC=13cm,∴△ABC的周长=AB+BC+AC=AB+BD+DC+AC=AB+AB+AC=2AB+AC,∵AE=3cm,∴AC=2AE=6cm,∴△ABC的周长=2AB+6=13+6=19cm．故选:D．
answer:19
correct answer:19
total:796  correct:209  acc:0.26256281407035176
------------------------------------------------------------------------------------
ID:796
picturePath:image/1244.png
answer:The answer is A.
ID:797
picturePath:image/1247.png
answer:30°
answer:30
correct answer:60
total:798  correct:209  acc:0.2619047619047619
------------------------------------------------------------------------------------
ID:798
picturePath:image/1248.png
answer:30°
answer:30
correct answer:60
total:799  correct:209  acc:0.2615769712140175
------------------------------------------------------------------------------------
ID:799
picturePath:image/1249.png
answer:This is a geometry problem involving a triangle and its incircle. The incircle of a triangle is the circle that is tangent to all three sides of the triangle. Given that the incircle is tangent to sides AB, BC, and CA at points D, E, and F respectively, and that AD = 2, we can use the fact that the radius of the incircle is equal to the distance from the vertex of the triangle to the point of tangency of the incircle with the opposite side. Since AD is the radius of the incircle, DE is the distance from vertex A to the point of tangency of the incircle with side AB. Similarly, EF is the distance from vertex B to the point of tangency of the incircle with side BC, and FD is the distance from vertex C to the point of tangency of the incircle with side CA. Since the incircle is tangent to all three sides, DE + EF + FD = BC = 5. Therefore, the perimeter of triangle ABC is 2(DE + EF + FD) = 2(5) = 10. The answer is A.
answer:10
correct answer:14
total:800  correct:209  acc:0.26125
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ID:800
picturePath:image/1254.png
answer:7
answer:7
correct answer:8
total:801  correct:209  acc:0.26092384519350814
------------------------------------------------------------------------------------
ID:801
picturePath:image/1261.png
answer:60
answer:60
correct answer:90
total:802  correct:209  acc:0.26059850374064836
------------------------------------------------------------------------------------
ID:802
picturePath:image/1264.png
answer:This is a right triangle problem. We can use the Pythagorean theorem to solve it. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, AB is the hypotenuse, and BD and DC are the other two sides. So, we have:

AB^2 = BD^2 + DC^2

Substituting the given values, we get:

AB^2 = 120^2 + 60^2
AB^2 = 14400 + 3600
AB^2 = 18000

Taking the square root of both sides, we get:

AB = √18000
AB ≈ 135

Therefore, the width of the river AB is approximately 135 meters.
answer:135
correct answer:100
total:803  correct:209  acc:0.2602739726027397
------------------------------------------------------------------------------------
ID:803
picturePath:image/1265.png
answer:60
answer:60
correct answer:40
total:804  correct:209  acc:0.2599502487562189
------------------------------------------------------------------------------------
ID:804
picturePath:image/1266.png
answer:50
answer:50
correct answer:70
total:805  correct:209  acc:0.2596273291925466
------------------------------------------------------------------------------------
ID:805
picturePath:image/1267.png
answer:110°
answer:110
correct answer:55
total:806  correct:209  acc:0.25930521091811415
------------------------------------------------------------------------------------
ID:806
picturePath:image/1269.png
answer:根据对顶角相等，可以得出∠2=∠1=30°。故选A。
answer:30
correct answer:150
total:807  correct:209  acc:0.2589838909541512
------------------------------------------------------------------------------------
ID:807
picturePath:image/1270.png
answer:解:∵圆柱的高是8cm,底面周长是30cm,∴圆柱的底面半径是15÷2π≈5cm,∴从点A爬到点B的最短路程是:π×5×8=40π≈125.7cm；答:从点A爬到点B的最短路程是125.7cm．
answer:125.7
correct answer:17
total:808  correct:209  acc:0.25866336633663367
------------------------------------------------------------------------------------
ID:808
picturePath:image/1272.png
answer:20°
answer:20
correct answer:80
total:809  correct:209  acc:0.2583436341161928
------------------------------------------------------------------------------------
ID:809
picturePath:image/1275.png
answer:Since \(AB \parallel CD\), we can use the corresponding angles postulate, which states that if two lines are parallel, then the angles formed by the transversal and each line are congruent. Therefore, \(∠1 = ∠2\). Given that \(∠2 = 125°\), we can conclude that \(∠1 = 125°\).
answer:125
correct answer:55
total:810  correct:209  acc:0.2580246913580247
------------------------------------------------------------------------------------
ID:810
picturePath:image/1277.png
answer:15°
answer:15
correct answer:55
total:811  correct:209  acc:0.25770653514180025
------------------------------------------------------------------------------------
ID:811
picturePath:image/1278.png
answer:40°
answer:40
correct answer:35
total:812  correct:209  acc:0.25738916256157635
------------------------------------------------------------------------------------
ID:812
picturePath:image/1279.png
answer:∠COE=\frac{1}{2}∠AOD=\frac{1}{2}×(180°-∠BOC)=\frac{1}{2}×(180°-70°)=65°.
answer:65
correct answer:145
total:813  correct:209  acc:0.2570725707257073
------------------------------------------------------------------------------------
ID:813
picturePath:image/1280.png
answer:60
answer:60
correct answer:90
total:814  correct:209  acc:0.25675675675675674
------------------------------------------------------------------------------------
ID:814
picturePath:image/1281.png
answer:The answer is 80.
answer:80
correct answer:70
total:815  correct:209  acc:0.25644171779141106
------------------------------------------------------------------------------------
ID:815
picturePath:image/1284.png
answer:8
answer:8
correct answer:4
total:816  correct:209  acc:0.25612745098039214
------------------------------------------------------------------------------------
ID:816
picturePath:image/1286.png
answer:100
answer:100
correct answer:80
total:817  correct:209  acc:0.2558139534883721
------------------------------------------------------------------------------------
ID:817
picturePath:image/1287.png
answer:根据题意，DE垂直平分AC，∠A=30°，∠ACB=80°，则∠AEC=∠A=30°，∠ACE=180°-∠ACB-∠AEC=180°-80°-30°=70°。又因为DE垂直平分AC，所以∠BCE=∠ACE=70°。故选：B。
answer:70
correct answer:50
total:818  correct:209  acc:0.2555012224938875
------------------------------------------------------------------------------------
ID:818
picturePath:image/1288.png
answer:The answer is \frac{1}{2}AC\cdotBD=\frac{1}{2}×10×6=30.
answer:30
correct answer:15
total:819  correct:209  acc:0.25518925518925517
------------------------------------------------------------------------------------
ID:819
picturePath:image/1292.png
answer:This is a geometric figure with labeled points A, B, C, D, and E. The figure shows a triangle ABC with angle ACB labeled as 45 degrees. A line segment AD is drawn from point A to point D, and it intersects BC at point D. Another line segment CE is drawn from point C to point E, and it intersects AB at point E. The length of CE is labeled as equal to AB. The angle ACE is labeled as 20 degrees. The question asks for the measure of angle B.
answer:20
correct answer:65
total:820  correct:209  acc:0.2548780487804878
------------------------------------------------------------------------------------
ID:820
picturePath:image/1293.png
answer:25°
answer:25
correct answer:25
total:821  correct:210  acc:0.2557856272838002
------------------------------------------------------------------------------------
ID:821
picturePath:image/1295.png
answer:60
answer:60
correct answer:70
total:822  correct:210  acc:0.25547445255474455
------------------------------------------------------------------------------------
ID:822
picturePath:image/1297.png
answer:解:∵AB=AC,∴⁀{AB}=⁀{AC},∴∠B=(180°-2∠CAB)=130°,∴∠ADC=\frac{1}{2}∠B=65°．故选:B．
answer:65
correct answer:115
total:823  correct:210  acc:0.2551640340218712
------------------------------------------------------------------------------------
ID:823
picturePath:image/1298.png
answer:解:∵扇形纸扇完全打开后,外侧两竹条AB,AC的夹角为150°,∴∠BAC=150°,∵弧BC长为50πcm,∴\frac{150×π×AB}{180}=50π,解得:AB=40cm．故选:A．
answer:40
correct answer:60
total:824  correct:210  acc:0.25485436893203883
------------------------------------------------------------------------------------
ID:824
picturePath:image/1299.png
answer:The answer is A.
ID:825
picturePath:image/1300.png
answer:20√2
answer:2
correct answer:240
total:826  correct:210  acc:0.2542372881355932
------------------------------------------------------------------------------------
ID:826
picturePath:image/1301.png
answer:5
answer:5
correct answer:2
total:827  correct:210  acc:0.2539298669891173
------------------------------------------------------------------------------------
ID:827
picturePath:image/1302.png
answer:由全等三角形的性质可得∠A=∠B=30°,∠C=∠D=20°,则∠COD=180°-30°-20°=130°.故选C
answer:130
correct answer:80
total:828  correct:210  acc:0.2536231884057971
------------------------------------------------------------------------------------
ID:828
picturePath:image/1303.png
answer:20
answer:20
correct answer:20
total:829  correct:211  acc:0.2545235223160434
------------------------------------------------------------------------------------
ID:829
picturePath:image/1307.png
answer:12
answer:12
correct answer:12.5
total:830  correct:211  acc:0.25421686746987954
------------------------------------------------------------------------------------
ID:830
picturePath:image/1308.png
answer:∠2 = ∠1 = 35°
answer:35
correct answer:145
total:831  correct:211  acc:0.2539109506618532
------------------------------------------------------------------------------------
ID:831
picturePath:image/1309.png
answer:20
answer:20
correct answer:20
total:832  correct:212  acc:0.2548076923076923
------------------------------------------------------------------------------------
ID:832
picturePath:image/1310.png
answer:根据平行线的性质，两直线平行，同位角相等。因此，∠2=∠1=50°。
answer:50
correct answer:130
total:833  correct:212  acc:0.2545018007202881
------------------------------------------------------------------------------------
ID:833
picturePath:image/1311.png
answer:解:∵CD为⊙O的直径,且CD⊥弦AB,∴⁀{BC}=⁀{BC},∴∠B=\frac{1}{2}∠AOC=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:65
total:834  correct:212  acc:0.2541966426858513
------------------------------------------------------------------------------------
ID:834
picturePath:image/1312.png
answer:36
answer:36
correct answer:36
total:835  correct:213  acc:0.25508982035928146
------------------------------------------------------------------------------------
ID:835
picturePath:image/1315.png
answer:∠ACB=\frac{1}{2}∠AOB=\frac{1}{2}×72°=36°.
answer:36
correct answer:36
total:836  correct:214  acc:0.25598086124401914
------------------------------------------------------------------------------------
ID:836
picturePath:image/1316.png
answer:25°
answer:25
correct answer:130
total:837  correct:214  acc:0.25567502986857826
------------------------------------------------------------------------------------
ID:837
picturePath:image/1317.png
answer:由于AB∥EF∥CD，根据平行线的性质，∠ABC和∠CEF互为内错角，所以∠ABC=∠CEF。已知∠ABC=46°，∠CEF=160°，所以∠CEF=∠ABC=46°。又因为∠ABC和∠BCE是邻补角，所以∠BCE=180°-∠ABC=180°-46°=134°。所以，∠BCE等于134°。
answer:134
correct answer:26
total:838  correct:214  acc:0.2553699284009546
------------------------------------------------------------------------------------
ID:838
picturePath:image/1322.png
answer:解:∵四边形ABCD为⊙O的内接四边形,∴∠A=∠BCD=120°,
由圆周角定理得,∠BOD=2∠A=240°,
故选:C．
answer:240
correct answer:120
total:839  correct:214  acc:0.2550655542312277
------------------------------------------------------------------------------------
ID:839
picturePath:image/1323.png
answer:25
answer:25
correct answer:75
total:840  correct:214  acc:0.25476190476190474
------------------------------------------------------------------------------------
ID:840
picturePath:image/1325.png
answer:This is a geometric figure with three points labeled A, B, and C, and two lines labeled D and E. The lines intersect at point D, and there is an angle labeled ∠CAE with a measure of 16 degrees. The question is asking for the measure of angle B.
answer:16
correct answer:37
total:841  correct:214  acc:0.2544589774078478
------------------------------------------------------------------------------------
ID:841
picturePath:image/1326.png
answer:20
answer:20
correct answer:14
total:842  correct:214  acc:0.25415676959619954
------------------------------------------------------------------------------------
ID:842
picturePath:image/1329.png
answer:∠2 = ∠1 = 60°
answer:60
correct answer:120
total:843  correct:214  acc:0.2538552787663108
------------------------------------------------------------------------------------
ID:843
picturePath:image/1330.png
answer:90°
answer:90
correct answer:45
total:844  correct:214  acc:0.25355450236966826
------------------------------------------------------------------------------------
ID:844
picturePath:image/1331.png
answer:60°
answer:60
correct answer:60
total:845  correct:215  acc:0.25443786982248523
------------------------------------------------------------------------------------
ID:845
picturePath:image/1332.png
answer:解:∵∠A=60°,∠1=85°,∴∠3=180°-∠A-∠1=180°-60°-85°=35°,∴∠2=∠3=35°．故选:A．
answer:35
correct answer:35
total:846  correct:216  acc:0.2553191489361702
------------------------------------------------------------------------------------
ID:846
picturePath:image/1334.png
answer:\( \text{A} \) 81 \( \text{B} \) 144 \( \text{C} \) 169 \( \text{D} \) 196 \( \text{E} \) 324.
answer:324
correct answer:25
total:847  correct:216  acc:0.2550177095631641
------------------------------------------------------------------------------------
ID:847
picturePath:image/1336.png
answer:30
answer:30
correct answer:30
total:848  correct:217  acc:0.25589622641509435
------------------------------------------------------------------------------------
ID:848
picturePath:image/1337.png
answer:7
answer:7
correct answer:8
total:849  correct:217  acc:0.25559481743227325
------------------------------------------------------------------------------------
ID:849
picturePath:image/1341.png
answer:Since ∠ABC=∠ACB and ∠A=50°, we know that ∠A+∠ABC+∠ACB=180°. Therefore, ∠ABC+∠ACB=180°-50°=130°. 

Since ∠1=∠2, we know that ∠1+∠BPC+∠2=180°. Therefore, ∠BPC=180°-(∠1+∠2)=180°-(∠1+∠1)=180°-2∠1.

Since ∠1=∠2, we know that ∠BPC=180°-2∠1=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-2(∠1)=180°-
answer:180
correct answer:115
total:850  correct:217  acc:0.25529411764705884
------------------------------------------------------------------------------------
ID:850
picturePath:image/1342.png
answer:解:如图所示:∠1=56°∴∠3=56°∴∠2=180°-56°×2=68°.故选B.
answer:68
correct answer:68
total:851  correct:218  acc:0.25616921269095183
------------------------------------------------------------------------------------
ID:851
picturePath:image/1344.png
answer:根据三角形内角和定理，∠BOE = 180° - ∠AOE - ∠DOE = 180° - 24° - 72° = 84°.
answer:84
correct answer:156
total:852  correct:218  acc:0.25586854460093894
------------------------------------------------------------------------------------
ID:852
picturePath:image/1345.png
answer:40
answer:40
correct answer:40
total:853  correct:219  acc:0.2567409144196952
------------------------------------------------------------------------------------
ID:853
picturePath:image/1346.png
answer:∠2 = 180° - ∠1 = 180° - 72° = 108°.
answer:108
correct answer:108
total:854  correct:220  acc:0.2576112412177986
------------------------------------------------------------------------------------
ID:854
picturePath:image/1349.png
answer:The answer is 60.
answer:60
correct answer:60
total:855  correct:221  acc:0.2584795321637427
------------------------------------------------------------------------------------
ID:855
picturePath:image/1350.png
answer:This is a question about geometry. To solve it, we need to use the properties of angles in a triangle and the fact that the sum of the angles in a triangle is 180 degrees. Let's break down the information given:

∠APB = 126°
∠AQF = 100°
∠APB and ∠AQF are alternate interior angles, so they are congruent. Therefore, ∠APB = ∠AQF = 126°.

Now, let's find ∠A and ∠F:

∠A + ∠B + ∠C = 180° (Sum of angles in a triangle)
∠A + ∠F + ∠C = 180° (Sum of angles in a triangle)
∠A + ∠F = 180° - ∠C (Subtracting ∠C from both sides of the equation)
∠A + ∠F = 180° - 126° (Substituting ∠APB with ∠AQF)
∠A + ∠F = 54° (Simplifying the equation)

Therefore, ∠A - ∠F = 54° - 54° = 0°.

The answer is: 0
answer:0
correct answer:46
total:856  correct:221  acc:0.25817757009345793
------------------------------------------------------------------------------------
ID:856
picturePath:image/1354.png
answer:65
answer:65
correct answer:115
total:857  correct:221  acc:0.25787631271878647
------------------------------------------------------------------------------------
ID:857
picturePath:image/1355.png
answer:The answer is 72.
answer:72
correct answer:108
total:858  correct:221  acc:0.25757575757575757
------------------------------------------------------------------------------------
ID:858
picturePath:image/1356.png
answer:解:由勾股定理得:AB=√{52+122}=13(m),则大树在折断前高为13+12=25(m)．故选:B．
answer:25
correct answer:18
total:859  correct:221  acc:0.2572759022118743
------------------------------------------------------------------------------------
ID:859
picturePath:image/1358.png
answer:∠2=\frac{1}{2}∠1=\frac{1}{2}×105°=52.5°.
answer:52.5
correct answer:75
total:860  correct:221  acc:0.2569767441860465
------------------------------------------------------------------------------------
ID:860
picturePath:image/1359.png
answer:6厘米.
answer:6
correct answer:3
total:861  correct:221  acc:0.25667828106852497
------------------------------------------------------------------------------------
ID:861
picturePath:image/1360.png
answer:解:∵△ABC≌△ADC,∠ABC=118°,∠DAC=40°,∴∠ACD=∠B=118°,∴∠BCD=180°-∠ACD=180°-118°=62°,故选:B．
answer:62
correct answer:44
total:862  correct:221  acc:0.25638051044083526
------------------------------------------------------------------------------------
ID:862
picturePath:image/1362.png
answer:解:如图,∵∠1=34°,∴∠3=∠1+30°=64°,∵直尺对边平行,∴∠2=∠3=64°．故选:B．
answer:64
correct answer:124
total:863  correct:221  acc:0.25608342989571264
------------------------------------------------------------------------------------
ID:863
picturePath:image/1366.png
answer:根据三角形内角和定理，∠4 = 180° - ∠1 - ∠2 - ∠3 = 180° - 90° - 90° - 42° = 58°.
answer:58
correct answer:138
total:864  correct:221  acc:0.25578703703703703
------------------------------------------------------------------------------------
ID:864
picturePath:image/1368.png
answer:解:∵∠ABO=30°,∠ADO=20°,∴∠BOD=∠ABO+∠ADO=30°+20°=50°,∴∠BAD=\frac{1}{2}∠BOD=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:50
total:865  correct:221  acc:0.25549132947976877
------------------------------------------------------------------------------------
ID:865
picturePath:image/1369.png
answer:a∥b,∠3=135°∴∠2=∠3=135°∵∠1+∠2=180°∴∠1=180°-∠2=180°-135°=45°.故选A
answer:45
correct answer:45
total:866  correct:222  acc:0.25635103926096997
------------------------------------------------------------------------------------
ID:866
picturePath:image/1371.png
answer:30
answer:30
correct answer:30
total:867  correct:223  acc:0.25720876585928487
------------------------------------------------------------------------------------
ID:867
picturePath:image/1374.png
answer:∠BOE = 180° - ∠1 - ∠2 = 180° - 95° - 53° = 32°.
answer:32
correct answer:32
total:868  correct:224  acc:0.25806451612903225
------------------------------------------------------------------------------------
ID:868
picturePath:image/1377.png
answer:19
answer:19
correct answer:10
total:869  correct:224  acc:0.25776754890678943
------------------------------------------------------------------------------------
ID:869
picturePath:image/1380.png
answer:解:∵弦AC∥半径OB,∴∠AOC=∠BOC=48°,∵OA=OC,∴∠OAC=\frac{180°-∠AOC}{2}=60°,∴∠OAB=∠OAC-∠AOC=60°-48°=12°．故选:A．
answer:12
correct answer:24
total:870  correct:224  acc:0.2574712643678161
------------------------------------------------------------------------------------
ID:870
picturePath:image/1382.png
answer:The answer is A.
ID:871
picturePath:image/1388.png
answer:The answer is 30.
answer:30
correct answer:30
total:872  correct:225  acc:0.2580275229357798
------------------------------------------------------------------------------------
ID:872
picturePath:image/1390.png
answer:Since AB is parallel to CD, the corresponding angles are congruent. Therefore, ∠BOC = ∠B = 33°. Since ∠AOB is a right angle, ∠AOC = 180° - ∠AOB - ∠BOC = 180° - 90° - 33° = 57°.
answer:57
correct answer:57
total:873  correct:226  acc:0.2588774341351661
------------------------------------------------------------------------------------
ID:873
picturePath:image/1392.png
answer:40°
answer:40
correct answer:130
total:874  correct:226  acc:0.2585812356979405
------------------------------------------------------------------------------------
ID:874
picturePath:image/1393.png
answer:根据三角形的一个外角等于与它不相邻的两个内角的和，可得∠BCD=180°-∠1-∠D=180°-78°-(180°-78°)=180°-78°-180°+78°=180°-180°+78°-78°=78°．故选:A．
answer:78
correct answer:102
total:875  correct:226  acc:0.2582857142857143
------------------------------------------------------------------------------------
ID:875
picturePath:image/1397.png
answer:3
answer:3
correct answer:8
total:876  correct:226  acc:0.2579908675799087
------------------------------------------------------------------------------------
ID:876
picturePath:image/1398.png
answer:The answer is 4.
answer:4
correct answer:4
total:877  correct:227  acc:0.2588369441277081
------------------------------------------------------------------------------------
ID:877
picturePath:image/1399.png
answer:This is a geometry problem involving angles and parallel lines. Given that AB is parallel to CD, we can use the property of parallel lines that corresponding angles are congruent. Therefore, ∠α = ∠ECD and ∠β = ∠DCB. Since the sum of the angles in a triangle is 180°, we can find the measure of ∠γ by subtracting the measures of ∠α and ∠β from 180°. 

∠γ = 180° - ∠α - ∠β
∠γ = 180° - 130° - 80°
∠γ = 70°

So, the measure of ∠γ is 70°.
answer:70
correct answer:30
total:878  correct:227  acc:0.25854214123006836
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ID:878
picturePath:image/1400.png
answer:由题意得,∠BAC=∠ADE=90°,∠E=45°,∠C=30°,∵AE∥BC,∴∠C+∠AED=∠BAC=90°,∴∠AED=90°-30°=60°,∴∠AFD=∠AED-∠E=60°-45°=15°.故选B.
answer:15
correct answer:75
total:879  correct:227  acc:0.2582480091012514
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ID:879
picturePath:image/1401.png
answer:60°
answer:60
correct answer:70
total:880  correct:227  acc:0.25795454545454544
------------------------------------------------------------------------------------
ID:880
picturePath:image/1404.png
answer:解:∵△ABC沿DE、EF翻折,顶点A,B均落在点O处,且EA与EB重合于线段EO,∴∠AED=∠BEF,∵∠CDO+∠CFO=106°,∴∠AED+∠BEF+∠C=106°,∴∠C=106°-∠AED-∠BEF,∵∠AED=∠BEF,∴∠C=106°-2∠AED,∵∠AED+∠C=180°,∴∠C=106°-2∠AED=180°-∠AED,∴∠AED=74°,∴∠C=106°-74°=32°．故选:B．
answer:32
correct answer:37
total:881  correct:227  acc:0.2576617480136209
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ID:881
picturePath:image/1405.png
answer:6cm
answer:6
correct answer:8
total:882  correct:227  acc:0.25736961451247165
------------------------------------------------------------------------------------
ID:882
picturePath:image/1406.png
answer:解:∵a∥b,∠1=35°,∴∠3=∠1=35°,∵∠2+∠3=90°,∴∠2=90°-35°=55°．故选:A．
answer:55
correct answer:125
total:883  correct:227  acc:0.2570781426953567
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ID:883
picturePath:image/1408.png
answer:根据三角形内角和定理，∠A+∠BDC+∠1+∠2=180°，将∠BDC=120°和∠1+∠2=55°代入，可得∠A+55°=180°-120°，解得∠A=65°.
answer:65
correct answer:65
total:884  correct:228  acc:0.2579185520361991
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ID:884
picturePath:image/1411.png
answer:根据对顶角相等，可以得出∠2=∠1=130°，∠3=∠4，再根据邻补角的定义，可以得出∠2+∠3=180°-∠4=180°-130°=50°,故选B
answer:50
correct answer:100
total:885  correct:228  acc:0.2576271186440678
------------------------------------------------------------------------------------
ID:885
picturePath:image/1412.png
answer:This is a geometry problem that involves angle relationships in a triangle. Given that CD is parallel to AB, we can use the corresponding angles postulate, which states that if two lines are parallel, then the angles that are formed by the transversal and each of the parallel lines are congruent. Therefore, ∠BCD = ∠B = 65°. Since ∠ACD and ∠BCD are alternate interior angles, they are congruent, so ∠ACD = ∠BCD = 40°. The sum of the angles in a triangle is 180°, so ∠ACB = 180° - ∠ACD - ∠BCD = 180° - 40° - 65° = 75°. Therefore, the answer is D.
answer:75
correct answer:75
total:886  correct:229  acc:0.2584650112866817
------------------------------------------------------------------------------------
ID:886
picturePath:image/1414.png
answer:解:如图,∵∠α=43°,∴∠1=43°,∠2=45°,∴∠3=180°-43°-45°=92°,∴∠β=90°-92°=92°-90°=2°．故选:B．
answer:2
correct answer:47
total:887  correct:229  acc:0.25817361894024804
------------------------------------------------------------------------------------
ID:887
picturePath:image/1416.png
answer:The answer is 80.
answer:80
correct answer:80
total:888  correct:230  acc:0.25900900900900903
------------------------------------------------------------------------------------
ID:888
picturePath:image/1417.png
answer:50°
answer:50
correct answer:50
total:889  correct:231  acc:0.25984251968503935
------------------------------------------------------------------------------------
ID:889
picturePath:image/1418.png
answer:60°
answer:60
correct answer:105
total:890  correct:231  acc:0.2595505617977528
------------------------------------------------------------------------------------
ID:890
picturePath:image/1420.png
answer:20
answer:20
correct answer:26
total:891  correct:231  acc:0.25925925925925924
------------------------------------------------------------------------------------
ID:891
picturePath:image/1422.png
answer:70°
answer:70
correct answer:105
total:892  correct:231  acc:0.2589686098654709
------------------------------------------------------------------------------------
ID:892
picturePath:image/1424.png
answer:80m
answer:80
correct answer:80
total:893  correct:232  acc:0.2597984322508399
------------------------------------------------------------------------------------
ID:893
picturePath:image/1425.png
answer:The answer is C.
ID:894
picturePath:image/1427.png
answer:6
answer:6
correct answer:48
total:895  correct:232  acc:0.25921787709497207
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ID:895
picturePath:image/1428.png
answer:The answer is 65°.
answer:65
correct answer:25
total:896  correct:232  acc:0.25892857142857145
------------------------------------------------------------------------------------
ID:896
picturePath:image/1434.png
answer:The answer is 22.
answer:22
correct answer:13
total:897  correct:232  acc:0.25863991081382387
------------------------------------------------------------------------------------
ID:897
picturePath:image/1435.png
answer:100
answer:100
correct answer:10
total:898  correct:232  acc:0.2583518930957684
------------------------------------------------------------------------------------
ID:898
picturePath:image/1437.png
answer:∠BOC = 360° - ∠COE - ∠COD - ∠BOD = 360° - 140° - ∠COD - (180° - ∠COE) = 360° - 140° - ∠COD - 180° + 140° = 360° - ∠COD - 180° + 140° = 180° - ∠COD + 140° = 320° - ∠COD.
answer:320
correct answer:80
total:899  correct:232  acc:0.25806451612903225
------------------------------------------------------------------------------------
ID:899
picturePath:image/1439.png
answer:解:∵D是AC的中点,E是BC的中点,∴DE是△ABC的中位线,∴DE=\frac{1}{2}AB,∵DE=18m,∴AB=2DE=36(m)．故选:B．
answer:36
correct answer:36
total:900  correct:233  acc:0.2588888888888889
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ID:900
picturePath:image/1440.png
answer:解:∵DE是AC的垂直平分线,AC=6cm,∴DA=DC,∵△ABD的周长为13cm,∴AB+BD+AD=AB+BD+DC=AB+BC=13cm,∴△ABC的周长=AB+BC+AC=13+6=19(cm)．故选:B．
answer:19
correct answer:19
total:901  correct:234  acc:0.2597114317425083
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ID:901
picturePath:image/1443.png
answer:解:∵四边形ABCD是平行四边形,∴∠DAB=∠C,AD∥BC,∴∠DAE=∠C,∵AE平分∠DAB,∴∠DAE=∠C=\frac{1}{2}∠DAB,∵∠AED=26°,∴∠DAB=2∠AED=52°,∴∠C=52°,故选:B．
answer:52
correct answer:52
total:902  correct:235  acc:0.26053215077605324
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ID:902
picturePath:image/1444.png
answer:由AB∥CD,∠ABE=120°,∠C=25°,根据三角形的一个外角等于与它不相邻的两个内角的和,可得∠α=∠ABE-∠C=120°-25°=95°.
answer:95
correct answer:85
total:903  correct:235  acc:0.2602436323366556
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ID:903
picturePath:image/1445.png
answer:The answer is 20.
answer:20
correct answer:20
total:904  correct:236  acc:0.2610619469026549
------------------------------------------------------------------------------------
ID:904
picturePath:image/1446.png
answer:60
answer:60
correct answer:60
total:905  correct:237  acc:0.261878453038674
------------------------------------------------------------------------------------
ID:905
picturePath:image/1447.png
answer:The answer is 50°.
answer:50
correct answer:30
total:906  correct:237  acc:0.26158940397350994
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ID:906
picturePath:image/1448.png
answer:解:∵位似图形由三角尺与其灯光照射下的中心投影组成,相似比为2:5,且三角尺的一边长为8cm,∴投影三角形的对应边长为8×\frac{2}{5}=3.2cm．故选:B．
answer:3.2
correct answer:20
total:907  correct:237  acc:0.2613009922822492
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ID:907
picturePath:image/1451.png
answer:解:∵AB∥CO,∠B=22°,∴∠COB=∠B=22°,∴∠A=∠COB=22°．故选:B．
answer:22
correct answer:44
total:908  correct:237  acc:0.2610132158590308
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ID:908
picturePath:image/1453.png
answer:Since AB is parallel to CD, we can use the corresponding angles postulate, which states that if two lines are parallel, then the angles formed by the transversal and each line are congruent. Therefore, ∠BAC = ∠ACD = 55°.
answer:55
correct answer:55
total:909  correct:238  acc:0.26182618261826185
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ID:909
picturePath:image/1455.png
answer:The answer is A.
ID:910
picturePath:image/1456.png
answer:解:如图,∵a∥b,∴∠1=∠3=80°,∵∠2+∠3=90°,∴∠2=90°-80°=10°．故选:B．
answer:10
correct answer:50
total:911  correct:238  acc:0.26125137211855104
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ID:911
picturePath:image/1457.png
answer:50°
answer:50
correct answer:80
total:912  correct:238  acc:0.26096491228070173
------------------------------------------------------------------------------------
ID:912
picturePath:image/1458.png
answer:根据圆的内接四边形的性质，∠BOD等于四边形ABCD的外角和，即∠BOD=360°-∠DCE=360°-70°=290°。
answer:290
correct answer:140
total:913  correct:238  acc:0.2606790799561884
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ID:913
picturePath:image/1459.png
answer:15
answer:15
correct answer:24
total:914  correct:238  acc:0.2603938730853392
------------------------------------------------------------------------------------
ID:914
picturePath:image/1460.png
answer:16+2+2=20(cm)
answer:20
correct answer:20
total:915  correct:239  acc:0.2612021857923497
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ID:915
picturePath:image/1463.png
answer:This is a geometry problem involving circles, angles, and lines. Given that point A, B, and C are on the circumference of circle O, and angle AOC is an inscribed angle, we can use the property that the measure of an inscribed angle is half the measure of its intercepted arc. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB. Since angle AOC is an inscribed angle and angle AOCB is an exterior angle, we can use the property that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. Therefore, angle AOCB = angle AOC + angle AOB.
ID:916
picturePath:image/1464.png
answer:解:如图,∵AB||CD,∴∠1=∠3=22°,∵∠D=45°,∴∠2=180°-45°-22°=113°．故选:B．
answer:113
correct answer:23
total:917  correct:239  acc:0.2606324972737186
------------------------------------------------------------------------------------
ID:917
picturePath:image/1466.png
answer:30
answer:30
correct answer:48
total:918  correct:239  acc:0.26034858387799564
------------------------------------------------------------------------------------
ID:918
picturePath:image/1468.png
answer:解:如图,连接OA,∵OA=OB,∴∠OAB=∠B=25°,∴∠OAD=∠OAB+∠B=50°,∵AD是⊙O的切线,∴∠OAD=90°,∴∠D=90°-∠OAD=90°-50°=40°．故选:B．
answer:40
correct answer:40
total:919  correct:240  acc:0.2611534276387378
------------------------------------------------------------------------------------
ID:919
picturePath:image/1470.png
answer:∠4 = 180° - ∠1 - ∠2 - ∠3 = 180° - 36° - 36° - 140° = 0°.
answer:0
correct answer:40
total:920  correct:240  acc:0.2608695652173913
------------------------------------------------------------------------------------
ID:920
picturePath:image/1471.png
answer:40°
answer:40
correct answer:65
total:921  correct:240  acc:0.26058631921824105
------------------------------------------------------------------------------------
ID:921
picturePath:image/1472.png
answer:The answer is 38.
answer:38
correct answer:22
total:922  correct:240  acc:0.2603036876355748
------------------------------------------------------------------------------------
ID:922
picturePath:image/1475.png
answer:0.5m
answer:0.5
correct answer:1
total:923  correct:240  acc:0.2600216684723727
------------------------------------------------------------------------------------
ID:923
picturePath:image/1476.png
answer:30°
answer:30
correct answer:30
total:924  correct:241  acc:0.26082251082251084
------------------------------------------------------------------------------------
ID:924
picturePath:image/1478.png
answer:解:∵AB为⊙O的直径,∴∠ACB=90°,∵∠ABD=20°,∴∠ACD=∠ABD=20°,∴∠BCD=∠ACB-∠ACD=90°-20°=70°．故选:D．
answer:70
correct answer:70
total:925  correct:242  acc:0.2616216216216216
------------------------------------------------------------------------------------
ID:925
picturePath:image/1481.png
answer:解:∵PA、PB分别与⊙O相切于点A、B,∴∠PAO=∠PBO=90°,∴∠AOB=180°-∠P=180°-80°=100°,∴∠C=\frac{1}{2}∠AOB=\frac{1}{2}×100°=50°．故选:B．
answer:50
correct answer:50
total:926  correct:243  acc:0.26241900647948163
------------------------------------------------------------------------------------
ID:926
picturePath:image/1482.png
answer:解:∵CA、CB分别与⊙O相切于A、B两点,∴∠OAC=∠OBC=90°,∵∠C=56°,∴∠BOC=360°-90°-90°-56°=124°,∴∠D=\frac{1}{2}∠BOC=62°．故选:B．
answer:62
correct answer:62
total:927  correct:244  acc:0.2632146709816613
------------------------------------------------------------------------------------
ID:927
picturePath:image/1483.png
answer:20
answer:20
correct answer:50
total:928  correct:244  acc:0.2629310344827586
------------------------------------------------------------------------------------
ID:928
picturePath:image/1484.png
answer:∠ABC = 80° + 25° = 105°.
answer:105
correct answer:55
total:929  correct:244  acc:0.26264800861141013
------------------------------------------------------------------------------------
ID:929
picturePath:image/1485.png
answer:60°
answer:60
correct answer:120
total:930  correct:244  acc:0.2623655913978495
------------------------------------------------------------------------------------
ID:930
picturePath:image/1487.png
answer:100
answer:100
correct answer:80
total:931  correct:244  acc:0.2620837808807734
------------------------------------------------------------------------------------
ID:931
picturePath:image/1489.png
answer:根据菱形的性质，对角线互相垂直平分，所以AO=OC=3cm，BO=OD=4cm，根据勾股定理，AB=√{3²+4²}=5cm，同理，AD=√{4²+3²}=5cm，BC=√{4²+3²}=5cm，CD=√{3²+4²}=5cm，所以菱形ABCD的边长为5cm。
answer:5
correct answer:5
total:932  correct:245  acc:0.2628755364806867
------------------------------------------------------------------------------------
ID:932
picturePath:image/1490.png
answer:30°
answer:30
correct answer:30
total:933  correct:246  acc:0.26366559485530544
------------------------------------------------------------------------------------
ID:933
picturePath:image/1491.png
answer:4.2米
answer:4.2
correct answer:6.5
total:934  correct:246  acc:0.2633832976445396
------------------------------------------------------------------------------------
ID:934
picturePath:image/1494.png
answer:解:∵∠OAB=35°,∴∠AOB=2∠OAB=2×35°=70°,∴∠C=\frac{1}{2}∠AOB=\frac{1}{2}×70°=35°．故选:A．
answer:35
correct answer:55
total:935  correct:246  acc:0.2631016042780749
------------------------------------------------------------------------------------
ID:935
picturePath:image/1495.png
answer:40°
answer:40
correct answer:130
total:936  correct:246  acc:0.26282051282051283
------------------------------------------------------------------------------------
ID:936
picturePath:image/1496.png
answer:解:∵∠A=28°,∠BFC=92°,∴∠B=∠BFC-∠A=92°-28°=64°,∵∠B=∠C,∴∠BDC=∠B=64°．故选:B．
answer:64
correct answer:60
total:937  correct:246  acc:0.26254002134471716
------------------------------------------------------------------------------------
ID:937
picturePath:image/1497.png
answer:65
answer:65
correct answer:120
total:938  correct:246  acc:0.2622601279317697
------------------------------------------------------------------------------------
ID:938
picturePath:image/1499.png
answer:This is a geometry problem involving an isosceles trapezoid and its diagonals. Since the diagonals of an isosceles trapezoid intersect at the midpoint of the longer base, and the diagonals are perpendicular bisectors of each other, we can conclude that triangle PFC is a 30-60-90 triangle. In such a triangle, the angle opposite the side that is twice the length of the shorter leg is 60 degrees. Therefore, angle FPC is 60 degrees.
answer:60
correct answer:55
total:939  correct:246  acc:0.26198083067092653
------------------------------------------------------------------------------------
ID:939
picturePath:image/1502.png
answer:∠A的度数等于∠CEF的度数，即∠A=110°。
answer:110
correct answer:70
total:940  correct:246  acc:0.26170212765957446
------------------------------------------------------------------------------------
ID:940
picturePath:image/1504.png
answer:40°
answer:40
correct answer:110
total:941  correct:246  acc:0.2614240170031881
------------------------------------------------------------------------------------
ID:941
picturePath:image/1505.png
answer:40°
answer:40
correct answer:80
total:942  correct:246  acc:0.2611464968152866
------------------------------------------------------------------------------------
ID:942
picturePath:image/1506.png
answer:60°
answer:60
correct answer:150
total:943  correct:246  acc:0.2608695652173913
------------------------------------------------------------------------------------
ID:943
picturePath:image/1509.png
answer:20m
answer:20
correct answer:20
total:944  correct:247  acc:0.2616525423728814
------------------------------------------------------------------------------------
ID:944
picturePath:image/1510.png
answer:60
answer:60
correct answer:20
total:945  correct:247  acc:0.2613756613756614
------------------------------------------------------------------------------------
ID:945
picturePath:image/1514.png
answer:30°
answer:30
correct answer:15
total:946  correct:247  acc:0.26109936575052856
------------------------------------------------------------------------------------
ID:946
picturePath:image/1516.png
answer:The answer is D.
ID:947
picturePath:image/1518.png
answer:100
answer:100
correct answer:80
total:948  correct:247  acc:0.26054852320675104
------------------------------------------------------------------------------------
ID:948
picturePath:image/1520.png
answer:Since DE is perpendicular to AB, we know that ∠AED is a right angle, so ∠AED = 90°. Since the sum of the angles in a triangle is 180°, we can find ∠ADE by subtracting ∠AED and ∠EDA from 180°: 180° - 90° - 35° = 55°. Since opposite angles in a parallelogram are congruent, we know that ∠C is congruent to ∠ADE, so ∠C = 55°.
answer:55
correct answer:55
total:949  correct:248  acc:0.2613277133825079
------------------------------------------------------------------------------------
ID:949
picturePath:image/1521.png
answer:100°
answer:100
correct answer:20
total:950  correct:248  acc:0.26105263157894737
------------------------------------------------------------------------------------
ID:950
picturePath:image/1522.png
answer:65
answer:65
correct answer:65
total:951  correct:249  acc:0.2618296529968454
------------------------------------------------------------------------------------
ID:951
picturePath:image/1523.png
answer:The answer is 35°.
answer:35
correct answer:55
total:952  correct:249  acc:0.2615546218487395
------------------------------------------------------------------------------------
ID:952
picturePath:image/1524.png
answer:根据题意，AB∥CD，∠ACD=40°，则∠ABC=∠ACD=40°。又因为AE平分∠CAB，所以∠CAE=∠ABC/2=20°。在△AEC中，∠AEC=180°-∠A-∠CAE=180°-20°-20°=140°。故选B。
answer:140
correct answer:70
total:953  correct:249  acc:0.26128016789087094
------------------------------------------------------------------------------------
ID:953
picturePath:image/1525.png
answer:54.
answer:54
correct answer:36
total:954  correct:249  acc:0.2610062893081761
------------------------------------------------------------------------------------
ID:954
picturePath:image/1526.png
answer:3,200米
answer:200
correct answer:2000
total:955  correct:249  acc:0.2607329842931937
------------------------------------------------------------------------------------
ID:955
picturePath:image/1528.png
answer:The answer is 65.
answer:65
correct answer:105
total:956  correct:249  acc:0.2604602510460251
------------------------------------------------------------------------------------
ID:956
picturePath:image/1530.png
answer:解:∵AB是⊙O直径,∴∠BOC=180°,∵∠AOC=130°,∴∠BOA=180°-130°=50°,∴∠D=\frac{1}{2}∠BOA=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:25
total:957  correct:250  acc:0.2612330198537095
------------------------------------------------------------------------------------
ID:957
picturePath:image/1531.png
answer:解:∵四边形ABCD是菱形,∴AC=BD=6,∴S~△ABD~=\frac{1}{2}AC•BD=\frac{1}{2}×6×6=18．故选:A．
answer:18
correct answer:12
total:958  correct:250  acc:0.2609603340292276
------------------------------------------------------------------------------------
ID:958
picturePath:image/1532.png
answer:根据全等三角形的性质,∠ABO=∠D=80°,∠BOC=∠DOC=70°,所以∠B=180°-80°-70°=30°.
answer:30
correct answer:30
total:959  correct:251  acc:0.26173096976016685
------------------------------------------------------------------------------------
ID:959
picturePath:image/1533.png
answer:50°
answer:50
correct answer:45
total:960  correct:251  acc:0.26145833333333335
------------------------------------------------------------------------------------
ID:960
picturePath:image/1534.png
answer:60°
answer:60
correct answer:30
total:961  correct:251  acc:0.2611862643080125
------------------------------------------------------------------------------------
ID:961
picturePath:image/1535.png
answer:The answer is D.
ID:962
picturePath:image/1536.png
answer:∠A = ∠C (对顶角) - ∠F (余角) = 70° - 30° = 40°.
answer:40
correct answer:40
total:963  correct:252  acc:0.2616822429906542
------------------------------------------------------------------------------------
ID:963
picturePath:image/1537.png
answer:30°
answer:30
correct answer:150
total:964  correct:252  acc:0.26141078838174275
------------------------------------------------------------------------------------
ID:964
picturePath:image/1538.png
answer:1.5米
answer:1.5
correct answer:2
total:965  correct:252  acc:0.261139896373057
------------------------------------------------------------------------------------
ID:965
picturePath:image/1540.png
answer:解:如图,∵AB||CD,∴∠1=∠3=48°,∵∠2+∠3+∠4=90°,∠4=60°,∴∠2=90°-∠3=90°-48°=42°,故选:B．
answer:42
correct answer:102
total:966  correct:252  acc:0.2608695652173913
------------------------------------------------------------------------------------
ID:966
picturePath:image/1541.png
answer:∠2 = ∠1 = 110°.
answer:110
correct answer:70
total:967  correct:252  acc:0.26059979317476734
------------------------------------------------------------------------------------
ID:967
picturePath:image/1543.png
answer:由DE∥BC,可得∠DEC=∠ACF=140°,∴∠AED=180°-140°=40°,∵∠ADE=105°,∴∠A=180°-105°-40°=35°.故选B.
answer:35
correct answer:35
total:968  correct:253  acc:0.26136363636363635
------------------------------------------------------------------------------------
ID:968
picturePath:image/1545.png
answer:The answer is A.
ID:969
picturePath:image/1547.png
answer:∠2 = ∠BAC - ∠1 = 78° - 42° = 36°.
answer:36
correct answer:60
total:970  correct:253  acc:0.26082474226804125
------------------------------------------------------------------------------------
ID:970
picturePath:image/1548.png
answer:解:∵四边形ABCD是平行四边形,∴AB=CD,AD=BC,OA=OC,∵□ABCD的周长为16cm,∴AB+BC=8cm,∵OE⊥AC,∴AE=CE,∴△DCE的周长为:DE+CE+CD=DE+AE+CD=AD+CD=AB+BC=8cm．故选:B．
answer:8
correct answer:8
total:971  correct:254  acc:0.2615859938208033
------------------------------------------------------------------------------------
ID:971
picturePath:image/1549.png
answer:根据圆周角定理，同弧所对的圆周角相等，因此∠AOC=2∠ABC。将∠ABC=70°代入，得到∠AOC=2×70°=140°。所以，∠AOC的度数等于140°。
answer:140
correct answer:140
total:972  correct:255  acc:0.2623456790123457
------------------------------------------------------------------------------------
ID:972
picturePath:image/1552.png
answer:60
answer:60
correct answer:25
total:973  correct:255  acc:0.2620760534429599
------------------------------------------------------------------------------------
ID:973
picturePath:image/1553.png
answer:当双翼收起时,可以通过闸机的物体的最大宽度是双翼边缘AC和BD之间的距离,即54cm-10cm=44cm.
answer:44
correct answer:64
total:974  correct:255  acc:0.2618069815195072
------------------------------------------------------------------------------------
ID:974
picturePath:image/1554.png
answer:解:∵圆锥的底面半径r=6,高h=8,∴母线l=√{h^{2}+r^{2}}=√{8^{2}+6^{2}}=10,∴圆锥的侧面积=π×6×10=60π．故选:A．
answer:60
correct answer:60
total:975  correct:256  acc:0.26256410256410256
------------------------------------------------------------------------------------
ID:975
picturePath:image/1555.png
answer:7
answer:7
correct answer:16
total:976  correct:256  acc:0.26229508196721313
------------------------------------------------------------------------------------
ID:976
picturePath:image/1556.png
answer:根据题意,菱形ABCD的对角线AC和BD相交于点O,且点M是边AB的中点,则OM是△ABC的中位线,∴OM=\frac{1}{2}AC,∵OM=5cm,∴AC=2OM=10cm,∴菱形ABCD的周长=4AC=40cm.
answer:40
correct answer:40
total:977  correct:257  acc:0.263050153531218
------------------------------------------------------------------------------------
ID:977
picturePath:image/1558.png
answer:The answer is 40.
answer:40
correct answer:20
total:978  correct:257  acc:0.26278118609406953
------------------------------------------------------------------------------------
ID:978
picturePath:image/1559.png
answer:∠2 = 180° - ∠1 = 180° - 115° = 65°.
answer:65
correct answer:65
total:979  correct:258  acc:0.2635342185903984
------------------------------------------------------------------------------------
ID:979
picturePath:image/1561.png
answer:30
answer:30
correct answer:48
total:980  correct:258  acc:0.26326530612244897
------------------------------------------------------------------------------------
ID:980
picturePath:image/1562.png
answer:解:∵BD、CE是△ABC的中线,∴AD=BD,AE=CE,∵点F、G分别是OB、OC的中点,∴FG=\frac{1}{2}OC,FG=\frac{1}{2}OB,∴FG=\frac{1}{2}AD,FG=\frac{1}{2}AE,∴FG=\frac{1}{2}AD=\frac{1}{2}AE,∴AD=AE,∵AO=3cm,BC=4cm,∴四边形DEFG的周长是:DE+EF+FG+GE=DE+EF+FG+AE=DE+AE+EF+FG=AD+BC=3+4=7cm．故选:A．
answer:7
correct answer:7
total:981  correct:259  acc:0.2640163098878695
------------------------------------------------------------------------------------
ID:981
picturePath:image/1563.png
answer:Since line a is parallel to line b, the corresponding angles ∠1 and ∠2 are congruent. Therefore, the degree of ∠2 is also 55°.
answer:55
correct answer:125
total:982  correct:259  acc:0.2637474541751528
------------------------------------------------------------------------------------
ID:982
picturePath:image/1564.png
answer:20
answer:20
correct answer:13
total:983  correct:259  acc:0.2634791454730417
------------------------------------------------------------------------------------
ID:983
picturePath:image/1566.png
answer:解:∵a∥b,∠1=36°,∴∠3=∠1=36°,∵AB⊥BC,∴∠2+∠3=90°,∴∠2=90°-36°=54°．故选:B．
answer:54
correct answer:54
total:984  correct:260  acc:0.26422764227642276
------------------------------------------------------------------------------------
ID:984
picturePath:image/1568.png
answer:根据线平行，内错角相等，可得∠BEC=∠DCE=144°,故选C
answer:144
correct answer:34
total:985  correct:260  acc:0.2639593908629442
------------------------------------------------------------------------------------
ID:985
picturePath:image/1569.png
answer:The answer is 50.
answer:50
correct answer:130
total:986  correct:260  acc:0.26369168356997974
------------------------------------------------------------------------------------
ID:986
picturePath:image/1570.png
answer:根据平行四边形的性质，对角相等，即∠A=∠C。已知∠A+∠C=100°，所以∠A=∠C=50°。在平行四边形中，两条对角线互相平分，所以∠B=180°-2∠A=180°-2×50°=80°。所以，∠B的度数是80°。
answer:80
correct answer:130
total:987  correct:260  acc:0.2634245187436677
------------------------------------------------------------------------------------
ID:987
picturePath:image/1571.png
answer:1.5
answer:1.5
correct answer:23
total:988  correct:260  acc:0.2631578947368421
------------------------------------------------------------------------------------
ID:988
picturePath:image/1576.png
answer:Since \(AB \parallel CD\), we have \(\angle B = \angle DCE\) by alternate interior angles. Since \(CD = CE\), we also have \(\angle DCE = \angle DEC\) by vertical angles. Therefore, \(\angle B = \angle DEC\). Since the sum of the angles in a triangle is \(180^\circ\), we have \(\angle B + \angle DEC + \angle CDE = 180^\circ\). Substituting \(\angle B = \angle DEC\) and \(\angle CDE = 75^\circ\), we have \(\angle B + \angle B + 75^\circ = 180^\circ\). Simplifying, we have \(\angle B = \frac{180^\circ - 75^\circ}{2} = \boxed{57^\circ}\). The answer is: 57
answer:57
correct answer:30
total:989  correct:260  acc:0.26289180990899896
------------------------------------------------------------------------------------
ID:989
picturePath:image/1579.png
answer:60°
answer:60
correct answer:120
total:990  correct:260  acc:0.26262626262626265
------------------------------------------------------------------------------------
ID:990
picturePath:image/1581.png
answer:The answer is 75°.
answer:75
correct answer:75
total:991  correct:261  acc:0.26337033299697277
------------------------------------------------------------------------------------
ID:991
picturePath:image/1582.png
answer:解:∵AE=10,BE=24,∴AB=√{A^{2}+B^{2}}=√{10^{2}+24^{2}}=26,∵此图是由四个全等的直角三角形拼接而成,∴EF=\frac{1}{2}AB=13．故选:B．
answer:13
correct answer:14
total:992  correct:261  acc:0.26310483870967744
------------------------------------------------------------------------------------
ID:992
picturePath:image/1585.png
answer:40°
answer:40
correct answer:90
total:993  correct:261  acc:0.2628398791540785
------------------------------------------------------------------------------------
ID:993
picturePath:image/1586.png
answer:If m is parallel to n and ∠1 is 105°, then ∠2 is also 105° because corresponding angles are congruent when two lines are parallel.
answer:105
correct answer:75
total:994  correct:261  acc:0.2625754527162978
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ID:994
picturePath:image/1589.png
answer:解:如图,连接OA、OB,∵PA、PB分别与⊙O相切于点A、B,∴∠PAO=∠PBO=90°,∴∠AOB=360°-90°-90°-50°=130°,∴∠C=\frac{1}{2}∠AOB=\frac{1}{2}×130°=65°．故选:B．
answer:65
correct answer:65
total:995  correct:262  acc:0.2633165829145729
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ID:995
picturePath:image/1590.png
answer:根据题意,DE是AC边的垂直平分线,则AE=CE,又∵BC=8厘米,AB=10厘米,∴△EBC的周长=BC+CE+EB=BC+AE+EB=BC+AC+EB=BC+AB+EB=BC+AB+AC=8+10+8=26(厘米).故选B.
answer:26
correct answer:18
total:996  correct:262  acc:0.26305220883534136
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ID:996
picturePath:image/1591.png
answer:Since line a is parallel to line b, we have \angle BCD = \angle 1 = 145^\circ. Since AB = AC, we have \angle B = \angle C = \frac{1}{2}(180^\circ - \angle A) = \frac{1}{2}(180^\circ - 30^\circ) = 75^\circ. Therefore, \angle 2 = 180^\circ - \angle BCD - \angle B = 180^\circ - 145^\circ - 75^\circ = \boxed{60^\circ}.The answer is: 60
answer:60
correct answer:40
total:997  correct:262  acc:0.26278836509528586
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ID:997
picturePath:image/1593.png
answer:∠AED = ∠C (对顶角相等)；∠AED = 180°-∠CAB (两直线平行,同旁内角互补)；∠CAB = ∠BAC + ∠C (三角形的一个外角等于和它不相邻的两个内角之和)；∠BAC = ∠AED - ∠C (两直线平行,同旁内角互补)；∠AED = 70° (已知)；∠C = 70° (已知)；∠BAC = ∠AED - ∠C = 70° - 70° = 0° (计算)；∠CAB = ∠BAC + ∠C = 0° + 70° = 70° (计算)；∠AED = 180°-∠CAB = 180°-70° = 110° (计算)；故选B.
answer:110
correct answer:125
total:998  correct:262  acc:0.2625250501002004
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ID:998
picturePath:image/1594.png
answer:解:∵四边形ABCD是平行四边形,∴AB=DC,AD=BC,OA=OC,∵▱ABCD的周长为32cm,∴AD+DC=16cm,∵OE⊥AC,∴AE=CE,∴△DCE的周长为:DE+CE+CD=DE+AE+CD=AD+DC=16cm．故选:B．
answer:16
correct answer:16
total:999  correct:263  acc:0.26326326326326327
------------------------------------------------------------------------------------
ID:999
picturePath:image/1595.png
answer:根据平行四边形的性质，对角线互相平分，∠BAD的平分线交BC于点E，∠ABC的平分线交AD于点F，连接EF，则AE=BE，AB=CD，BF=CF，∵BF=12，AB=10，∴CF=12，CD=10，∴BE=CF-BF=12-12=0，∴AE=BE=0，故选：B．
answer:0
correct answer:16
total:1000  correct:263  acc:0.263
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ID:1000
picturePath:image/1596.png
answer:The answer is 14.
answer:14
correct answer:11
total:1001  correct:263  acc:0.2627372627372627
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ID:1001
picturePath:image/1602.png
answer:The answer is A.
ID:1002
picturePath:image/1603.png
answer:76
answer:76
correct answer:76
total:1003  correct:264  acc:0.26321036889332006
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ID:1003
picturePath:image/1605.png
answer:20
answer:20
correct answer:15
total:1004  correct:264  acc:0.26294820717131473
------------------------------------------------------------------------------------
ID:1004
picturePath:image/1607.png
answer:64°
answer:64
correct answer:58
total:1005  correct:264  acc:0.2626865671641791
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ID:1005
picturePath:image/1608.png
answer:解:∵a∥b,∴∠1=∠2=65°,∵∠1+∠2+∠3=180°,∴∠3=180°-∠1-∠2=180°-65°-65°=50°,故选:B．
answer:50
correct answer:105
total:1006  correct:264  acc:0.2624254473161034
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ID:1006
picturePath:image/1609.png
answer:50°
answer:50
correct answer:25
total:1007  correct:264  acc:0.26216484607745777
------------------------------------------------------------------------------------
ID:1007
picturePath:image/1610.png
answer:解:∵BF平分∠ABC,CF平分∠ACB,∠BFC=115°,∴∠ABC+∠ACB=2∠BFC=230°,∴∠A=180°-(∠ABC+∠ACB)=180°-230°=50°．故选:A．
answer:50
correct answer:50
total:1008  correct:265  acc:0.2628968253968254
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ID:1008
picturePath:image/1611.png
answer:60°
answer:60
correct answer:60
total:1009  correct:266  acc:0.26362735381565905
------------------------------------------------------------------------------------
ID:1009
picturePath:image/1612.png
answer:10\sqrt{3}m.
answer:3
correct answer:13
total:1010  correct:266  acc:0.2633663366336634
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ID:1010
picturePath:image/1613.png
answer:90°-30°=60°
answer:60
correct answer:60
total:1011  correct:267  acc:0.26409495548961426
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ID:1011
picturePath:image/1614.png
answer:根据题意,点O到AB、BC、AC的距离相等,则∠BOC=\frac{1}{2}(180°-∠A)=\frac{1}{2}(180°-60°)=60°.故选B.
answer:60
correct answer:120
total:1012  correct:267  acc:0.26383399209486164
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ID:1012
picturePath:image/1615.png
answer:解:∵四边形ABCD是平行四边形,∴AD∥BC,∴∠A+∠C=180°,∵∠A=125°,∴∠C=55°,∵CE⊥AB,∴在Rt△CBE中,∠BCE=180°-90°-55°=35°．故选:B．
answer:35
correct answer:35
total:1013  correct:268  acc:0.26456071076011844
------------------------------------------------------------------------------------
ID:1013
picturePath:image/1616.png
answer:The answer is A.
ID:1014
picturePath:image/1617.png
answer:20
answer:20
correct answer:35
total:1015  correct:268  acc:0.26403940886699506
------------------------------------------------------------------------------------
ID:1015
picturePath:image/1618.png
answer:The answer is 50.
answer:50
correct answer:20
total:1016  correct:268  acc:0.2637795275590551
------------------------------------------------------------------------------------
ID:1016
picturePath:image/1619.png
answer:∵b∥c,a⊥b,∴a⊥c,∴∠1+∠2=90°,∵∠1=130°,∴∠2=90°-∠1=90°-130°=-40°,∵∠2为三角形的外角,∴∠2=-40°+180°=140°,故选B.
answer:140
correct answer:40
total:1017  correct:268  acc:0.2635201573254671
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ID:1017
picturePath:image/1622.png
answer:24
answer:24
correct answer:12
total:1018  correct:268  acc:0.2632612966601179
------------------------------------------------------------------------------------
ID:1018
picturePath:image/1626.png
answer:40
answer:40
correct answer:25
total:1019  correct:268  acc:0.26300294406280667
------------------------------------------------------------------------------------
ID:1019
picturePath:image/1627.png
answer:根据矩形的性质，对角线互相平分，所以∠AOB=∠BOC=55°。又因为四边形ABCD是矩形，所以∠AOD=∠BOC=55°。故选B。
answer:55
correct answer:110
total:1020  correct:268  acc:0.2627450980392157
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ID:1020
picturePath:image/1629.png
answer:根据勾股定理，在Rt△ABC中，AB²+BC²=AC²，即20²+BC²=90²，解得BC=60米；在Rt△BDE中，BD²+DE²=BE²，即30²+DE²=60²，解得DE=30米。因此，河宽BC为60米。
answer:60
correct answer:40
total:1021  correct:268  acc:0.2624877571008815
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ID:1021
picturePath:image/1632.png
answer:75
answer:75
correct answer:35
total:1022  correct:268  acc:0.2622309197651663
------------------------------------------------------------------------------------
ID:1022
picturePath:image/1633.png
answer:30
answer:30
correct answer:70
total:1023  correct:268  acc:0.2619745845552297
------------------------------------------------------------------------------------
ID:1023
picturePath:image/1634.png
answer:The answer is 52°.
answer:52
correct answer:38
total:1024  correct:268  acc:0.26171875
------------------------------------------------------------------------------------
ID:1024
picturePath:image/1636.png
answer:40
answer:40
correct answer:40
total:1025  correct:269  acc:0.2624390243902439
------------------------------------------------------------------------------------
ID:1025
picturePath:image/1638.png
answer:解:∵∠AOB=100°,∴∠ACB=\frac{1}{2}∠AOB=50°．故选:B．
answer:50
correct answer:130
total:1026  correct:269  acc:0.2621832358674464
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ID:1026
picturePath:image/1639.png
answer:5
answer:5
correct answer:3
total:1027  correct:269  acc:0.2619279454722493
------------------------------------------------------------------------------------
ID:1027
picturePath:image/1641.png
answer:30°
answer:30
correct answer:30
total:1028  correct:270  acc:0.26264591439688717
------------------------------------------------------------------------------------
ID:1028
picturePath:image/1642.png
answer:The answer is 62°.
answer:62
correct answer:118
total:1029  correct:270  acc:0.26239067055393583
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ID:1029
picturePath:image/1644.png
answer:解:∵△ADE∽△ABC,∴\frac{AD}{AB}=\frac{DE}{BC},∵AD=1,BD=2,∴\frac{1}{1+2}=\frac{1}{3},∴△ADE与△ABC的相似比为1:3．故选:A．
answer:3
correct answer:3
total:1030  correct:271  acc:0.26310679611650484
------------------------------------------------------------------------------------
ID:1030
picturePath:image/1647.png
answer:根据平行线的性质，∠B和∠E是对顶角，所以∠B=∠E=45°。又因为BE交CD于点F，所以∠D=∠B=45°。故选A。
answer:45
correct answer:24
total:1031  correct:271  acc:0.26285160038797284
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ID:1031
picturePath:image/1649.png
answer:根据三角形内角和定理，∠ACD = 180° - ∠A - ∠B = 180° - 62° - 38° = 80°.
answer:80
correct answer:100
total:1032  correct:271  acc:0.2625968992248062
------------------------------------------------------------------------------------
ID:1032
picturePath:image/1650.png
answer:100
answer:100
correct answer:70
total:1033  correct:271  acc:0.2623426911907067
------------------------------------------------------------------------------------
ID:1033
picturePath:image/1651.png
answer:4
answer:4
correct answer:5
total:1034  correct:271  acc:0.2620889748549323
------------------------------------------------------------------------------------
ID:1034
picturePath:image/1653.png
answer:解:∵OB=OC,∠OCB=40°,∴∠OBC=∠OCB=40°,∴∠BOC=180°-∠OBC-∠OCB=100°,∴∠A=\frac{1}{2}∠BOC=50°．故选:B．
answer:50
correct answer:50
total:1035  correct:272  acc:0.26280193236714977
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ID:1035
picturePath:image/1654.png
answer:解:∵AD是⊙O的直径,∴∠ACD=90°,∵∠ABC=30°,∴∠CAD=90°-∠ABC=90°-30°=60°．故选:D．
answer:60
correct answer:60
total:1036  correct:273  acc:0.2635135135135135
------------------------------------------------------------------------------------
ID:1036
picturePath:image/1656.png
answer:The answer is \frac{2\sqrt{3}}{3}
answer:3
correct answer:3
total:1037  correct:274  acc:0.26422372227579555
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ID:1037
picturePath:image/1657.png
answer:解:∵∠BAC与∠ABC的平分线相交于点I,∴∠IAB=∠IAC,∠ICB=∠IBC,∵∠BAC=120°,∴∠IAB+∠IBC=60°,∵AB=BC,∴⁀{IB}=⁀{IC},∴∠IBD=∠ICD,∴△IBD≌△ICD(SAS),∴BD=CD,∵DI是△ABD的中位线,∴DI=\frac{1}{2}AB,∵AB=BC,∴DI=\frac{1}{2}BC,∴DI是△ABC的中位线,∴DI=\frac{1}{2}AC,∴DI=\frac{1}{2}AB=\frac{1}{2}BC=\frac{1}{2}AC,∵AB=BC=AC,∴DI=\frac{1}{2}AB=\frac{1}{2}BC=\frac{1}{2}AC=\frac{1}{2}BC=\frac{1}{2}AC=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{1}{2}AB=\frac{
answer:2
correct answer:3
total:1038  correct:274  acc:0.26396917148362237
------------------------------------------------------------------------------------
ID:1038
picturePath:image/1660.png
answer:解:∵点D、E分别是⊙O的内接正三角形ABC的AB、AC边上的中点,⊙O的半径为2,∴DE是△ABC的中位线,∴DE=\frac{1}{2}BC,∵△ABC是等边三角形,∴BC=3,∴DE=\frac{1}{2}BC=1.5．故选:A．
answer:1.5
correct answer:3
total:1039  correct:274  acc:0.26371511068334935
------------------------------------------------------------------------------------
ID:1039
picturePath:image/1661.png
answer:20
answer:20
correct answer:20
total:1040  correct:275  acc:0.2644230769230769
------------------------------------------------------------------------------------
ID:1040
picturePath:image/1662.png
answer:解:∵∠BAC=60°,∴∠BOC=2∠BAC=120°,∵OB=OC,∴△OBC是等边三角形,∴BC=OB=2．故选:A．
answer:2
correct answer:3
total:1041  correct:275  acc:0.2641690682036503
------------------------------------------------------------------------------------
ID:1041
picturePath:image/1663.png
answer:The answer is 45°.
answer:45
correct answer:45
total:1042  correct:276  acc:0.2648752399232246
------------------------------------------------------------------------------------
ID:1042
picturePath:image/1665.png
answer:40
answer:40
correct answer:60
total:1043  correct:276  acc:0.26462128475551294
------------------------------------------------------------------------------------
ID:1043
picturePath:image/1666.png
answer:解:∵弧CB=弧BD,∴∠CAB=∠DBA=24°,∵AB是⊙O的直径,∴∠ADB=90°,∴∠ABD=90°-∠DBA=90°-24°=66°．故选:D．
answer:66
correct answer:66
total:1044  correct:277  acc:0.2653256704980843
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ID:1044
picturePath:image/1668.png
answer:70
answer:70
correct answer:75
total:1045  correct:277  acc:0.26507177033492824
------------------------------------------------------------------------------------
ID:1045
picturePath:image/1669.png
answer:根据题意，∵⊙O与△ABC三边均相交,在三边上截得的线段DE=FG=HK,∠A=50°,∴∠BOC=2∠A=100°．故选：B．
answer:100
correct answer:115
total:1046  correct:277  acc:0.26481835564053535
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ID:1046
picturePath:image/1670.png
answer:解:∵游乐园的大观览车半径为25米,已知观览车绕圆心O顺时针做匀速运动,旋转一周用12分钟,∴观览车每秒旋转\frac{360°}{12}=\frac{30°}{1}°,∴4分钟后,观览车旋转了4×\frac{30°}{1}=120°,∴经过4分钟后,此人距地面CD的高度是:25×2=50(米)．故选:A．
answer:50
correct answer:2
total:1047  correct:277  acc:0.26456542502387775
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ID:1047
picturePath:image/1672.png
answer:20
answer:20
correct answer:50
total:1048  correct:277  acc:0.2643129770992366
------------------------------------------------------------------------------------
ID:1048
picturePath:image/1673.png
answer:解:∵直线l交圆O于A、B两点,且将圆O分成3:1两段,∴∠AOB=180°×\frac{3}{4}=135°,∵OA=OB=2cm,∴△OAB是等腰三角形,∴AB=OA=2cm,∴S△OAB=\frac{1}{2}×2×2=2cm²．故选:A．
answer:2
correct answer:2
total:1049  correct:278  acc:0.2650142993326978
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ID:1049
picturePath:image/1674.png
answer:100°
answer:100
correct answer:80
total:1050  correct:278  acc:0.26476190476190475
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ID:1050
picturePath:image/1675.png
answer:解:∵⁀{BC}=⁀{CD}=⁀{DE},∠BOC=40°,∴∠BOE=∠COE=\frac{1}{2}∠BOC=20°,∵AB是⊙O的直径,∴∠AOE=90°,∴∠AOE=90°-20°=70°．故选:B．
answer:70
correct answer:60
total:1051  correct:278  acc:0.26450999048525214
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ID:1051
picturePath:image/1678.png
answer:The answer is 74.
answer:74
correct answer:106
total:1052  correct:278  acc:0.26425855513307983
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ID:1052
picturePath:image/1680.png
answer:100°
answer:100
correct answer:115
total:1053  correct:278  acc:0.26400759734093066
------------------------------------------------------------------------------------
ID:1053
picturePath:image/1683.png
answer:解:∵半径OD⊥弦BC,∴BD=CD,∵∠BAC=60°,∴∠BOD=30°,∵⊙O的半径为2,∴OD=\frac{1}{2}OB=1．故选:A．
answer:1
correct answer:1
total:1054  correct:279  acc:0.2647058823529412
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ID:1054
picturePath:image/1684.png
answer:25
answer:25
correct answer:40
total:1055  correct:279  acc:0.26445497630331755
------------------------------------------------------------------------------------
ID:1055
picturePath:image/1686.png
answer:∠AOD = ∠AOC/2 = 70°/2 = 35°.
answer:35
correct answer:40
total:1056  correct:279  acc:0.26420454545454547
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ID:1056
picturePath:image/1687.png
answer:根据圆周角定理，同弧所对的圆周角的度数相等，所以∠AOB=2∠A=60°．故选C．
answer:60
correct answer:120
total:1057  correct:279  acc:0.2639545884578997
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ID:1057
picturePath:image/1689.png
answer:The answer is 60.
answer:60
correct answer:120
total:1058  correct:279  acc:0.2637051039697543
------------------------------------------------------------------------------------
ID:1058
picturePath:image/1691.png
answer:5.
answer:5
correct answer:1
total:1059  correct:279  acc:0.26345609065155806
------------------------------------------------------------------------------------
ID:1059
picturePath:image/1693.png
answer:根据圆周角定理，同弧所对的圆周角相等，⁀{BD}=⁀{CD}，则∠BOD=∠COD=60°，∴∠AOC=2∠COD=120°．故选：D．
answer:120
correct answer:60
total:1060  correct:279  acc:0.2632075471698113
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ID:1060
picturePath:image/1694.png
answer:20°
answer:20
correct answer:50
total:1061  correct:279  acc:0.2629594721960415
------------------------------------------------------------------------------------
ID:1061
picturePath:image/1695.png
answer:50°
answer:50
correct answer:50
total:1062  correct:280  acc:0.263653483992467
------------------------------------------------------------------------------------
ID:1062
picturePath:image/1697.png
answer:40°
answer:40
correct answer:40
total:1063  correct:281  acc:0.264346190028222
------------------------------------------------------------------------------------
ID:1063
picturePath:image/1700.png
answer:50°
answer:50
correct answer:50
total:1064  correct:282  acc:0.2650375939849624
------------------------------------------------------------------------------------
ID:1064
picturePath:image/1702.png
answer:The answer is A.
ID:1065
picturePath:image/1704.png
answer:2\sqrt{3}-2
answer:2
correct answer:1
total:1066  correct:282  acc:0.26454033771106944
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ID:1066
picturePath:image/1705.png
answer:8√3
answer:3
correct answer:2
total:1067  correct:282  acc:0.2642924086223055
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ID:1067
picturePath:image/1706.png
answer:20°
answer:20
correct answer:3
total:1068  correct:282  acc:0.2640449438202247
------------------------------------------------------------------------------------
ID:1068
picturePath:image/1707.png
answer:52
answer:52
correct answer:37
total:1069  correct:282  acc:0.2637979420018709
------------------------------------------------------------------------------------
ID:1069
picturePath:image/1708.png
answer:This is a geometry problem involving a circle, a point outside the circle, and a line that intersects the circle at two points. The given information includes the radius of the circle, the distance from the point outside the circle to the center of the circle, and the fact that the line intersects the circle at a right angle. The question asks for the minimum value of the sum of the distances from points A and B to the center of the circle.

To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is the line segment connecting points A and B, and the other two sides are the distances from points A and B to the center of the circle.

Since the line intersects the circle at a right angle, the line segment connecting points A and B is the hypotenuse of a right triangle with one leg being the radius of the circle (2) and the other leg being the distance from the point outside the circle to the center of the circle (3.5). Applying the Pythagorean theorem, we have:

AB^2 = 2^2 + 3.5^2
AB^2 = 4 + 12.25
AB^2 = 16.25

Taking the square root of both sides, we get:

AB = √16.25
AB = 4√2

Therefore, the minimum value of the sum of the distances from points A and B to the center of the circle is 4√2.
answer:2
correct answer:3
total:1070  correct:282  acc:0.2635514018691589
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ID:1070
picturePath:image/1709.png
answer:90°-30°=60°, 故选:D．
answer:60
correct answer:60
total:1071  correct:283  acc:0.2642390289449113
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ID:1071
picturePath:image/1710.png
answer:根据圆周角定理，圆周角的度数等于它所对的弧的度数的一半。已知∠AOB是⊙O的圆心角，∠AOB=60°，则∠ACB的度数是60°÷2=30°。故选B。
answer:30
correct answer:30
total:1072  correct:284  acc:0.26492537313432835
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ID:1072
picturePath:image/1711.png
answer:∠ABC=\frac{1}{2}∠AOC=\frac{1}{2}×72°=36°.
answer:36
correct answer:36
total:1073  correct:285  acc:0.2656104380242311
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ID:1073
picturePath:image/1712.png
answer:110°
answer:110
correct answer:110
total:1074  correct:286  acc:0.26629422718808193
------------------------------------------------------------------------------------
ID:1074
picturePath:image/1713.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵AB=5√{3},∴AC=5,在Rt△ACB中,根据勾股定理,得:
BC=√{AB^{2}-AC^{2}}=√{(5√{3})^{2}-5^{2}}=5,
则∠ACB=∠A=90°-∠A=90°-60°=30°．
故选:B．
answer:30
correct answer:60
total:1075  correct:286  acc:0.266046511627907
------------------------------------------------------------------------------------
ID:1075
picturePath:image/1714.png
answer:根据圆周角定理，在同圆或等圆中，同弧或等弧所对的圆周角相等。因此，∠BOC = ∠ABO + ∠ACO = 22° + 42° = 64°。
answer:64
correct answer:128
total:1076  correct:286  acc:0.26579925650557623
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ID:1076
picturePath:image/1715.png
answer:解:∵∠AOC=140°,∴∠BOC=180°-140°=40°,∴∠BAC=\frac{1}{2}∠BOC=\frac{1}{2}×40°=20°．故选:A．
answer:20
correct answer:20
total:1077  correct:287  acc:0.26648096564531104
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ID:1077
picturePath:image/1716.png
answer:解:∵A、B、C是⊙O上三点,∠ACB=25°,∴∠AOB=2∠ACB=50°,∵AO=BO,∴∠BAO=∠ABO=\frac{180°-50°}{2}=65°．故选:A．
answer:65
correct answer:65
total:1078  correct:288  acc:0.26716141001855287
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ID:1078
picturePath:image/1719.png
answer:4
answer:4
correct answer:3
total:1079  correct:288  acc:0.2669138090824838
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ID:1079
picturePath:image/1721.png
answer:Since CD is perpendicular to AB, we know that ∠ADC = 90°. Since ∠CAB = 20°, we can find ∠AOC by subtracting ∠CAB from 180° (since the angles on a diameter add up to 180°): ∠AOC = 180° - ∠CAB = 180° - 20° = 160°. Finally, we can find ∠AOD by subtracting ∠AOC from 360° (since the angles on a circle add up to 360°): ∠AOD = 360° - ∠AOC = 360° - 160° = 200°. Therefore, the answer is B.
answer:200
correct answer:140
total:1080  correct:288  acc:0.26666666666666666
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ID:1080
picturePath:image/1722.png
answer:70
answer:70
correct answer:70
total:1081  correct:289  acc:0.26734505087881594
------------------------------------------------------------------------------------
ID:1081
picturePath:image/1723.png
answer:Since \( \overline{AB} \) is the diameter, \( \angle ACB = 90^\circ \). Given that \( \angle CAB = 23^\circ \), we can find \( \angle ABC \) by subtracting \( \angle CAB \) from \( 90^\circ \): \( 90^\circ - 23^\circ = \boxed{67^\circ} \).
answer:67
correct answer:67
total:1082  correct:290  acc:0.2680221811460259
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ID:1082
picturePath:image/1724.png
answer:根据圆周角定理，在同圆或等圆中，圆周角的度数等于它所对的圆心角的一半。因此，如果∠CAB是50°，那么圆心角∠BOC将是2∠CAB，即100°。
answer:100
correct answer:100
total:1083  correct:291  acc:0.26869806094182824
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ID:1083
picturePath:image/1726.png
answer:22
answer:22
correct answer:22
total:1084  correct:292  acc:0.2693726937269373
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ID:1084
picturePath:image/1727.png
answer:解:∵∠CBD=30°,∴⁀{BC}=2∠CBD=60°,∵AB是直径,∴∠ADB=90°,∴⁀{AD}=90°-60°=30°,∴CD=\frac{1}{2}AB•sin30°=4×\frac{1}{2}×\frac{√{3}}{2}=2√{3}．故选:D．
answer:3
correct answer:4
total:1085  correct:292  acc:0.26912442396313363
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ID:1085
picturePath:image/1729.png
answer:42°
answer:42
correct answer:48
total:1086  correct:292  acc:0.26887661141804786
------------------------------------------------------------------------------------
ID:1086
picturePath:image/1730.png
answer:解:∵∠C=30°,∴∠AOB=2∠C=60°,∵OA=OB,∴△AOB是等边三角形,∴AB=OA=3．故选:A．
answer:3
correct answer:3
total:1087  correct:293  acc:0.26954921803127874
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ID:1087
picturePath:image/1731.png
answer:解:∵∠ACB=112°,∴∠AOB=2∠ACB=224°,∴∠α=\frac{1}{2}∠AOB=112°．故选:B．
answer:112
correct answer:136
total:1088  correct:293  acc:0.2693014705882353
------------------------------------------------------------------------------------
ID:1088
picturePath:image/1732.png
answer:50°
answer:50
correct answer:30
total:1089  correct:293  acc:0.26905417814508725
------------------------------------------------------------------------------------
ID:1089
picturePath:image/1733.png
answer:8.
answer:8
correct answer:15
total:1090  correct:293  acc:0.26880733944954127
------------------------------------------------------------------------------------
ID:1090
picturePath:image/1734.png
answer:20°
answer:20
correct answer:50
total:1091  correct:293  acc:0.2685609532538955
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ID:1091
picturePath:image/1735.png
answer:64°
answer:64
correct answer:64
total:1092  correct:294  acc:0.2692307692307692
------------------------------------------------------------------------------------
ID:1092
picturePath:image/1736.png
answer:64°
answer:64
correct answer:64
total:1093  correct:295  acc:0.2698993595608417
------------------------------------------------------------------------------------
ID:1093
picturePath:image/1739.png
answer:根据圆周角定理,∠BAC=∠EAD,又∵∠BAC+∠EAD=180°,∴∠BAC=∠EAD=90°,∴DE是直径,∵DE=12,∴BC=2DE=24.故选B.
answer:24
correct answer:16
total:1094  correct:295  acc:0.2696526508226691
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ID:1094
picturePath:image/1740.png
answer:解:∵AO∥DC,∠AOD=20°,∴∠AOC=∠AOD=20°,∴∠BOC=2∠AOC=40°,∴∠B=\frac{1}{2}∠BOC=20°．故选:A．
answer:20
correct answer:80
total:1095  correct:295  acc:0.2694063926940639
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ID:1095
picturePath:image/1741.png
answer:解:∵∠BOC=2∠A,∠A=50°,∴∠BOC=100°,∴∠OBC=\frac{1}{2}∠BOC=50°．故选:A．
answer:50
correct answer:40
total:1096  correct:295  acc:0.26916058394160586
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ID:1096
picturePath:image/1742.png
answer:解:∵∠BOC=100°,∴∠AOB=50°,∵OA=OB,∴∠OAB=∠OBA=\frac{1}{2}(180°-50°)=65°,
故选:A．
answer:65
correct answer:50
total:1097  correct:295  acc:0.2689152233363719
------------------------------------------------------------------------------------
ID:1097
picturePath:image/1743.png
answer:31°
answer:31
correct answer:31
total:1098  correct:296  acc:0.26958105646630237
------------------------------------------------------------------------------------
ID:1098
picturePath:image/1744.png
answer:解:∵∠BOC与∠BAC是同弧所对的圆心角与圆周角,∠BOC=100°,∴∠BAC=\frac{1}{2}∠BOC=50°．故选:B．
answer:50
correct answer:50
total:1099  correct:297  acc:0.27024567788899
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ID:1099
picturePath:image/1745.png
answer:Since the sum of the angles in a triangle is 180 degrees, and the sum of the angles around a point on a circle is 360 degrees, we can deduce that the measure of each angle in the triangle is (360 - 300) / 3 = 60 degrees. Therefore, the measure of angle C is 60 degrees.
answer:60
correct answer:30
total:1100  correct:297  acc:0.27
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ID:1100
picturePath:image/1746.png
answer:50°
answer:50
correct answer:50
total:1101  correct:298  acc:0.2706630336058129
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ID:1101
picturePath:image/1747.png
answer:解:∵∠ABC与∠AOC是同弧所对的圆周角与圆心角,∠ABC=130°,∴∠AOC=2∠ABC=260°．故选:D．
answer:260
correct answer:100
total:1102  correct:298  acc:0.27041742286751363
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ID:1102
picturePath:image/1748.png
answer:The answer is 50°.
answer:50
correct answer:25
total:1103  correct:298  acc:0.2701722574796011
------------------------------------------------------------------------------------
ID:1103
picturePath:image/1749.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵∠CAB=40°,∴∠ABC=90°-40°=50°,∴∠ADC=∠ABC=50°．故选:B．
answer:50
correct answer:50
total:1104  correct:299  acc:0.2708333333333333
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ID:1104
picturePath:image/1750.png
answer:解:∵AB为⊙O的直径,∴∠ACB=90°,∵∠BAC=30°,∴∠B=90°-∠BAC=60°,∵⁀{AD}=⁀{CD},∴∠DAC=∠B=60°．故选:B．
answer:60
correct answer:30
total:1105  correct:299  acc:0.27058823529411763
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ID:1105
picturePath:image/1753.png
answer:解:∵AD是半圆的直径,∴∠ADB=90°,∵∠ADC=55°,∴∠ABD=90°-∠ADC=35°,∵点C是弧BD的中点,∴∠BCD=∠ABD=35°,∴∠BAD=∠BCD=35°．故选:B．
answer:35
correct answer:70
total:1106  correct:299  acc:0.27034358047016277
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ID:1106
picturePath:image/1754.png
answer:6√3
answer:3
correct answer:3
total:1107  correct:300  acc:0.27100271002710025
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ID:1107
picturePath:image/1755.png
answer:解:∵∠ACB=112°,∴∠AOB=2∠ACB=112°×2=224°,∴∠α=\frac{1}{2}∠AOB=\frac{1}{2}×224°=112°．故选:B．
answer:112
correct answer:136
total:1108  correct:300  acc:0.27075812274368233
------------------------------------------------------------------------------------
ID:1108
picturePath:image/1756.png
answer:60°
answer:60
correct answer:105
total:1109  correct:300  acc:0.27051397655545534
------------------------------------------------------------------------------------
ID:1109
picturePath:image/1757.png
answer:根据圆周角定理,在⊙O中,∠BAC=15°,∠ADC=20°,则∠ABO=∠ADC=20°,故选B.
answer:20
correct answer:55
total:1110  correct:300  acc:0.2702702702702703
------------------------------------------------------------------------------------
ID:1110
picturePath:image/1758.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵∠ADC=20°,∴∠CAB=90°-∠ADC=70°．故选:A．
answer:70
correct answer:70
total:1111  correct:301  acc:0.2709270927092709
------------------------------------------------------------------------------------
ID:1111
picturePath:image/1759.png
answer:25
answer:25
correct answer:40
total:1112  correct:301  acc:0.27068345323741005
------------------------------------------------------------------------------------
ID:1112
picturePath:image/1762.png
answer:6√2
answer:2
correct answer:2
total:1113  correct:302  acc:0.27133872416891286
------------------------------------------------------------------------------------
ID:1113
picturePath:image/1763.png
answer:解:∵⊙O是△ABC的外接圆,∠C=35°,∴∠AOB=2∠C=2×35°=70°,∵OA=OB,∴∠OBA=\frac{180°-∠AOB}{2}=\frac{180°-70°}{2}=55°．故选:A．
answer:55
correct answer:55
total:1114  correct:303  acc:0.2719928186714542
------------------------------------------------------------------------------------
ID:1114
picturePath:image/1764.png
answer:68
answer:68
correct answer:112
total:1115  correct:303  acc:0.2717488789237668
------------------------------------------------------------------------------------
ID:1115
picturePath:image/1765.png
answer:解:∵弦CD⊥AB于点E,∴⁀{AC}=⁀{BC},∴∠ABD=\frac{1}{2}∠AOC=\frac{1}{2}×110°=55°．故选:A．
answer:55
correct answer:55
total:1116  correct:304  acc:0.2724014336917563
------------------------------------------------------------------------------------
ID:1116
picturePath:image/1766.png
answer:30°
answer:30
correct answer:30
total:1117  correct:305  acc:0.2730528200537153
------------------------------------------------------------------------------------
ID:1117
picturePath:image/1767.png
answer:解:∵∠BOC=110°,∴∠D=\frac{1}{2}∠BOC=55°,∵AD∥OC,∴∠ABD=∠D=55°．故选:B．
answer:55
correct answer:20
total:1118  correct:305  acc:0.27280858676207514
------------------------------------------------------------------------------------
ID:1118
picturePath:image/1768.png
answer:20
answer:20
correct answer:25
total:1119  correct:305  acc:0.2725647899910634
------------------------------------------------------------------------------------
ID:1119
picturePath:image/1769.png
answer:20°
answer:20
correct answer:50
total:1120  correct:305  acc:0.27232142857142855
------------------------------------------------------------------------------------
ID:1120
picturePath:image/1771.png
answer:4cm
answer:4
correct answer:3
total:1121  correct:305  acc:0.272078501338091
------------------------------------------------------------------------------------
ID:1121
picturePath:image/1772.png
answer:50
answer:50
correct answer:50
total:1122  correct:306  acc:0.2727272727272727
------------------------------------------------------------------------------------
ID:1122
picturePath:image/1774.png
answer:The answer is 60.
answer:60
correct answer:60
total:1123  correct:307  acc:0.27337488869100623
------------------------------------------------------------------------------------
ID:1123
picturePath:image/1775.png
answer:根据圆周角定理，同弧所对的圆周角相等，因此∠BED=∠ACD=40°。故选A。
answer:40
correct answer:50
total:1124  correct:307  acc:0.27313167259786475
------------------------------------------------------------------------------------
ID:1124
picturePath:image/1777.png
answer:根据圆周角定理,在同圆或等圆中,相等的圆周角所对的弧相等,所对的圆心角也相等.已知∠ACB=50°,则∠AOC=50°,又∵∠AOP=55°,∴∠POB=∠AOP+∠AOC=55°+50°=105°.故选C.
answer:105
correct answer:45
total:1125  correct:307  acc:0.2728888888888889
------------------------------------------------------------------------------------
ID:1125
picturePath:image/1778.png
answer:3
answer:3
correct answer:5
total:1126  correct:307  acc:0.27264653641207814
------------------------------------------------------------------------------------
ID:1126
picturePath:image/1779.png
answer:20
answer:20
correct answer:50
total:1127  correct:307  acc:0.2724046140195209
------------------------------------------------------------------------------------
ID:1127
picturePath:image/1781.png
answer:The answer is 80.
answer:80
correct answer:50
total:1128  correct:307  acc:0.2721631205673759
------------------------------------------------------------------------------------
ID:1128
picturePath:image/1782.png
answer:解:∵直径AB垂直于弦CD,∴CE=DE,在Rt△OCE中,∠CAO=22.5°,OC=6,∴CE=OC•sin22.5°=6×\frac{1}{2}=3,∴CD=2CE=6．故选:B．
answer:6
correct answer:2
total:1129  correct:307  acc:0.27192205491585475
------------------------------------------------------------------------------------
ID:1129
picturePath:image/1783.png
answer:2.5
answer:2.5
correct answer:2
total:1130  correct:307  acc:0.27168141592920353
------------------------------------------------------------------------------------
ID:1130
picturePath:image/1784.png
answer:The answer is 90-35=55.
answer:55
correct answer:55
total:1131  correct:308  acc:0.2723253757736516
------------------------------------------------------------------------------------
ID:1131
picturePath:image/1785.png
answer:The answer is 40.
answer:40
correct answer:110
total:1132  correct:308  acc:0.27208480565371024
------------------------------------------------------------------------------------
ID:1132
picturePath:image/1787.png
answer:解:∵∠BOD=130°,∴∠AOD=50°,∵半径OD∥AC,∴∠A=∠AOD=50°,∵AB是直径,∴∠ACB=90°,∴∠B=∠ACB-∠A=90°-50°=40°．故选:B．
answer:40
correct answer:40
total:1133  correct:309  acc:0.2727272727272727
------------------------------------------------------------------------------------
ID:1133
picturePath:image/1788.png
answer:解:∵∠ADB与∠ABC是同弧所对的圆周角与圆心角,∠ADB=70°,∴∠ABC=2∠ADB=140°．故选:D．
answer:140
correct answer:20
total:1134  correct:309  acc:0.2724867724867725
------------------------------------------------------------------------------------
ID:1134
picturePath:image/1790.png
answer:This is a geometric figure with a semicircle and a line segment labeled with points A, B, C, D, and O. The line segment AB is the diameter of the semicircle, and point O is the center of the circle. The angle ∠BOC is given as 50 degrees. The question asks for the measure of angle ∠D.
answer:50
correct answer:115
total:1135  correct:309  acc:0.27224669603524226
------------------------------------------------------------------------------------
ID:1135
picturePath:image/1791.png
answer:解:∵∠O=50°,∴∠A=\frac{1}{2}∠O=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:25
total:1136  correct:310  acc:0.272887323943662
------------------------------------------------------------------------------------
ID:1136
picturePath:image/1792.png
answer:130°
answer:130
correct answer:25
total:1137  correct:310  acc:0.2726473175021988
------------------------------------------------------------------------------------
ID:1137
picturePath:image/1793.png
answer:解:∵∠AOB和∠COD互补,∴∠AOB+∠COD=180°,∵∠AOB=∠COD,∴∠AOB=90°,∵AB=2,CD=4,∴OA=OC=\frac{1}{2}AB=\frac{1}{2}×2=1,∴OB=√{OC²+CD²}=√{1²+4²}=√{17},∴圆O的半径为√{17}．故选:C．
answer:17
correct answer:5
total:1138  correct:310  acc:0.27240773286467485
------------------------------------------------------------------------------------
ID:1138
picturePath:image/1794.png
answer:解:∵AB是⊙O的直径,∴∠ADB=90°,∴∠A=90°-∠ABD=90°-50°=40°,∴∠C=∠A=40°．故选:B．
answer:40
correct answer:40
total:1139  correct:311  acc:0.2730465320456541
------------------------------------------------------------------------------------
ID:1139
picturePath:image/1795.png
answer:根据圆周角定理，∠ADC是直径所对的圆周角，所以∠ADC=90°。又因为∠A=60°，所以∠C=360°-∠A-∠ADC=360°-60°-90°=210°。所以，∠C的度数是210°。
answer:210
correct answer:30
total:1140  correct:311  acc:0.27280701754385966
------------------------------------------------------------------------------------
ID:1140
picturePath:image/1798.png
answer:解:∵弦CD⊥直径AB,∴∠AOD=90°,∵∠CAB=30°,∴∠BOD=60°,∵OD=2,∴BD=\frac{180°-60°}{60°}×2=2√3,∴CD=BD-OD=2√3-2．故选:A．
answer:2
correct answer:3
total:1141  correct:311  acc:0.27256792287467135
------------------------------------------------------------------------------------
ID:1141
picturePath:image/1799.png
answer:解:∵AB是的直径,∴∠ACB=90°,∵∠CAB=35°,∴∠ABC=90°-35°=55°,∴∠ADC=∠ABC=55°．故选:B．
answer:55
correct answer:55
total:1142  correct:312  acc:0.2732049036777583
------------------------------------------------------------------------------------
ID:1142
picturePath:image/1800.png
answer:解:∵AB是⊙O的弦,直径CD交AB于点E,∠C=15°,∴∠AEB=30°,∵AE=EB=3,∴△AEB为等边三角形,∴∠AEB=∠ABE=60°,∴∠BEA=90°,∴OE=\frac{1}{2}AB=1.5×3=4.5．故选:D．
answer:4.5
correct answer:3
total:1143  correct:312  acc:0.27296587926509186
------------------------------------------------------------------------------------
ID:1143
picturePath:image/1801.png
answer:解:∵直径AB⊥CD,∴⁀{BC}=⁀{BD},∴∠BAD=\frac{1}{2}∠BOC=\frac{1}{2}×48°=24°．故选:B．
answer:24
correct answer:24
total:1144  correct:313  acc:0.2736013986013986
------------------------------------------------------------------------------------
ID:1144
picturePath:image/1802.png
answer:解:∵AB⊥CD,∴∠BAC=90°,∵∠A=35°,∴∠B=90°-35°=55°．故选:B．
answer:55
correct answer:55
total:1145  correct:314  acc:0.274235807860262
------------------------------------------------------------------------------------
ID:1145
picturePath:image/1803.png
answer:解:∵∠BCD与∠BOD是同弧所对的圆周角与圆心角,∠BCD=140°,∴∠BOD=2∠BCD=2×140°=280°,∵∠BOD是圆周角,∴∠BOD=360°-280°=80°．故选:B．
answer:80
correct answer:80
total:1146  correct:315  acc:0.27486910994764396
------------------------------------------------------------------------------------
ID:1146
picturePath:image/1804.png
answer:解:∵OC⊥AB,∴⁀{AC}=⁀{BC},∴∠OBA=\frac{1}{2}∠ADC=\frac{1}{2}×35°=17.5°．故选:A．
answer:17.5
correct answer:20
total:1147  correct:315  acc:0.2746294681778553
------------------------------------------------------------------------------------
ID:1147
picturePath:image/1805.png
answer:∠APB=\frac{1}{2}∠AOB=\frac{1}{2}×40°=20°.
answer:20
correct answer:20
total:1148  correct:316  acc:0.27526132404181186
------------------------------------------------------------------------------------
ID:1148
picturePath:image/1806.png
answer:解:∵∠D=32°,∴∠B=∠D=32°,∴∠AOC=2∠B=2×32°=64°．故选:D．
answer:64
correct answer:64
total:1149  correct:317  acc:0.2758920800696258
------------------------------------------------------------------------------------
ID:1149
picturePath:image/1807.png
answer:解:∵∠BCD与∠BOD是同弧所对的圆周角与圆心角,∠BCD=120°,∴∠BOD=2∠BCD=240°．故选:D．
answer:240
correct answer:120
total:1150  correct:317  acc:0.2756521739130435
------------------------------------------------------------------------------------
ID:1150
picturePath:image/1808.png
answer:17
answer:17
correct answer:56
total:1151  correct:317  acc:0.27541268462206775
------------------------------------------------------------------------------------
ID:1151
picturePath:image/1809.png
answer:2×25°=50°.
answer:50
correct answer:100
total:1152  correct:317  acc:0.2751736111111111
------------------------------------------------------------------------------------
ID:1152
picturePath:image/1811.png
answer:30°
answer:30
correct answer:75
total:1153  correct:317  acc:0.27493495229835213
------------------------------------------------------------------------------------
ID:1153
picturePath:image/1812.png
answer:解:∵∠A与∠BOC所对的弧都是AC,∴∠BOC=2∠A=100°,∵OB=OC,∴∠OBC=(180°-100°)÷2=40°．故选:A．
answer:40
correct answer:40
total:1154  correct:318  acc:0.2755632582322357
------------------------------------------------------------------------------------
ID:1154
picturePath:image/1813.png
answer:�∠ADC=26°,∵OC⊥AB,∴∠AOB=90°,∴∠COB=90°-∠ADC=90°-26°=64°.
answer:64
correct answer:52
total:1155  correct:318  acc:0.2753246753246753
------------------------------------------------------------------------------------
ID:1155
picturePath:image/1814.png
answer:4
answer:4
correct answer:3
total:1156  correct:318  acc:0.2750865051903114
------------------------------------------------------------------------------------
ID:1156
picturePath:image/1815.png
answer:解:∵AB为⊙O的直径,∴∠ADB=90°,∵∠ADC=26°,∴∠ABD=90°-26°=64°,∴∠CAB=\frac{1}{2}∠ABD=32°．故选:B．
answer:32
correct answer:64
total:1157  correct:318  acc:0.2748487467588591
------------------------------------------------------------------------------------
ID:1157
picturePath:image/1819.png
answer:20
answer:20
correct answer:35
total:1158  correct:318  acc:0.27461139896373055
------------------------------------------------------------------------------------
ID:1158
picturePath:image/1820.png
answer:解:∵∠AOC=120°,∴∠BOC=180°-∠AOC=60°,∴∠BDC=\frac{1}{2}∠BOC=\frac{1}{2}×60°=30°,
故选:A．
answer:30
correct answer:30
total:1159  correct:319  acc:0.27523727351164795
------------------------------------------------------------------------------------
ID:1159
picturePath:image/1822.png
answer:60
answer:60
correct answer:15
total:1160  correct:319  acc:0.275
------------------------------------------------------------------------------------
ID:1160
picturePath:image/1823.png
answer:50°
answer:50
correct answer:130
total:1161  correct:319  acc:0.2747631352282515
------------------------------------------------------------------------------------
ID:1161
picturePath:image/1825.png
answer:根据圆周角定理，同弧所对的圆周角的度数等于它所对的圆心角度数的一半。因此，∠C的度数为\( \frac{1}{2}∠AOC \)。将∠AOC=80°代入，得到∠C的度数为\( \frac{1}{2}×80^{\circ}=40^{\circ} \)。故选B。
answer:40
correct answer:40
total:1162  correct:320  acc:0.27538726333907054
------------------------------------------------------------------------------------
ID:1162
picturePath:image/1827.png
answer:70
answer:70
correct answer:110
total:1163  correct:320  acc:0.2751504729148753
------------------------------------------------------------------------------------
ID:1163
picturePath:image/1828.png
answer:30°
answer:30
correct answer:150
total:1164  correct:320  acc:0.27491408934707906
------------------------------------------------------------------------------------
ID:1164
picturePath:image/1829.png
answer:解:∵BC是半圆O的直径,∴∠BOC=180°,∵∠A=70°,∴∠BOD=110°,∵OD=OE,∴∠DOE=55°,
故选:B．
answer:55
correct answer:40
total:1165  correct:320  acc:0.27467811158798283
------------------------------------------------------------------------------------
ID:1165
picturePath:image/1830.png
answer:The answer is 40.
answer:40
correct answer:50
total:1166  correct:320  acc:0.274442538593482
------------------------------------------------------------------------------------
ID:1166
picturePath:image/1832.png
answer:解:∵∠AOC=126°,∴∠BOC=180°-∠AOC=180°-126°=54°,∴∠CDB=\frac{1}{2}∠BOC=\frac{1}{2}×54°=27°．故选:B．
answer:27
correct answer:27
total:1167  correct:321  acc:0.2750642673521851
------------------------------------------------------------------------------------
ID:1167
picturePath:image/1834.png
answer:20
answer:20
correct answer:10
total:1168  correct:321  acc:0.2748287671232877
------------------------------------------------------------------------------------
ID:1168
picturePath:image/1835.png
answer:解:∵∠AOD=132°,∴∠BOD=180°-132°=48°,∴∠C=48°．故选:B．
answer:48
correct answer:24
total:1169  correct:321  acc:0.2745936698032506
------------------------------------------------------------------------------------
ID:1169
picturePath:image/1836.png
answer:解:∵∠ABC与∠AOC是同弧所对的圆周角与圆心角,∠ABC=130°,∴∠AOC=2∠ABC=260°．故选:D．
answer:260
correct answer:100
total:1170  correct:321  acc:0.2743589743589744
------------------------------------------------------------------------------------
ID:1170
picturePath:image/1838.png
answer:解:∵∠ABO=40°,∴∠AOC=2∠ABO=80°,∴∠C=\frac{1}{2}∠AOC=40°．故选:A．
answer:40
correct answer:50
total:1171  correct:321  acc:0.2741246797608881
------------------------------------------------------------------------------------
ID:1171
picturePath:image/1839.png
answer:解:∵∠ADC与∠CBA所对的弧都是AC,∴∠CBA=∠ADC=35°．故选:A．
answer:35
correct answer:35
total:1172  correct:322  acc:0.27474402730375425
------------------------------------------------------------------------------------
ID:1172
picturePath:image/1841.png
answer:This is a geometry problem involving a circle, a point outside the circle, and a line that intersects the circle at two points. The given information includes the radius of the circle, the distance from the point outside the circle to the center of the circle, and the fact that the line intersects the circle at a right angle. The question asks for the minimum value of the sum of the distances from points A and B to the center of the circle.

To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is the line segment connecting points A and B, and the other two sides are the distances from points A and B to the center of the circle.

Since the line intersects the circle at a right angle, the line segment connecting points A and B is a diameter of the circle. Therefore, the sum of the distances from points A and B to the center of the circle is equal to the diameter of the circle.

Given that the radius of the circle is 3, the diameter of the circle is 2 times the radius, which is 6. Therefore, the minimum value of the sum of the distances from points A and B to the center of the circle is 6.
answer:6
correct answer:4
total:1173  correct:322  acc:0.27450980392156865
------------------------------------------------------------------------------------
ID:1173
picturePath:image/1842.png
answer:解:∵AB是⊙O的直径,∴∠ADB=90°,∵∠ABD=30°,∴AD=\frac{1}{2}BO=\frac{1}{2}×4=2．故选:B．
answer:2
correct answer:4
total:1174  correct:322  acc:0.27427597955706984
------------------------------------------------------------------------------------
ID:1174
picturePath:image/1843.png
answer:2.5
answer:2.5
correct answer:2
total:1175  correct:322  acc:0.27404255319148935
------------------------------------------------------------------------------------
ID:1175
picturePath:image/1845.png
answer:50°
answer:50
correct answer:40
total:1176  correct:322  acc:0.27380952380952384
------------------------------------------------------------------------------------
ID:1176
picturePath:image/1846.png
answer:40°
answer:40
correct answer:20
total:1177  correct:322  acc:0.2735768903993203
------------------------------------------------------------------------------------
ID:1177
picturePath:image/1848.png
answer:3
answer:3
correct answer:5
total:1178  correct:322  acc:0.2733446519524618
------------------------------------------------------------------------------------
ID:1178
picturePath:image/1849.png
answer:解:∵∠ABC与∠AOC是同弧所对的圆周角与圆心角,∠ABC=130°,∴∠AOC=2∠ABC=260°．故选:D．
answer:260
correct answer:100
total:1179  correct:322  acc:0.2731128074639525
------------------------------------------------------------------------------------
ID:1179
picturePath:image/1850.png
answer:解:∵∠AOB是⊙O的圆心角,∠AOB=50°,∴∠ACB=\frac{1}{2}∠AOB=25°．故选:B．
answer:25
correct answer:25
total:1180  correct:323  acc:0.27372881355932205
------------------------------------------------------------------------------------
ID:1180
picturePath:image/1851.png
answer:30°
answer:30
correct answer:30
total:1181  correct:324  acc:0.2743437764606266
------------------------------------------------------------------------------------
ID:1181
picturePath:image/1852.png
answer:The answer is 50.
answer:50
correct answer:115
total:1182  correct:324  acc:0.27411167512690354
------------------------------------------------------------------------------------
ID:1182
picturePath:image/1853.png
answer:解:∵∠CBD=2∠BDC,∠BAC=44°,∴∠BDC=\frac{1}{2}∠BAC=\frac{1}{2}×44°=22°,∴∠CAD=∠CBD=2∠BDC=2×22°=44°．故选:A．
answer:44
correct answer:88
total:1183  correct:324  acc:0.2738799661876585
------------------------------------------------------------------------------------
ID:1183
picturePath:image/1854.png
answer:解:∵OC⊥AB,∴⁀{AC}=⁀{BC},∴∠BOC=2∠ADC,∵∠ADC=32°,∴∠BOC=64°,故选:D．
answer:64
correct answer:64
total:1184  correct:325  acc:0.27449324324324326
------------------------------------------------------------------------------------
ID:1184
picturePath:image/1857.png
answer:解:∵∠C=20°,∠B=50°,∴∠A=180°-∠B-∠C=180°-50°-20°=110°．故选:B．
answer:110
correct answer:30
total:1185  correct:325  acc:0.2742616033755274
------------------------------------------------------------------------------------
ID:1185
picturePath:image/1858.png
answer:12
answer:12
correct answer:6
total:1186  correct:325  acc:0.2740303541315346
------------------------------------------------------------------------------------
ID:1186
picturePath:image/1859.png
answer:54°
answer:54
correct answer:63
total:1187  correct:325  acc:0.2737994945240101
------------------------------------------------------------------------------------
ID:1187
picturePath:image/1860.png
answer:解:∵AB是圆O的直径,∴∠ADB=90°,∵AB=2DE,∴AD=DE,∴∠DAE=∠E=16°,∴∠AED=90°-16°=74°,∴∠ABC=∠AED=74°．故选:B．
answer:74
correct answer:24
total:1188  correct:325  acc:0.2735690235690236
------------------------------------------------------------------------------------
ID:1188
picturePath:image/1861.png
answer:根据圆周角定理，在同圆或等圆中，同弧或等弧所对的圆周角相等。因此，∠AOB = 2∠ACB = 2 × 22° = 44°.
answer:44
correct answer:44
total:1189  correct:326  acc:0.27417998317914216
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ID:1189
picturePath:image/1862.png
answer:解:∵AB是⊙O的直径,∴∠ADB=90°,∵四边形ABCD中,∠ACD=20°,∴∠BAD=90°-20°=70°．故选:B．
answer:70
correct answer:70
total:1190  correct:327  acc:0.2747899159663866
------------------------------------------------------------------------------------
ID:1190
picturePath:image/1864.png
answer:70
answer:70
correct answer:70
total:1191  correct:328  acc:0.2753988245172124
------------------------------------------------------------------------------------
ID:1191
picturePath:image/1865.png
answer:65
answer:65
correct answer:115
total:1192  correct:328  acc:0.2751677852348993
------------------------------------------------------------------------------------
ID:1192
picturePath:image/1866.png
answer:Since \( \overline{AB} \) is the diameter, it is perpendicular to \( \overline{CD} \), so \( \angle AOC = 90^\circ \). Since \( \angle AOC \) and \( \angle ADC \) are corresponding angles, they are congruent, so \( \angle AOC = \angle ADC = 54^\circ \). The sum of the angles in a triangle is 180 degrees, so \( \angle BAC = 180^\circ - \angle AOC - \angle ADC = 180^\circ - 90^\circ - 54^\circ = 36^\circ \). Since \( \overline{AB} \) is perpendicular to \( \overline{CD} \), \( \angle CAB \) is a right angle, so \( \angle CAB = 90^\circ \). The sum of the angles in a triangle is 180 degrees, so \( \angle BCA = 180^\circ - \angle CAB - \angle BAC = 180^\circ - 90^\circ - 36^\circ = 54^\circ \). Therefore, the answer is C.
answer:54
correct answer:36
total:1193  correct:328  acc:0.2749371332774518
------------------------------------------------------------------------------------
ID:1193
picturePath:image/1867.png
answer:解:∵AB是半圆O的直径,∴∠ACB=90°,∵点C为半圆O上一点,D是⁀{AC}的中点,∠DAC=40°,∴∠ADC=20°,∴∠CAB=90°-∠ADC=70°．故选:A．
answer:70
correct answer:10
total:1194  correct:328  acc:0.2747068676716918
------------------------------------------------------------------------------------
ID:1194
picturePath:image/1868.png
answer:解:∵AB为⊙O的直径,∴∠ACB=90°,∵∠CAB=20°,∴∠B=90°-20°=70°,∴∠CAD=∠B=70°．故选:B．
answer:70
correct answer:35
total:1195  correct:328  acc:0.2744769874476987
------------------------------------------------------------------------------------
ID:1195
picturePath:image/1870.png
answer:90°
answer:90
correct answer:75
total:1196  correct:328  acc:0.27424749163879597
------------------------------------------------------------------------------------
ID:1196
picturePath:image/1871.png
answer:72°/2=36°
answer:36
correct answer:36
total:1197  correct:329  acc:0.27485380116959063
------------------------------------------------------------------------------------
ID:1197
picturePath:image/1872.png
answer:根据圆周角定理，同弧所对的圆周角相等，因此∠OAC=∠AOB=50°。故选A。
answer:50
correct answer:15
total:1198  correct:329  acc:0.27462437395659434
------------------------------------------------------------------------------------
ID:1198
picturePath:image/1873.png
answer:The answer is A.
ID:1199
picturePath:image/1874.png
answer:The answer is 60.
answer:60
correct answer:120
total:1200  correct:329  acc:0.27416666666666667
------------------------------------------------------------------------------------
ID:1200
picturePath:image/1876.png
answer:解:∵BC是⊙O的直径,∴∠BAC=90°,∵∠D=36°,∴∠B=∠D=36°,∴∠BCA=90°-∠B=90°-36°=54°．故选:B．
answer:54
correct answer:54
total:1201  correct:330  acc:0.27477102414654453
------------------------------------------------------------------------------------
ID:1201
picturePath:image/1878.png
answer:40
answer:40
correct answer:80
total:1202  correct:330  acc:0.27454242928452577
------------------------------------------------------------------------------------
ID:1202
picturePath:image/1879.png
answer:解:∵∠AMD=100°,∠A=30°,∴∠ADC=∠AMD-∠A=100°-30°=70°,∴∠B=\frac{1}{2}∠ADC=\frac{1}{2}×70°=35°．故选:B．
answer:35
correct answer:50
total:1203  correct:330  acc:0.2743142144638404
------------------------------------------------------------------------------------
ID:1203
picturePath:image/1880.png
answer:72°
answer:72
correct answer:72
total:1204  correct:331  acc:0.2749169435215947
------------------------------------------------------------------------------------
ID:1204
picturePath:image/1881.png
answer:80°
answer:80
correct answer:80
total:1205  correct:332  acc:0.27551867219917014
------------------------------------------------------------------------------------
ID:1205
picturePath:image/1886.png
answer:60°
answer:60
correct answer:30
total:1206  correct:332  acc:0.2752902155887231
------------------------------------------------------------------------------------
ID:1206
picturePath:image/1887.png
answer:解:∵∠AOC=130°,∴∠BOC=180°-130°=50°,∴∠D=\frac{1}{2}∠BOC=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:25
total:1207  correct:333  acc:0.275890637945319
------------------------------------------------------------------------------------
ID:1207
picturePath:image/1888.png
answer:80°
answer:80
correct answer:80
total:1208  correct:334  acc:0.2764900662251656
------------------------------------------------------------------------------------
ID:1208
picturePath:image/1890.png
answer:20
answer:20
correct answer:25
total:1209  correct:334  acc:0.2762613730355666
------------------------------------------------------------------------------------
ID:1209
picturePath:image/1891.png
answer:70
answer:70
correct answer:100
total:1210  correct:334  acc:0.27603305785123966
------------------------------------------------------------------------------------
ID:1210
picturePath:image/1893.png
answer:解:∵AB是⊙O直径,∴∠ACB=90°,∵∠BAC=40°,∴∠B=90°-∠BAC=90°-40°=50°,∴∠ADC=∠B=50°．故选:B．
answer:50
correct answer:50
total:1211  correct:335  acc:0.27663088356729976
------------------------------------------------------------------------------------
ID:1211
picturePath:image/1896.png
answer:30°
answer:30
correct answer:30
total:1212  correct:336  acc:0.27722772277227725
------------------------------------------------------------------------------------
ID:1212
picturePath:image/1897.png
answer:解:∵∠BOC=40°,∴∠AOC=180°-40°=140°,∴∠D=\frac{1}{2}∠AOC=\frac{1}{2}×140°=70°．故选:B．
answer:70
correct answer:110
total:1213  correct:336  acc:0.27699917559769166
------------------------------------------------------------------------------------
ID:1213
picturePath:image/1900.png
answer:解:∵∠B=∠EAC,∴BE=CE,∵AE是直径,∴∠BAE=90°,∵BE=CE,∴△ABE≌△ACE(ASA),∴AC=AB,∵AE=10cm,∴AC=AB=5cm．故选:A．
answer:5
correct answer:2
total:1214  correct:336  acc:0.27677100494233936
------------------------------------------------------------------------------------
ID:1214
picturePath:image/1902.png
answer:40°
answer:40
correct answer:40
total:1215  correct:337  acc:0.2773662551440329
------------------------------------------------------------------------------------
ID:1215
picturePath:image/1903.png
answer:解:∵∠AOD=136°,∴∠BOD=44°,∴∠C=\frac{1}{2}∠BOD=22°．故选:B．
answer:22
correct answer:22
total:1216  correct:338  acc:0.2779605263157895
------------------------------------------------------------------------------------
ID:1216
picturePath:image/1904.png
answer:解:∵∠α=140°,∴∠BOC=140°,∴∠A=\frac{1}{2}∠BOC=\frac{1}{2}×140°=70°．故选:B．
answer:70
correct answer:110
total:1217  correct:338  acc:0.2777321281840592
------------------------------------------------------------------------------------
ID:1217
picturePath:image/1905.png
answer:Since ∠BOC and ∠BAC are both angles on the circumference of a circle, they are complementary angles. Therefore, ∠BAC = 180° - ∠BOC = 180° - 72° = 108°.
answer:108
correct answer:36
total:1218  correct:338  acc:0.277504105090312
------------------------------------------------------------------------------------
ID:1218
picturePath:image/1907.png
answer:解:∵AB为⊙O的直径,∴∠ACB=90°,∵∠CAB=20°,∴∠B=90°-∠CAB=70°,∴∠ADC=∠B=70°．故选:B．
answer:70
correct answer:110
total:1219  correct:338  acc:0.27727645611156687
------------------------------------------------------------------------------------
ID:1219
picturePath:image/1909.png
answer:解:∵∠AOC=140°,∴∠BOC=180°-140°=40°,∴∠BDC=\frac{1}{2}∠BOC=\frac{1}{2}×40°=20°．故选:A．
answer:20
correct answer:20
total:1220  correct:339  acc:0.2778688524590164
------------------------------------------------------------------------------------
ID:1220
picturePath:image/1913.png
answer:25
answer:25
correct answer:25
total:1221  correct:340  acc:0.2784602784602785
------------------------------------------------------------------------------------
ID:1221
picturePath:image/1914.png
answer:50°
answer:50
correct answer:25
total:1222  correct:340  acc:0.27823240589198034
------------------------------------------------------------------------------------
ID:1222
picturePath:image/1916.png
answer:解:∵∠BOD=130°,∴∠ACD=\frac{1}{2}∠BOD=\frac{1}{2}×130°=65°．故选:B．
answer:65
correct answer:25
total:1223  correct:340  acc:0.27800490596892885
------------------------------------------------------------------------------------
ID:1223
picturePath:image/1917.png
answer:解:∵AB是直径,∴∠ACB=90°,∵∠ABC=70°,∴∠CAB=90°-∠ABC=20°,∵CD=AD,∴∠DAB=∠CAB=20°,∴∠BAD=180°-∠DAB-∠ADB=180°-20°-20°=140°．故选:D．
answer:140
correct answer:35
total:1224  correct:340  acc:0.2777777777777778
------------------------------------------------------------------------------------
ID:1224
picturePath:image/1918.png
answer:This is a right triangle with a right angle at point C. We can use the Pythagorean theorem to find the length of CD. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, AB is the hypotenuse, and BC and CD are the other two sides. So, we have:

AB^2 = BC^2 + CD^2
10^2 = 6^2 + CD^2
100 = 36 + CD^2
64 = CD^2
CD = √64
CD = 8

So, the length of CD is 8.
answer:8
correct answer:4.8
total:1225  correct:340  acc:0.27755102040816326
------------------------------------------------------------------------------------
ID:1225
picturePath:image/1919.png
answer:17
answer:17
correct answer:56
total:1226  correct:340  acc:0.27732463295269166
------------------------------------------------------------------------------------
ID:1226
picturePath:image/1920.png
answer:This is a geometry problem involving circles, perpendicular lines, and angles. Since CD is perpendicular to AB, we know that angle AOC is a right angle (90 degrees). We are given that angle CAO is 70 degrees, so angle AOD is 180 - 90 - 70 = 20 degrees. Since AB is a diameter, angle BOD is also 90 degrees. Therefore, angle BOD is 90 - 20 = 70 degrees.The answer is: 70
answer:70
correct answer:140
total:1227  correct:340  acc:0.27709861450692747
------------------------------------------------------------------------------------
ID:1227
picturePath:image/1922.png
answer:36°
answer:36
correct answer:24
total:1228  correct:340  acc:0.2768729641693811
------------------------------------------------------------------------------------
ID:1228
picturePath:image/1923.png
answer:60
answer:60
correct answer:60
total:1229  correct:341  acc:0.2774613506916192
------------------------------------------------------------------------------------
ID:1229
picturePath:image/1924.png
answer:15m.
answer:15
correct answer:15
total:1230  correct:342  acc:0.2780487804878049
------------------------------------------------------------------------------------
ID:1230
picturePath:image/1925.png
answer:40
answer:40
correct answer:40
total:1231  correct:343  acc:0.2786352558895207
------------------------------------------------------------------------------------
ID:1231
picturePath:image/1926.png
answer:解:∵∠BOC与∠A、∠C是同弧所对的圆心角与圆周角,∠BOC=100°,∴∠A=\frac{1}{2}∠BOC=50°．
故选:A．
answer:50
correct answer:50
total:1232  correct:344  acc:0.2792207792207792
------------------------------------------------------------------------------------
ID:1232
picturePath:image/1927.png
answer:8
answer:8
correct answer:2
total:1233  correct:344  acc:0.27899432278994324
------------------------------------------------------------------------------------
ID:1233
picturePath:image/1929.png
answer:解:∵OC⊥AB,∴AC=BC,在Rt△OAC中,OC=5m,AC=CD=8m,∴AC=√{OC²+AC²}=√{5²+8²}=9(米),∴AB=2AC=2×9=18(米)．故选:D．
answer:18
correct answer:8
total:1234  correct:344  acc:0.2787682333873582
------------------------------------------------------------------------------------
ID:1234
picturePath:image/1930.png
answer:解:∵AB∥CD,∴△ABO∽△CD0,∴\frac{AB}{CD}=\frac{BO}{CO},即\frac{600}{x}=\frac{600}{650},解得x=650．故选:A．
answer:650
correct answer:200
total:1235  correct:344  acc:0.2785425101214575
------------------------------------------------------------------------------------
ID:1235
picturePath:image/1933.png
answer:解:∵∠α+∠β=180°,∠α=130°,∴∠β=180°-130°=50°．
故选:A．
answer:50
correct answer:40
total:1236  correct:344  acc:0.2783171521035599
------------------------------------------------------------------------------------
ID:1236
picturePath:image/1934.png
answer:4
answer:4
correct answer:3
total:1237  correct:344  acc:0.27809215844785773
------------------------------------------------------------------------------------
ID:1237
picturePath:image/1936.png
answer:解:∵AB是⊙O的直径,∴∠ADB=90°,∵∠ABD=56°,∴∠A=90°-56°=34°,∴∠BCD=∠A=34°．故选:B．
answer:34
correct answer:34
total:1238  correct:345  acc:0.27867528271405495
------------------------------------------------------------------------------------
ID:1238
picturePath:image/1937.png
answer:8.5
answer:8.5
correct answer:15
total:1239  correct:345  acc:0.2784503631961259
------------------------------------------------------------------------------------
ID:1239
picturePath:image/1939.png
answer:5
answer:5
correct answer:4
total:1240  correct:345  acc:0.2782258064516129
------------------------------------------------------------------------------------
ID:1240
picturePath:image/1940.png
answer:解:∵OB=5m,OC=3m,∴OA=√{OB²-OC²}=4(m),∴污水面宽AB=2OA=2×4=8(m)．故选:B．
answer:8
correct answer:8
total:1241  correct:346  acc:0.27880741337630943
------------------------------------------------------------------------------------
ID:1241
picturePath:image/1942.png
answer:解:∵AO=BO=50cm,CO=DO=30cm,∴AB=√{50²-30²}=40(cm),∵桌面a距离地面m为40cm高,∴tan∠AOB$=\frac{40}{40}=$1,∴∠AOB=45°．故选:D．
answer:45
correct answer:120
total:1242  correct:346  acc:0.2785829307568438
------------------------------------------------------------------------------------
ID:1242
picturePath:image/1944.png
answer:70
answer:70
correct answer:110
total:1243  correct:346  acc:0.27835880933226065
------------------------------------------------------------------------------------
ID:1243
picturePath:image/1946.png
answer:10
answer:10
correct answer:10
total:1244  correct:347  acc:0.27893890675241156
------------------------------------------------------------------------------------
ID:1244
picturePath:image/1947.png
answer:50°
answer:50
correct answer:50
total:1245  correct:348  acc:0.27951807228915665
------------------------------------------------------------------------------------
ID:1245
picturePath:image/1949.png
answer:5
answer:5
correct answer:3
total:1246  correct:348  acc:0.27929373996789725
------------------------------------------------------------------------------------
ID:1246
picturePath:image/1952.png
answer:解:∵OB=10,AB=16,∴OA=8,在Rt△AOB中,根据勾股定理得:BC=√{16^{2}-8^{2}}=12,∴水最大深度为:BC-OA=12-8=4．故选:C．
answer:4
correct answer:4
total:1247  correct:349  acc:0.27987169206094625
------------------------------------------------------------------------------------
ID:1247
picturePath:image/1955.png
answer:This is a geometric figure with three labeled points: A, B, and C. The figure shows a triangle with an angle labeled as 30 degrees at point C. There is also a line segment labeled as AD, which appears to be a perpendicular line from point D to the base BC of the triangle. The angle at point D is labeled as 20 degrees. The question is asking for the measure of angle BAC. To solve this, we can use the properties of triangles and the given information. Since AD is a perpendicular line to BC, we know that angle ADB is a right angle (90 degrees). We also know that angle C is 30 degrees and angle DAB is 20 degrees. Therefore, angle BAC can be found by subtracting the sum of angles C and DAB from 180 degrees (since the sum of angles in a triangle is 180 degrees). So, angle BAC = 180 - 30 - 20 = 130 degrees. Therefore, the correct answer is D.
answer:130
correct answer:80
total:1248  correct:349  acc:0.2796474358974359
------------------------------------------------------------------------------------
ID:1248
picturePath:image/1956.png
answer:15°
answer:15
correct answer:60
total:1249  correct:349  acc:0.27942353883106485
------------------------------------------------------------------------------------
ID:1249
picturePath:image/1957.png
answer:This is a geometric figure with labeled points A, B, C, D, E, and F. The figure is a right triangle with angle CAB equal to 90 degrees. The angle ABC is 72 degrees. AD is the angle bisector of angle CAB, intersecting BC at point D. CE is the altitude from C to AD in triangle ACD. The question is asking for the measure of angle ECD.
answer:72
correct answer:27
total:1250  correct:349  acc:0.2792
------------------------------------------------------------------------------------
ID:1250
picturePath:image/1958.png
answer:解:∵点M、N分别是弦AB、AC的中点,∴MN是△ABC的中位线,∴MN=\frac{1}{2}BC,∵∠C=45°,∴∠B=45°,∴BC=AB=10,∴MN=\frac{1}{2}BC=5．故选:A．
answer:5
correct answer:2
total:1251  correct:349  acc:0.2789768185451639
------------------------------------------------------------------------------------
ID:1251
picturePath:image/1960.png
answer:根据勾股定理，在直角三角形ABC中，AB² = AC² + BC²，即5² = 4² + 3²，解得AB = √(4² + 3²) = √(16 + 9) = √25 = 5。

又因为CD是AB边上的高，所以CD是△ABC的中位线，所以CD = 1/2 * AB = 1/2 * 5 = 2.5。

所以，CD = 2.5。
answer:2.5
correct answer:5
total:1252  correct:349  acc:0.2787539936102236
------------------------------------------------------------------------------------
ID:1252
picturePath:image/1961.png
answer:42°
answer:42
correct answer:48
total:1253  correct:349  acc:0.2785315243415802
------------------------------------------------------------------------------------
ID:1253
picturePath:image/1965.png
answer:S_{阴影}=S_{扇形}-S_{三角形}= \frac{90π×4^{2}}{360} - \frac{1}{2}×2×2= \frac{16π}{3} -2= \frac{16π}{3}- \frac{12π}{3}= \frac{4π}{3} $．  故选:D．
answer:3
correct answer:1
total:1254  correct:349  acc:0.2783094098883573
------------------------------------------------------------------------------------
ID:1254
picturePath:image/1966.png
answer:解:∵弦AB∥CD,∴∠C=∠ADB=30°,∵CD是半圆O的直径,∴OC⊥CD,∴∠OCD=90°,∴∠COD=60°,∵Sv=\frac{1}{2}×CD•OC•sin∠COD=\frac{1}{2}×6×3×\frac{√{3}}{2}=\frac{3√{3}}{2},故选:A．
answer:2
correct answer:2
total:1255  correct:350  acc:0.2788844621513944
------------------------------------------------------------------------------------
ID:1255
picturePath:image/1972.png
answer:The answer is \frac{1}{4}×2×2=\frac{1}{2}
answer:2
correct answer:1
total:1256  correct:350  acc:0.2786624203821656
------------------------------------------------------------------------------------
ID:1256
picturePath:image/1974.png
answer:6π
answer:6
correct answer:2
total:1257  correct:350  acc:0.27844073190135243
------------------------------------------------------------------------------------
ID:1257
picturePath:image/1977.png
answer:2.5
answer:2.5
correct answer:2.4
total:1258  correct:350  acc:0.27821939586645467
------------------------------------------------------------------------------------
ID:1258
picturePath:image/1979.png
answer:解:∵圆锥的底面半径为3,母线长为6,∴圆锥的底面周长=6π,圆锥的母线长=6,∴圆锥的侧面积=\frac{1}{2}×6π×6=9π．故选:A．
answer:9
correct answer:18
total:1259  correct:350  acc:0.2779984114376489
------------------------------------------------------------------------------------
ID:1259
picturePath:image/1980.png
answer:40°
answer:40
correct answer:40
total:1260  correct:351  acc:0.2785714285714286
------------------------------------------------------------------------------------
ID:1260
picturePath:image/1981.png
answer:解:∵⁀{AB}=⁀{BC}=⁀{CD},∴∠BDC=\frac{1}{2}∠BEC=\frac{1}{2}×110°=55°．故选:A．
answer:55
correct answer:35
total:1261  correct:351  acc:0.27835051546391754
------------------------------------------------------------------------------------
ID:1261
picturePath:image/1988.png
answer:0.5×π×1²=0.5πcm².
answer:0.5
correct answer:2
total:1262  correct:351  acc:0.27812995245641836
------------------------------------------------------------------------------------
ID:1262
picturePath:image/1989.png
answer:The answer is 20\sqrt{3}cm.
answer:3
correct answer:2
total:1263  correct:351  acc:0.27790973871733965
------------------------------------------------------------------------------------
ID:1263
picturePath:image/1991.png
answer:12πcm.
answer:12
correct answer:4
total:1264  correct:351  acc:0.2776898734177215
------------------------------------------------------------------------------------
ID:1264
picturePath:image/1992.png
answer:解:∵AB是直径,∴∠ADB=90°,∵∠DEA=30°,∴∠AED=60°,∴AE=AD,∵AB=AC=2,∴AD=1,∴CD=1．故选:A．
answer:1
correct answer:3
total:1265  correct:351  acc:0.2774703557312253
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ID:1265
picturePath:image/1994.png
answer:解:∵圆锥的底面半径OB=6cm,高OC=8cm,∴圆锥的母线长为√{62+82}=10(cm),∴这个圆锥漏斗的侧面积=\frac{1}{2}•2π•6•10=60π(cm2)．故选:A．
answer:2
correct answer:2
total:1266  correct:352  acc:0.27804107424960506
------------------------------------------------------------------------------------
ID:1266
picturePath:image/1997.png
answer:解:圆锥的母线长为24cm,底面半径为10cm,∴圆锥的底面周长=2π×10=20πcm,∴这张扇形纸板的弧长=圆锥的底面周长=20πcm,∴扇形纸板的面积=\frac{20π×24}{2}=240πcm^{2}．故选:A．
answer:2
correct answer:2
total:1267  correct:353  acc:0.2786108918705604
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ID:1267
picturePath:image/1998.png
answer:解:∵圆锥的母线长是2.5米,底面半径为2米,∴底面周长=2π×2=4π,∴扇形的弧长=圆锥的底面周长=4π,∴扇形的面积=\frac{4π×2.5}{2}=5π,即做成这把遮阳伞需要布料的面积是5π平方米．故选:C．
answer:5
correct answer:5
total:1268  correct:354  acc:0.27917981072555204
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ID:1268
picturePath:image/2004.png
answer:1/4.
ID:1269
picturePath:image/2005.png
answer:S_{阴影} = S_{圆ABC} + S_{圆AC} - S_{△ABC} = \frac{1}{2} \times 2π + \frac{1}{2} \times 2π - \frac{1}{2} \times \frac{1}{2} \times 2π = \frac{3}{2} \times \frac{1}{2} \times 2π = \frac{3}{4} \times 2π = \frac{3}{2}π
ID:1270
picturePath:image/2006.png
answer:解:∵OB=6cm,OC=8cm,∴圆锥的母线长为√{62+82}=10(cm),∴这个圆锥的侧面积是=\frac{1}{2}•2π•6•10=60π(cm2)．故选:A．
answer:2
correct answer:2
total:1271  correct:355  acc:0.27930763178599527
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ID:1271
picturePath:image/2007.png
answer:2πr=2π*4=8π.
answer:8
correct answer:2
total:1272  correct:355  acc:0.27908805031446543
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ID:1272
picturePath:image/2008.png
answer:解:∵圆锥的底面半径为2,母线长为6,∴圆锥的底面周长=2π×2=4π,∴侧面积=\frac{1}{2}×4π×6=6π．故选:C．
answer:6
correct answer:12
total:1273  correct:355  acc:0.2788688138256088
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ID:1273
picturePath:image/2009.png
answer:3.14cm
answer:3.14
correct answer:15
total:1274  correct:355  acc:0.27864992150706436
------------------------------------------------------------------------------------
ID:1274
picturePath:image/2010.png
answer:解:∵圆锥的底面半径r=3,高h=4,∴圆锥的母线l=√{h^{2}+r^{2}}=√{4^{2}+3^{2}}=5,∴圆锥的侧面积=\frac{1}{2}•2π•3•5=15π．故选:A．
answer:15
correct answer:15
total:1275  correct:356  acc:0.2792156862745098
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ID:1275
picturePath:image/2012.png
answer:解:∵圆锥的母线长为10cm,其侧面展开图是圆心角为216°的扇形,∴展开扇形的弧长为:\frac{216π×10}{180}=12πcm,即圆锥的底面周长为12πcm,∴圆锥的底面半径为:12π÷2π=6cm,∴圆锥的高为:√{10^{2}-6^{2}}=8(cm)故选:B．
answer:8
correct answer:8
total:1276  correct:357  acc:0.2797805642633229
------------------------------------------------------------------------------------
ID:1276
picturePath:image/2013.png
answer:15
answer:15
correct answer:15
total:1277  correct:358  acc:0.2803445575567737
------------------------------------------------------------------------------------
ID:1277
picturePath:image/2015.png
answer:The side area of a cone is given by the formula πrl, where r is the radius of the base and l is the slant height. In this case, the radius of the base is half the length of the hypotenuse of the right triangle, which is the square root of the sum of the squares of the lengths of the legs (AC and BC). The slant height is the length of the hypotenuse of the right triangle, which is given as 13 cm. Substituting these values into the formula, we get π(6.5)(13) = 153π. Therefore, the side area of the cone is 153π square centimeters.
answer:153
correct answer:2
total:1278  correct:358  acc:0.28012519561815336
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ID:1278
picturePath:image/2018.png
answer:20.
answer:20
correct answer:10
total:1279  correct:358  acc:0.2799061767005473
------------------------------------------------------------------------------------
ID:1279
picturePath:image/2020.png
answer:S = \frac{1}{2} \cdot l \cdot h = \frac{1}{2} \cdot 13 \cdot 12 = 78 \text{cm}^2.
answer:2
correct answer:65
total:1280  correct:358  acc:0.2796875
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ID:1280
picturePath:image/2024.png
answer:2√{3}
answer:3
correct answer:3
total:1281  correct:359  acc:0.28024980483996875
------------------------------------------------------------------------------------
ID:1281
picturePath:image/2025.png
answer:The side area of a cone can be calculated using the formula: side area = πrl, where r is the radius of the base and l is the slant height of the cone. Given that the radius of the base is 3 cm and the slant height is 5 cm, the side area can be calculated as: side area = π(3)(5) = 15π square centimeters.
answer:15
correct answer:2
total:1282  correct:359  acc:0.2800312012480499
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ID:1282
picturePath:image/2029.png
answer:The side area of a cone is given by the formula πrl, where r is the radius and l is the slant height. In this case, the radius is half the length of the base (BC), which is 2, and the slant height is the hypotenuse of the right triangle, which can be found using the Pythagorean theorem: AB^2 + BC^2 = AC^2. Substituting the given values, we get 5^2 + 4^2 = AC^2, so AC = √(5^2 + 4^2) = √(25 + 16) = √41. Therefore, the side area of the cone is π(2)(√41) = 2π√41 square units.
answer:41
correct answer:15
total:1283  correct:359  acc:0.2798129384255651
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ID:1283
picturePath:image/2030.png
answer:60°
answer:60
correct answer:180
total:1284  correct:359  acc:0.279595015576324
------------------------------------------------------------------------------------
ID:1284
picturePath:image/2032.png
answer:解:∵扇形ABE的弧长为AB=16,∴圆锥底面周长为16π,∴圆锥底面半径为\frac{16π}{2π}=8．故选:B．
answer:8
correct answer:4
total:1285  correct:359  acc:0.2793774319066148
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ID:1285
picturePath:image/2035.png
answer:132
answer:132
correct answer:64
total:1286  correct:359  acc:0.27916018662519443
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ID:1286
picturePath:image/2037.png
answer:6
answer:6
correct answer:3
total:1287  correct:359  acc:0.2789432789432789
------------------------------------------------------------------------------------
ID:1287
picturePath:image/2039.png
answer:6cm
answer:6
correct answer:2
total:1288  correct:359  acc:0.27872670807453415
------------------------------------------------------------------------------------
ID:1288
picturePath:image/2040.png
answer:解:∵圆锥的母线长AB=10cm,高AO=6cm,∴圆锥的底面半径OB=5cm,∴圆锥的底面周长=2πr=2π×5=10π(cm),圆锥的侧面积=\frac{1}{2}×10π×6=30π(cm²),圆锥的底面积=πr²=π×5²=25π(cm²),∴圆锥的面积=30π+25π=55π(cm²)．故选:C．
answer:55
correct answer:2
total:1289  correct:359  acc:0.27851047323506595
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ID:1289
picturePath:image/2043.png
answer:6cm.
answer:6
correct answer:2
total:1290  correct:359  acc:0.27829457364341087
------------------------------------------------------------------------------------
ID:1290
picturePath:image/2044.png
answer:8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√{3}/3=8√
answer:8
correct answer:15
total:1291  correct:359  acc:0.27807900852052675
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ID:1291
picturePath:image/2045.png
answer:6cm.
answer:6
correct answer:2
total:1292  correct:359  acc:0.27786377708978327
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ID:1292
picturePath:image/2046.png
answer:This is a geometry problem involving a quadrilateral and parallel lines. Since DC is parallel to AB, we can use the property of similar triangles to find the length of BD. Triangles ACD and ABD are similar, so their corresponding sides are proportional. We know that BC = 2, AB = AC = AD = 3, and DC is parallel to AB. Therefore, triangle ACD is similar to triangle ABD, and we can set up the proportion:

\[\frac{AD}{BD} = \frac{AC}{BC}\]

Substituting the given values, we have:

\[\frac{3}{BD} = \frac{3}{2}\]

Cross-multiplying, we get:

\[3 \times 2 = 3 \times BD\]

\[6 = 3BD\]

Dividing both sides by 3, we find:

\[BD = \frac{6}{3} = 2\]

Therefore, the length of BD is 2. The answer is A.
answer:2
correct answer:2
total:1293  correct:360  acc:0.27842227378190254
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ID:1293
picturePath:image/2050.png
answer:解:设这个纸帽的底面半径为xcm,圆锥的母线长为16cm,则扇形的弧长为π×16=16π(cm),根据题意,得
\frac{16π}{180}=\frac{90πx}{180},
解得x=8．
故选:A．
answer:8
correct answer:4
total:1294  correct:360  acc:0.2782071097372488
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ID:1294
picturePath:image/2053.png
answer:6.
answer:6
correct answer:5
total:1295  correct:360  acc:0.277992277992278
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ID:1295
picturePath:image/2054.png
answer:解:纸杯的侧面积为:π×5×15=75πcm^{2}．故选:A．
answer:2
correct answer:2
total:1296  correct:361  acc:0.2785493827160494
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ID:1296
picturePath:image/2056.png
answer:48πcm².
answer:48
correct answer:2
total:1297  correct:361  acc:0.27833461835003853
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ID:1297
picturePath:image/2057.png
answer:解:∵AB,AC为⊙O的两条弦,OD⊥AB于D,OE⊥AC于E,∴⁀{AD}=⁀{BD},⁀{AE}=⁀{CE},∴∠DOE=∠BOC=160°．故选:D．
answer:160
correct answer:100
total:1298  correct:361  acc:0.2781201848998459
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ID:1298
picturePath:image/2061.png
answer:6cm.
answer:6
correct answer:3
total:1299  correct:361  acc:0.2779060816012317
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ID:1299
picturePath:image/2062.png
answer:解:∵圆锥的底面半径为5,母线长为8,∴圆锥的底面周长=10π,高=8√{3},∴该圆锥的侧面积=\frac{1}{2}×10π×8√{3}=40√{3}π．故选:D．
answer:3
correct answer:40
total:1300  correct:361  acc:0.2776923076923077
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ID:1300
picturePath:image/2064.png
answer:2.
answer:2
correct answer:4
total:1301  correct:361  acc:0.27747886241352804
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ID:1301
picturePath:image/2066.png
answer:解:∵扇形的弧长为:\frac{60π×1}{180}=15π,∴圆锥的底面圆的周长为15π,∴圆锥的底面圆的半径为:\frac{15π}{2π}=7.5cm．故选:A．
answer:7.5
correct answer:6
total:1302  correct:361  acc:0.2772657450076805
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ID:1302
picturePath:image/2067.png
answer:解:∵PA和PB分别切⊙O于A和B点,∴∠OAP=∠OBP=90°,∵∠APB=60°,∴∠AOB=120°,∵圆锥的底面周长为6π,∴圆锥的底面半径为3,∴圆锥的高为:√{3^{2}-3^{2}}×\frac{120π}{180}=3√{3}．故选:C．
answer:3
correct answer:2
total:1303  correct:361  acc:0.27705295471987723
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ID:1303
picturePath:image/2068.png
answer:40°
answer:40
correct answer:40
total:1304  correct:362  acc:0.2776073619631902
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ID:1304
picturePath:image/2069.png
answer:解:圆锥的底面周长为:2π×10=20π(cm),∵圆锥的底面周长等于扇形的弧长,∴扇形的弧长为20πcm,∴扇形的面积为:\frac{20π×24}{360}=8π(cm²)．故选:A．
answer:8
correct answer:2
total:1305  correct:362  acc:0.27739463601532566
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ID:1305
picturePath:image/2071.png
answer:解:圆锥的底面周长=扇形的弧长=AB=16,
所以圆锥的底面半径r=\frac{AB}{2π}=\frac{16}{2π}=2(π≈3.14),
故选:B．
answer:3.14
correct answer:4
total:1306  correct:362  acc:0.2771822358346095
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ID:1306
picturePath:image/2076.png
answer:2.5*2*π+2*2π=5π+4π=9π(平方米)
answer:9
correct answer:5
total:1307  correct:362  acc:0.2769701606732976
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ID:1307
picturePath:image/2079.png
answer:解:∵∠APD=∠B+∠A,∠APD=76°,∠A=40°,∴∠B=76°-40°=36°,
故选:B．
answer:36
correct answer:36
total:1308  correct:363  acc:0.2775229357798165
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ID:1308
picturePath:image/2081.png
answer:解:∵小圆的半径为1cm,∴扇形的弧长为2π,∴圆锥的底面圆的半径为\frac{2π}{2π}=1cm,∴圆锥的母线长为√{2^{2}+1^{2}}=√{5}(cm),∴圆锥的底面周长为2π×1=2π(cm),∴圆锥的侧面积=\frac{1}{2}×2π×1×√{5}=2√{5}(cm^{2}),∴圆锥的全面积为2√{5}+π(cm^{2})．故选:D．
answer:2
correct answer:3
total:1309  correct:363  acc:0.2773109243697479
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ID:1309
picturePath:image/2085.png
answer:解:圆锥底面周长为:2πr=2×3.14×6=12.56cm,∵圆锥的侧面展开扇形的弧长等于圆锥底面周长,∴2πrl=12.56,即r=\frac{12.56}{2π×10}=\frac{6.28}{\pi}≈2cm．故选:A．
answer:2
correct answer:6
total:1310  correct:363  acc:0.2770992366412214
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ID:1310
picturePath:image/2088.png
answer:圆锥的底面积公式为：πr²，其中r为圆锥的底面半径。由于圆锥的母线长为5cm，根据勾股定理，圆锥的底面半径为√{5²-4²}=3cm。因此，圆锥的底面积为π×3²=9πcm²。
answer:9
correct answer:2
total:1311  correct:363  acc:0.2768878718535469
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ID:1311
picturePath:image/2089.png
answer:解:∵扇形AOB的半径为3cm,圆心角的度数为120°,∴扇形的弧长AOB=\frac{120π×3}{180}=2π,∴圆锥的底面圆的半径为r=\frac{2π}{2π}=\frac{1}{2}(cm),∴圆锥的母线长l=3cm,∴圆锥的侧面积=\frac{1}{2}×2π×3×\frac{1}{2}=1.5π(cm^{2})．故选:A．
answer:2
correct answer:2
total:1312  correct:364  acc:0.2774390243902439
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ID:1312
picturePath:image/2093.png
answer:解:∵AB=50cm,AD=14cm,∴BC=AB-AD=50-14=36(cm),∵AE=AF,DE=DF,∴∠AEP=∠AEP,∴AP平分∠BAC,∴∠BAP=\frac{1}{2}∠BAC=\frac{1}{2}×180°=90°,∴四边形ABEP是矩形,∴PE=AB=50cm,∴DE=PE-AD=50-14=36(cm),∴制作一把这样的纸伞需要油纸的面积=\frac{1}{2}×AB×DE=\frac{1}{2}×50×36=600(cm2)．故选:B．
answer:2
correct answer:2
total:1313  correct:365  acc:0.277989337395278
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ID:1313
picturePath:image/2096.png
answer:6cm.
answer:6
correct answer:2
total:1314  correct:365  acc:0.2777777777777778
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ID:1314
picturePath:image/2098.png
answer:解:设圆锥的底面半径为r,则扇形的弧长为2πr,∵圆锥的底面周长等于扇形的弧长,∴2πr=2π×8,解得:r=4．故圆锥的底面半径为4cm．故选:B．
answer:4
correct answer:2
total:1315  correct:365  acc:0.27756653992395436
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ID:1315
picturePath:image/2099.png
answer:6
answer:6
correct answer:2
total:1316  correct:365  acc:0.2773556231003039
------------------------------------------------------------------------------------
ID:1316
picturePath:image/2100.png
answer:圆锥的底面半径为\frac{1.5}{\pi}=0.5m, 所以圆锥的底面周长为2π×0.5=πm, 圆锥的侧面积=\frac{1}{2}×π×0.5×2.5=3.75πm², 所以做这把遮阳伞需用布料的面积是3.75πm².
answer:3.75
correct answer:2
total:1317  correct:365  acc:0.2771450265755505
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ID:1317
picturePath:image/2102.png
answer:The answer is 100.
answer:100
correct answer:100
total:1318  correct:366  acc:0.27769347496206376
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ID:1318
picturePath:image/2104.png
answer:圆锥的侧面积是底面周长乘以高，底面周长是$2\pi r$，高是$40-10=30$，所以侧面积是$2\pi \times 10 \times 30 = 600\pi$。
answer:600
correct answer:400
total:1319  correct:366  acc:0.27748294162244125
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ID:1319
picturePath:image/2105.png
answer:解:∵圆锥的母线长为50cm,高为30cm,∴圆锥的底面圆的半径为25cm,∴底面周长=2π×25=50π(cm),∴需要的铁皮面积=\frac{1}{2}×50π×30=750π(cm^{2})．故选:A．
answer:2
correct answer:2000
total:1320  correct:366  acc:0.2772727272727273
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ID:1320
picturePath:image/2106.png
answer:6cm.
answer:6
correct answer:2
total:1321  correct:366  acc:0.27706283118849356
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ID:1321
picturePath:image/2110.png
answer:解:∵在Rt△ABC中,∠ACB=90°,AC=4,BC=3,∴AB=√{AC^{2}+BC^{2}}=√{4^{2}+3^{2}}=5,∴该旋转体的侧面积=\frac{1}{2}×2π×5×4=10π．故选:C．
answer:10
correct answer:15
total:1322  correct:366  acc:0.27685325264750377
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ID:1322
picturePath:image/2116.png
answer:1.5π×15²+π×15×20=450π+300π=750π(平方厘米)；答:至少需要750π平方厘米.
answer:750
correct answer:2
total:1323  correct:366  acc:0.2766439909297052
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ID:1323
picturePath:image/2119.png
answer:圆锥的侧面积公式为：πrl，其中r为底面半径，l为母线长。已知母线长为2，底面圆的周长为3π，所以底面半径r=3π/(2π)=3/2。将r=3/2代入公式中，得到圆锥的侧面积为：π×2×(3/2)=3π。故选：B。
answer:3
correct answer:3
total:1324  correct:367  acc:0.277190332326284
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ID:1324
picturePath:image/2121.png
answer:6π.
answer:6
correct answer:2
total:1325  correct:367  acc:0.2769811320754717
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ID:1325
picturePath:image/2122.png
answer:6.
answer:6
correct answer:3
total:1326  correct:367  acc:0.27677224736048267
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ID:1326
picturePath:image/2124.png
answer:The answer is A.
ID:1327
picturePath:image/2125.png
answer:圆锥的侧面积公式为：πrl，其中r为底面半径，l为母线长。在这个问题中，底面半径OB=9，高OC=12，所以母线长为√{OB²+OC²}=√{9²+12²}=15。将r=9，l=15代入公式，得到侧面积为：π×9×15=135π。所以，这个圆锥形纸帽的侧面积为135π平方厘米。
answer:135
correct answer:135
total:1328  correct:368  acc:0.27710843373493976
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ID:1328
picturePath:image/2126.png
answer:解:∵扇形的面积=\frac{120π×6^{2}}{360}=12π,∴圆锥的侧面积=\frac{1}{2}×12π=6π．故选:A．
answer:6
correct answer:12
total:1329  correct:368  acc:0.2768999247554552
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ID:1329
picturePath:image/2129.png
answer:解:∵∠BAC=90°,AB=2,BC=5,∴AC=√{AB^{2}+BC^{2}}=√{2^{2}+5^{2}}=√{29},∴圆锥的侧面积=\frac{1}{2}•2π•5=5π,底面积为π•2^{2}=4π,∴圆锥的表面积为5π+4π=9π．故选:B．
answer:9
correct answer:14
total:1330  correct:368  acc:0.27669172932330827
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ID:1330
picturePath:image/2130.png
answer:360°
answer:360
correct answer:216
total:1331  correct:368  acc:0.2764838467317806
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ID:1331
picturePath:image/2132.png
answer:6cm.
answer:6
correct answer:2
total:1332  correct:368  acc:0.27627627627627627
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ID:1332
picturePath:image/2134.png
answer:6cm
answer:6
correct answer:2
total:1333  correct:368  acc:0.2760690172543136
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ID:1333
picturePath:image/2138.png
answer:解:∵∠C=90°,AB=5,AC=4,∴BC=√{AB^{2}-AC^{2}}=√{5^{2}-4^{2}}=3,∴圆锥的底面周长=2πr=2π×3=6π,∴圆锥的侧面积=\frac{1}{2}×6π×4=6π,故选:A．
answer:6
correct answer:15
total:1334  correct:368  acc:0.27586206896551724
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ID:1334
picturePath:image/2142.png
answer:0.5
answer:0.5
correct answer:4
total:1335  correct:368  acc:0.27565543071161047
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ID:1335
picturePath:image/2143.png
answer:解:∵圆锥的底面圆的半径为8,母线长为15,∴圆锥的底面周长=2×8π=16π,∴扇形的圆心角α的度数=圆锥的底面周长÷圆锥的母线长=16π÷15π=4(度)．故选:A．
answer:4
correct answer:192
total:1336  correct:368  acc:0.2754491017964072
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ID:1336
picturePath:image/2145.png
answer:解:∵圆锥的母线长是25cm,高是24cm,∴圆锥的底面半径是:25÷2=12.5(cm),∴圆锥的侧面积是:12.5π×24=300π(cm^{2})．故选:A．
answer:2
correct answer:2
total:1337  correct:369  acc:0.27599102468212416
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ID:1337
picturePath:image/2149.png
answer:解:∵圆锥的母线为10,∴底面圆的半径为5,∵底面圆的直径为12,∴底面圆的半径为6,∴圆锥的底面周长=12π,∴圆锥的侧面积=\frac{1}{2}×12π×10=60π．故选:A．
answer:60
correct answer:60
total:1338  correct:370  acc:0.2765321375186846
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ID:1338
picturePath:image/2153.png
answer:解:∵圆锥的底面半径OA=2,母线AB=3,∴圆锥的底面周长=2π×2=4π,∴这个圆锥的侧面积=\frac{1}{2}×4π×3=6π．故选:C．
answer:6
correct answer:6
total:1339  correct:371  acc:0.2770724421209858
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ID:1339
picturePath:image/2155.png
answer:解:圆锥的侧面积=\frac{1}{2}•2π•15•20=300π(cm^{2})；故选:A．
answer:2
correct answer:2
total:1340  correct:372  acc:0.27761194029850744
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ID:1340
picturePath:image/2156.png
answer:解:∵∠ACB与∠AOB是同弧所对的圆心角与圆周角,∠ACB=60°,∴∠AOB=60°．故选:A．
answer:60
correct answer:120
total:1341  correct:372  acc:0.27740492170022374
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ID:1341
picturePath:image/2158.png
answer:The answer is 20.
answer:20
correct answer:24
total:1342  correct:372  acc:0.27719821162444114
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ID:1342
picturePath:image/2159.png
answer:6cm
answer:6
correct answer:1
total:1343  correct:372  acc:0.27699180938198065
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ID:1343
picturePath:image/2161.png
answer:解:∵OA=30cm,∠AOB=120°,∴△OAB是等边三角形,∴OA=OB=AB=30cm,∴烟囱帽的底面圆的半径为:15cm．故选:B．
answer:15
correct answer:10
total:1344  correct:372  acc:0.2767857142857143
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ID:1344
picturePath:image/2164.png
answer:解:圆锥的底面周长=2πr=2π×3=6π,∵圆锥的母线长是5米,∴圆锥的侧面积=\frac{1}{2}lr=\frac{1}{2}×6π×5=15π,即做这把遮阳伞需用布料的面积是15π平方米．故选:A．
answer:15
correct answer:15
total:1345  correct:373  acc:0.27732342007434946
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ID:1345
picturePath:image/2165.png
answer:解:纸杯的侧面积为:2π×5×15=150πcm^{2}．故选:A．
answer:2
correct answer:75
total:1346  correct:373  acc:0.2771173848439822
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ID:1346
picturePath:image/2167.png
answer:解:∵AM的垂直平分线与⊙A交于B、C两点,∴BC=AB=AC,∵∠MBF=20°,∴∠MAB=∠MBF=20°,∴∠MBA=180°-20°-20°=140°,∴∠DMF=\frac{1}{2}∠MBA=\frac{1}{2}×140°=70°．故选:B．
answer:70
correct answer:30
total:1347  correct:373  acc:0.2769116555308092
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ID:1347
picturePath:image/2168.png
answer:解:∵圆锥的底面圆的直径AC=12分米,∴圆锥的底面圆的半径=6分米,∵伞骨AB长为9分米,∴圆锥的母线长=√{9²+6²}=15(分米),∴圆锥的侧面积=\frac{1}{2}•2π•6•15=90(平方分米)．答:制作这样的一把雨伞至少需要绸布面料90平方分米．故选:B．
answer:90
correct answer:54
total:1348  correct:373  acc:0.2767062314540059
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ID:1348
picturePath:image/2171.png
answer:2.5π×2=5π(平方米)
answer:5
correct answer:5
total:1349  correct:374  acc:0.27724240177909565
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ID:1349
picturePath:image/2172.png
answer:16πcm²
answer:16
correct answer:2
total:1350  correct:374  acc:0.277037037037037
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ID:1350
picturePath:image/2174.png
answer:400.
answer:400
correct answer:2
total:1351  correct:374  acc:0.2768319763138416
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ID:1351
picturePath:image/2175.png
answer:4cm.
answer:4
correct answer:2
total:1352  correct:374  acc:0.27662721893491127
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ID:1352
picturePath:image/2176.png
answer:解:∵扇形的圆心角α的度数为120°,∴弧长AB=\frac{120π×30}{180}=20π,∵扇形的半径AB为30cm,∴圆锥的底面周长为20π,∴圆锥的底面半径OB=\frac{20π}{2π}=10cm．故选:B．
answer:10
correct answer:10
total:1353  correct:375  acc:0.2771618625277162
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ID:1353
picturePath:image/2177.png
answer:解:∵圆锥的底面半径为6,母线长为10,∴圆锥的底面周长=12π×6=72π,∴圆锥的侧面展开扇形的弧长=圆锥的底面周长=72π,∴圆锥的侧面积=\frac{1}{2}×72π×10=360π．故选:A．
answer:360
correct answer:60
total:1354  correct:375  acc:0.27695716395864106
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ID:1354
picturePath:image/2178.png
answer:解:圆锥的母线长为2米,底面半径为1米,则底面周长=2π=2×3.14=6.28(m),高为√{2^{2}-1^{2}}=√{3^{2}-1^{2}}=√{2^{2}}=2(m),所以这个圆锥的侧面积=\frac{1}{2}×6.28×2=6.28(m^{2})．故选:A．
answer:2
correct answer:2
total:1355  correct:376  acc:0.2774907749077491
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ID:1355
picturePath:image/2182.png
answer:解:∵圆锥的母线长是13cm,高是12cm,∴圆锥的底面圆的半径是:13÷2=6(厘米),∴该圆锥的底面圆的面积是π×6²=36π(平方厘米)．故选:A．
answer:36
correct answer:2
total:1356  correct:376  acc:0.27728613569321536
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ID:1356
picturePath:image/2183.png
answer:4√{3}
answer:3
correct answer:3
total:1357  correct:377  acc:0.2778187177597642
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ID:1357
picturePath:image/2185.png
answer:The side area of a cone is given by the formula πrl, where r is the radius and l is the slant height. In this case, the radius is the length of the hypotenuse of the right triangle, which can be found using the Pythagorean theorem: r = √(AC² + BC²) = √(4² + 5²) = √(16 + 25) = √41. The slant height is the length of the hypotenuse, which is √(AC² + BC²) = √(4² + 5²) = √(16 + 25) = √41. Therefore, the side area of the cone is πrl = π√41 * √41 = 41π.
answer:41
correct answer:15
total:1358  correct:377  acc:0.2776141384388807
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ID:1358
picturePath:image/2188.png
answer:The side area of the cone can be calculated by the formula πrl, where r is the radius and l is the slant height of the cone. In this case, the radius is half of the base AB, which is 2.5, and the slant height is the hypotenuse of the right triangle ABC, which can be calculated using the Pythagorean theorem: l = √{AB² + AC²} = √{5² + 4²} = √{41}. Therefore, the side area of the cone is π * 2.5 * √{41} = 6.25√{41}π.
answer:41
correct answer:15
total:1359  correct:377  acc:0.27740986019131714
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ID:1359
picturePath:image/2190.png
answer:2\*π\*1=2π
answer:2
correct answer:2
total:1360  correct:378  acc:0.27794117647058825
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ID:1360
picturePath:image/2191.png
answer:解:∵圆锥的母线长是13cm,高是12cm,∴圆锥的底面圆的半径是:13÷2=6(厘米),∴圆锥的侧面积=π×6×12=72π(平方厘米)．故选:A．
answer:72
correct answer:2
total:1361  correct:378  acc:0.2777369581190301
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ID:1361
picturePath:image/2193.png
answer:2π+2π×1×2=6π
answer:6
correct answer:5
total:1362  correct:378  acc:0.2775330396475771
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ID:1362
picturePath:image/2194.png
answer:解:∵圆锥的底面半径为9cm,高为12cm,∴圆锥的母线长为√{9²+12²}=15(cm)∵圆锥的底面周长为2πR=2π×9=18π(cm),∴扇形薄纸板的弧长为18π,∴该扇形薄纸板的圆心角为\frac{18π×180}{180\pi}=180°．故选:A．
answer:180
correct answer:216
total:1363  correct:378  acc:0.2773294203961849
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ID:1363
picturePath:image/2195.png
answer:6
answer:6
correct answer:6
total:1364  correct:379  acc:0.27785923753665687
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ID:1364
picturePath:image/2196.png
answer:Since line AB is parallel to CD, we have \angle A + \angle C = 180^\circ. Substituting the given values, we get 40^\circ + 80^\circ = 120^\circ. Since the sum of the angles in a triangle is 180^\circ, we have \angle E + 40^\circ + 80^\circ = 180^\circ. Solving for \angle E, we get \angle E = 180^\circ - 40^\circ - 80^\circ = \boxed{60^\circ}.The answer is: 60
answer:60
correct answer:40
total:1365  correct:379  acc:0.27765567765567767
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ID:1365
picturePath:image/2201.png
answer:The answer is \frac{1}{2}
answer:2
correct answer:3
total:1366  correct:379  acc:0.2774524158125915
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ID:1366
picturePath:image/2203.png
answer:12cm
answer:12
correct answer:4
total:1367  correct:379  acc:0.27724945135332846
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ID:1367
picturePath:image/2206.png
answer:The answer is \frac{2}{\sqrt{3}}.
answer:3
correct answer:1
total:1368  correct:379  acc:0.277046783625731
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ID:1368
picturePath:image/2207.png
answer:圆锥的侧面积公式为：πrl，其中r为底面半径，l为母线长。将给定的值代入公式，得到：π×1×3=3π。所以，这个圆锥的侧面积为3π。
answer:3
correct answer:3
total:1369  correct:380  acc:0.27757487216946675
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ID:1369
picturePath:image/2208.png
answer:40°
answer:40
correct answer:70
total:1370  correct:380  acc:0.2773722627737226
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ID:1370
picturePath:image/2212.png
answer:解:圆锥的母线长为15cm,底面半径为6cm,则扇形的弧长为:2π×6=12π(cm),根据题意得:扇形面积=\frac{12π×15}{2}=90π(cm^{2})．故选:A．
answer:2
correct answer:90
total:1371  correct:380  acc:0.27716994894237784
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ID:1371
picturePath:image/2213.png
answer:6√3.
answer:3
correct answer:2
total:1372  correct:380  acc:0.27696793002915454
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ID:1372
picturePath:image/2214.png
answer:解:∵圆锥的底面圆的直径为5cm,∴底面半径为2.5cm,∵母线为8cm,∴圆锥的底面周长=2π×2.5=5πcm,∴圆锥部分的面积=\frac{1}{2}×5π×8=20π(cm^{2})．故选:A．
answer:2
correct answer:2
total:1373  correct:381  acc:0.27749453750910413
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ID:1373
picturePath:image/2219.png
answer:解:设底面半径为r,则扇形的弧长为:\frac{144π×10}{180}=8π,所以2πr=8π,解得:r=4．故选:B．
answer:4
correct answer:4
total:1374  correct:382  acc:0.2780203784570597
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ID:1374
picturePath:image/2220.png
answer:3.14×1²+2×3.14×1×2=3.14+6.28=9.42(平方米)
answer:9.42
correct answer:2
total:1375  correct:382  acc:0.2778181818181818
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ID:1375
picturePath:image/2221.png
answer:解:∵OB=6cm,OC=8cm,∴BC=√{OC^{2}+OB^{2}}=10(cm),∴这个圆锥漏斗的侧面积=\frac{1}{2}×10×8=40(cm^{2})．故选:B．
answer:2
correct answer:2
total:1376  correct:383  acc:0.27834302325581395
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ID:1376
picturePath:image/2222.png
answer:圆锥的侧面积公式为：πrl，其中r为底面半径，l为母线长。将已知的数值代入公式，得到：π×3×4=12π。因此，该圆锥的侧面积为12π平方厘米。
answer:12
correct answer:12
total:1377  correct:384  acc:0.2788671023965142
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ID:1377
picturePath:image/2223.png
answer:80°
answer:80
correct answer:80
total:1378  correct:385  acc:0.2793904208998549
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ID:1378
picturePath:image/2228.png
answer:Since OC || AB, we have \angle AOC = \angle A = 20^\circ. 

Since OC is the radius of the circle, we have \angle OAC = \angle OCA = 90^\circ. 

Therefore, \angle 1 = \angle OAC - \angle AOC = 90^\circ - 20^\circ = \boxed{70^\circ}.The answer is: 70
answer:70
correct answer:60
total:1379  correct:385  acc:0.27918781725888325
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ID:1379
picturePath:image/2230.png
answer:圆锥的侧面积公式为：πrl，其中r为底面半径，l为母线长。根据题意，底面半径r=3cm，高l=4cm，代入公式得：π×3×4=12πcm²。故选：B。
answer:12
correct answer:2
total:1380  correct:385  acc:0.27898550724637683
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ID:1380
picturePath:image/2231.png
answer:圆锥形帐篷顶的母线长AB=10cm,底面半径长BO=5m,则底面周长为2πr=2π×5=10πcm,侧面积=\frac{1}{2}×2πr×AB=\frac{1}{2}×2π×5×10=50πcm².
answer:50
correct answer:2
total:1381  correct:385  acc:0.278783490224475
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ID:1381
picturePath:image/2232.png
answer:圆锥的侧面积公式为：πrl，其中r为底面半径，l为母线长。将给定的数值代入公式，得到：π×5cm×10cm=50πcm²。所以，这个圆锥的侧面积为50πcm²。
answer:50
correct answer:2
total:1382  correct:385  acc:0.2785817655571635
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ID:1382
picturePath:image/2236.png
answer:6π。
answer:6
correct answer:6
total:1383  correct:386  acc:0.27910339840925524
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ID:1383
picturePath:image/2237.png
answer:解:∵∠ACB=45°,∴∠AOB=90°,∵OA=OB,∴△AOB是等腰三角形,∴AO=BO=\frac{1}{2}AB=\frac{1}{2}×10=5,∴AD=2AO=10．故选:A．
answer:10
correct answer:2
total:1384  correct:386  acc:0.27890173410404623
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ID:1384
picturePath:image/2238.png
answer:解:圆锥的侧面积=\frac{1}{2}•2π•8•20=160π(cm^{2})．答:制作这样的一顶圣诞帽至少需要的纸板面积是160πcm^{2}．故选:A．
answer:2
correct answer:2
total:1385  correct:387  acc:0.2794223826714801
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ID:1385
picturePath:image/2239.png
answer:圆锥的侧面积公式为：πrl，其中r为圆锥底面半径，l为圆锥母线长。

已知圆锥的母线AB长为5，高AO为4，根据勾股定理，可以得到圆锥底面半径r的值为√{5²-4²}=√{9}=3。

将r=3，l=5代入圆锥的侧面积公式πrl中，可以得到该圆锥的侧面积为：π×3×5=15π。

因此，该圆锥的侧面积为15π。
answer:15
correct answer:15
total:1386  correct:388  acc:0.27994227994227994
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ID:1386
picturePath:image/2243.png
answer:8.
answer:8
correct answer:5
total:1387  correct:388  acc:0.27974044700793077
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ID:1387
picturePath:image/2247.png
answer:解:∵⊙O的半径OC垂直于弦AB,∠AOC=50°,∴∠AOB=2∠AOC=100°,∴∠CDB=\frac{1}{2}∠AOB=50°．故选:A．
answer:50
correct answer:25
total:1388  correct:388  acc:0.27953890489913547
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ID:1388
picturePath:image/2250.png
answer:解:∵圆锥形烟囱帽的底面直径为80,母线长为50,∴底面半径为40,∴烟囱帽的侧面积=\frac{1}{2}•2π•40•50=π×40²=1600π．故选:A．
answer:1600
correct answer:2000
total:1389  correct:388  acc:0.279337652987761
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ID:1389
picturePath:image/2253.png
answer:The answer is 28.
answer:28
correct answer:48
total:1390  correct:388  acc:0.27913669064748203
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ID:1390
picturePath:image/2257.png
answer:This is a geometry problem involving circles and triangles. Given that \( \angle C = 90^\circ \) and \( CP \) is the diameter of the circle, we know that \( \angle CPD = 90^\circ \). This means that \( DP \) is the height of the triangle \( ACD \) and \( BP \) is the projection of \( DP \) onto \( AB \). The length of \( BP \) is minimized when \( DP \) is perpendicular to \( AB \), which occurs when \( D \) is the foot of the altitude from \( C \) to \( AB \). Since \( AC = 6 \), \( AB \) is the hypotenuse of right triangle \( ACD \), so \( AB = \sqrt{AC^2 + CD^2} = \sqrt{6^2 + CD^2} \). Since \( DP \) is the height of \( ACD \), we have \( CD = \sqrt{AC^2 - DP^2} = \sqrt{6^2 - 2^2} = 2\sqrt{5} \). Therefore, \( AB = \sqrt{6^2 + (2\sqrt{5})^2} = \sqrt{6^2 + 20} = \sqrt{36 + 20} = \sqrt{56} = 2\sqrt{14} \). Thus, the length of \( AB \) is \( 2\sqrt{14} \).
answer:14
correct answer:13
total:1391  correct:388  acc:0.27893601725377426
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ID:1391
picturePath:image/2277.png
answer:解:∵Rt△ABC的斜边AB=5,一条直角边AC=3,∴BC=√{5^{2}-3^{2}}=4,∴圆锥的底面周长=8π,∴圆锥的侧面积=\frac{1}{2}×8π×4=8π．故选:A．
answer:8
correct answer:15
total:1392  correct:388  acc:0.27873563218390807
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ID:1392
picturePath:image/2278.png
answer:140°
answer:140
correct answer:140
total:1393  correct:389  acc:0.27925340990667624
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ID:1393
picturePath:image/2279.png
answer:解:∵Rt△ABC的斜边AB=13cm,一条直角边AC=5cm,∴另一直角边BC=12cm,∴圆锥的底面圆的半径为6cm,母线长为12cm,∴这个圆锥的表面积=\frac{1}{2}×2π×6×12+π×6^{2}=144π+36π=180π≈5664(cm^{2})．故选:A．
answer:2
correct answer:90
total:1394  correct:389  acc:0.27905308464849354
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ID:1394
picturePath:image/2280.png
answer:本题考查圆锥的侧面展开图．圆锥的底面周长为$2πr=2π×3=6πcm$，所以圆锥的侧面展开图的面积为$S=\frac{1}{2}×6π×8=24πcm^{2}$．故选：$B$．
answer:2
correct answer:2
total:1395  correct:390  acc:0.27956989247311825
------------------------------------------------------------------------------------
ID:1395
picturePath:image/2282.png
answer:解:∵圆锥的母线长为30cm,底面直径为80cm,∴圆锥的底面半径为40cm,∴圆锥的侧面积=\frac{1}{2}×2π×40×30=400π(cm^{2})．故选:A．
answer:2
correct answer:2
total:1396  correct:391  acc:0.2800859598853868
------------------------------------------------------------------------------------
ID:1396
picturePath:image/2284.png
answer:12
answer:12
correct answer:10.2
total:1397  correct:391  acc:0.27988546886184684
------------------------------------------------------------------------------------
ID:1397
picturePath:image/2286.png
answer:解:∵竹竿与点A相距8m,与旗杆相距22m,∴竹竿与点A、旗杆构成两个相似三角形,∴\frac{旗杆的高度}{8}=\frac{3.2}{22},解得:旗杆的高度=16m．故选:A．
answer:16
correct answer:12
total:1398  correct:391  acc:0.27968526466380544
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ID:1398
picturePath:image/2287.png
answer:This is a right triangle PTQ with QS = 60m and ST = 120m. According to the Pythagorean theorem, PT^2 = QS^2 + ST^2, so PT = √(60^2 + 120^2) = √(3600 + 14400) = √18000 = 180m. The width of the river is PQ, which is PT - QS = 180 - 60 = 120m. Therefore, the width of the river is 120m. The answer is A.
answer:120
correct answer:120
total:1399  correct:392  acc:0.2802001429592566
------------------------------------------------------------------------------------
ID:1399
picturePath:image/2288.png
answer:2.1
answer:2.1
correct answer:3.5
total:1400  correct:392  acc:0.28
------------------------------------------------------------------------------------
ID:1400
picturePath:image/2291.png
answer:解:∵DF∥BC,∴△DEF∽△ABC,∴\frac{DE}{AB}=\frac{EF}{BC},即\frac{0.5}{AB}=\frac{0.25}{20},解得AB=10米．故选:A．
answer:10
correct answer:11.5
total:1401  correct:392  acc:0.2798001427551749
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ID:1401
picturePath:image/2292.png
answer:5
answer:5
correct answer:3
total:1402  correct:392  acc:0.2796005706134094
------------------------------------------------------------------------------------
ID:1402
picturePath:image/2293.png
answer:4.2cm
answer:4.2
correct answer:2.16
total:1403  correct:392  acc:0.2794012829650748
------------------------------------------------------------------------------------
ID:1403
picturePath:image/2294.png
answer:20m
answer:20
correct answer:6
total:1404  correct:392  acc:0.2792022792022792
------------------------------------------------------------------------------------
ID:1404
picturePath:image/2295.png
answer:解:∵AB∥DE,∴△ABD∽△ECD,∴\frac{AB}{DE}=\frac{CD}{BD},即\frac{AB}{3.2}=\frac{1.6}{7.8+3.2},∴AB≈12.5(米)．故选:B．
answer:12.5
correct answer:3.9
total:1405  correct:392  acc:0.27900355871886123
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ID:1405
picturePath:image/2296.png
answer:22m
answer:22
correct answer:9
total:1406  correct:392  acc:0.27880512091038406
------------------------------------------------------------------------------------
ID:1406
picturePath:image/2297.png
answer:解:∵DE∥AB,∴△DEF∽△ABC,∴\frac{EF}{AC}=\frac{DE}{AB},即\frac{15}{1.5}=\frac{30}{AB},解得AB=10．故选:B．
answer:10
correct answer:5
total:1407  correct:392  acc:0.27860696517412936
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ID:1407
picturePath:image/2298.png
answer:3.6米
answer:3.6
correct answer:6
total:1408  correct:392  acc:0.2784090909090909
------------------------------------------------------------------------------------
ID:1408
picturePath:image/2300.png
answer:16米
answer:16
correct answer:5
total:1409  correct:392  acc:0.2782114975159688
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ID:1409
picturePath:image/2301.png
answer:解:∵∠AOB=96°,∴∠ACB=\frac{1}{2}∠AOB=48°．故选:B．
answer:48
correct answer:132
total:1410  correct:392  acc:0.27801418439716313
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ID:1410
picturePath:image/2302.png
answer:60
answer:60
correct answer:48
total:1411  correct:392  acc:0.27781715095676823
------------------------------------------------------------------------------------
ID:1411
picturePath:image/2303.png
answer:3.5
answer:3.5
correct answer:8
total:1412  correct:392  acc:0.2776203966005666
------------------------------------------------------------------------------------
ID:1412
picturePath:image/2304.png
answer:为了估计河的宽度，我们可以使用勾股定理来计算河的对岸点A到点D的距离。设点A到点D的距离为x米。

根据勾股定理，我们有：
AB² = AD² - BD²
AC² = AD² - DC²
AD² = AB² + BD²
AD² = AC² + DC²

将BD=120m, DC=40m, AB⊥BC, 以及AB² = AD² - BD²和AC² = AD² - DC²代入上式，我们得到：
x² = 120² + 40²
x² = 14400 + 1600
x² = 16000
x = √16000
x ≈ 127m

因此，这条河的大致宽度是127米。
answer:127
correct answer:90
total:1413  correct:392  acc:0.2774239207360226
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ID:1413
picturePath:image/2305.png
answer:根据题意，我们可以得出以下信息：

- AO = 18cm
- BO = 21cm
- OC = 6cm
- OD = 7cm
- CD = 5cm

由于OC和OD是AO和BO的垂直平分线，CD是AB的垂直平分线，我们可以得出：

- AB = 2OC = 2 * 6cm = 12cm
- AB = 2OD = 2 * 7cm = 14cm
- AB = 2CD = 2 * 5cm = 10cm

因此，AB的值为12cm。
answer:12
correct answer:15
total:1414  correct:392  acc:0.27722772277227725
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ID:1414
picturePath:image/2306.png
answer:解:∵DE∥AB,∴△DEF∽△ABC,∴\frac{EF}{AC}=\frac{DE}{AB},即\frac{25}{1.6}=\frac{50}{AB},解得AB=40．故选:B．
answer:40
correct answer:6.6
total:1415  correct:392  acc:0.27703180212014133
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ID:1415
picturePath:image/2307.png
answer:解:∵DE⊥BC,∴△ADE∽△CBE,∴\frac{AB}{BC}=\frac{DE}{CE},∵AB=2m,BC=14m,DE=1.5m,∴\frac{2}{14}=\frac{1.5}{CE},解得:CE=9.6m,∴CD=CE-DE=9.6-1.5=8.1m．故选:A．
answer:8.1
correct answer:12
total:1416  correct:392  acc:0.2768361581920904
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ID:1416
picturePath:image/2308.png
answer:tanα=\frac{AC}{CD}=\frac{3}{11}
answer:11
correct answer:7
total:1417  correct:392  acc:0.2766407904022583
------------------------------------------------------------------------------------
ID:1417
picturePath:image/2309.png
answer:40
answer:40
correct answer:40
total:1418  correct:393  acc:0.2771509167842031
------------------------------------------------------------------------------------
ID:1418
picturePath:image/2310.png
answer:解:∵AB⊥BC,EC⊥BC,∴△ABD∽△ECD,∴\frac{AB}{EC}=\frac{BD}{CD},即\frac{AB}{50}=\frac{120}{60},解得:AB=100米．故选:B．
answer:100
correct answer:100
total:1419  correct:394  acc:0.2776603241719521
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ID:1419
picturePath:image/2311.png
answer:The answer is D.
ID:1420
picturePath:image/2312.png
answer:解:∵∠AOB=140°,∴∠AMO=\frac{1}{2}∠AOB=\frac{1}{2}×140°=70°,∵点M是⁀{AB}的中点,∴∠BOM=∠AMO=70°,∴∠N=180°-∠BOM=180°-70°=110°,
故选:A．
answer:110
correct answer:35
total:1421  correct:394  acc:0.2772695285010556
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ID:1421
picturePath:image/2313.png
answer:解:∵AD⊥DE,DE∥BC,∴△ADE∽△ABC,∴\frac{AD}{AB}=\frac{DE}{BC},即\frac{12}{AB}=\frac{40}{24},解得AB=20．故选:B．
answer:20
correct answer:18
total:1422  correct:394  acc:0.2770745428973277
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ID:1422
picturePath:image/2314.png
answer:70米
answer:70
correct answer:70
total:1423  correct:395  acc:0.27758257203092057
------------------------------------------------------------------------------------
ID:1423
picturePath:image/2315.png
answer:根据题意，学生站在点C处时，他头顶端的影子与旗杆的影子重合在点A处，说明此时学生、点A、点C、旗杆构成一个直角三角形，且∠ACB=90°。

设旗杆的高度为x米，则根据相似三角形的性质，有：

\[\frac{AC}{AB} = \frac{CB}{BC}\]

将已知的数值代入上式，得到：

\[\frac{1.6}{x} = \frac{20}{20+2}\]

解得：

\[x = \frac{1.6 \times 22}{20} = 19.2\]

因此，旗杆的高度为19.2米。

故选：B。
answer:19.2
correct answer:17.6
total:1424  correct:395  acc:0.2773876404494382
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ID:1424
picturePath:image/2316.png
answer:20√{3}m
answer:3
correct answer:3
total:1425  correct:396  acc:0.27789473684210525
------------------------------------------------------------------------------------
ID:1425
picturePath:image/2317.png
answer:The answer is 18.
answer:18
correct answer:10.8
total:1426  correct:396  acc:0.2776998597475456
------------------------------------------------------------------------------------
ID:1426
picturePath:image/2318.png
answer:2.6
answer:2.6
correct answer:4.8
total:1427  correct:396  acc:0.2775052557813595
------------------------------------------------------------------------------------
ID:1427
picturePath:image/2319.png
answer:0.15
answer:0.15
correct answer:0.5
total:1428  correct:396  acc:0.2773109243697479
------------------------------------------------------------------------------------
ID:1428
picturePath:image/2320.png
answer:10.8
answer:10.8
correct answer:6.4
total:1429  correct:396  acc:0.27711686494051785
------------------------------------------------------------------------------------
ID:1429
picturePath:image/2321.png
answer:6.
answer:6
correct answer:12
total:1430  correct:396  acc:0.27692307692307694
------------------------------------------------------------------------------------
ID:1430
picturePath:image/2322.png
answer:12m
answer:12
correct answer:12
total:1431  correct:397  acc:0.2774283717679944
------------------------------------------------------------------------------------
ID:1431
picturePath:image/2323.png
answer:解:∵AB为⊙O直径,∴∠ADB=90°,∵∠DAB=20°,∴∠ABD=90°-20°=70°,∴∠ACD=∠ABD=70°．故选:B．
answer:70
correct answer:110
total:1432  correct:397  acc:0.2772346368715084
------------------------------------------------------------------------------------
ID:1432
picturePath:image/2324.png
answer:3.6
answer:3.6
correct answer:8.5
total:1433  correct:397  acc:0.27704117236566644
------------------------------------------------------------------------------------
ID:1433
picturePath:image/2327.png
answer:0.6m
answer:0.6
correct answer:0.3
total:1434  correct:397  acc:0.27684797768479774
------------------------------------------------------------------------------------
ID:1434
picturePath:image/2329.png
answer:2.4m
answer:2.4
correct answer:10
total:1435  correct:397  acc:0.27665505226480835
------------------------------------------------------------------------------------
ID:1435
picturePath:image/2330.png
answer:The answer is 12.
answer:12
correct answer:10
total:1436  correct:397  acc:0.2764623955431755
------------------------------------------------------------------------------------
ID:1436
picturePath:image/2331.png
answer:解:∵EF∥AB,∴△EFG∽△ABD,∴\frac{EF}{AB}=\frac{FG}{BD},即\frac{60}{AB}=\frac{30}{8},解得AB=120cm．故选:A．
answer:120
correct answer:5.6
total:1437  correct:397  acc:0.2762700069589422
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ID:1437
picturePath:image/2332.png
answer:The answer is 88.
answer:88
correct answer:88
total:1438  correct:398  acc:0.2767732962447844
------------------------------------------------------------------------------------
ID:1438
picturePath:image/2333.png
answer:解:∵BC∥PQ,∴△ABC∽△APQ,∴\frac{AB}{AP}=\frac{BC}{PQ},∵AB:AP=2:5,BC=20cm,∴\frac{2}{5}=\frac{20}{PQ},解得PQ=40cm．故选:D．
answer:40
correct answer:50
total:1439  correct:398  acc:0.27658095899930507
------------------------------------------------------------------------------------
ID:1439
picturePath:image/2334.png
answer:3.6米
answer:3.6
correct answer:4.5
total:1440  correct:398  acc:0.2763888888888889
------------------------------------------------------------------------------------
ID:1440
picturePath:image/2335.png
answer:4.5
answer:4.5
correct answer:4
total:1441  correct:398  acc:0.2761970853573907
------------------------------------------------------------------------------------
ID:1441
picturePath:image/2337.png
answer:根据题意，AB∥DE，所以△ADE∽△ABC，则\frac{DE}{DC}=\frac{AC}{AB}，即\frac{DE}{30}=\frac{5}{6}，解得DE=25m.
answer:25
correct answer:36
total:1442  correct:398  acc:0.27600554785020803
------------------------------------------------------------------------------------
ID:1442
picturePath:image/2338.png
answer:20mm
answer:20
correct answer:3
total:1443  correct:398  acc:0.2758142758142758
------------------------------------------------------------------------------------
ID:1443
picturePath:image/2339.png
answer:0.25m
answer:0.25
correct answer:8
total:1444  correct:398  acc:0.2756232686980609
------------------------------------------------------------------------------------
ID:1444
picturePath:image/2340.png
answer:解:∵AB∥CD,∴△ABC∽△ADE,∴\frac{AB}{AD}=\frac{BC}{DE},即\frac{10}{AD}=\frac{30}{20},解得AD=15m．故选:B．
answer:15
correct answer:20
total:1445  correct:398  acc:0.2754325259515571
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ID:1445
picturePath:image/2342.png
answer:根据相似三角形的性质，我们可以得到：

\[\frac{AB}{BE} = \frac{CD}{DE}\]

\[\frac{2}{1.5} = \frac{CD}{DE}\]

\[\frac{4}{3} = \frac{CD}{DE}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{CD}{8 - CD}\]

\[\frac{4}{3} = \frac{
answer:3
correct answer:7.5
total:1446  correct:398  acc:0.2752420470262794
------------------------------------------------------------------------------------
ID:1446
picturePath:image/2343.png
answer:60
answer:60
correct answer:30
total:1447  correct:398  acc:0.2750518313752592
------------------------------------------------------------------------------------
ID:1447
picturePath:image/2344.png
answer:2.5m
answer:2.5
correct answer:3.5
total:1448  correct:398  acc:0.27486187845303867
------------------------------------------------------------------------------------
ID:1448
picturePath:image/2346.png
answer:4.8米
answer:4.8
correct answer:4.2
total:1449  correct:398  acc:0.274672187715666
------------------------------------------------------------------------------------
ID:1449
picturePath:image/2347.png
answer:5
answer:5
correct answer:2.4
total:1450  correct:398  acc:0.27448275862068966
------------------------------------------------------------------------------------
ID:1450
picturePath:image/2348.png
answer:4.8
answer:4.8
correct answer:4.45
total:1451  correct:398  acc:0.2742935906271537
------------------------------------------------------------------------------------
ID:1451
picturePath:image/2349.png
answer:This is a right triangle problem. We can use the Pythagorean theorem to solve it. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is the distance between the flagpole and the point where the shadows meet, which is 8m + 22m = 30m. The other two sides are the height of the flagpole and the length of the shadow it casts, which is 3.2m. So, we have:

30^2 = 3.2^2 + h^2

where h is the height of the flagpole. Solving for h, we get:

h = √(30^2 - 3.2^2) = √(900 - 9.6) = √890.4 ≈ 29.5m

Therefore, the height of the flagpole is approximately 29.5m. The answer is:
answer:29.5
correct answer:12
total:1452  correct:398  acc:0.2741046831955923
------------------------------------------------------------------------------------
ID:1452
picturePath:image/2350.png
answer:0.25m
answer:0.25
correct answer:8
total:1453  correct:398  acc:0.27391603578802476
------------------------------------------------------------------------------------
ID:1453
picturePath:image/2351.png
answer:8
answer:8
correct answer:3.2
total:1454  correct:398  acc:0.2737276478679505
------------------------------------------------------------------------------------
ID:1454
picturePath:image/2355.png
answer:1.8
answer:1.8
correct answer:6
total:1455  correct:398  acc:0.27353951890034367
------------------------------------------------------------------------------------
ID:1455
picturePath:image/2356.png
answer:This is a geometry problem involving shadows and parallel lines. Since AB is parallel to CD and the light source P is directly above AB, the shadows CD and AB are also parallel. Therefore, triangle PAB is similar to triangle PCD by the AA (Angle-Angle) similarity criterion. This means that the corresponding sides of the two triangles are proportional. 

We can set up a proportion using the given information: 

\[\frac{AB}{CD} = \frac{PA}{PC}\]

Substituting the given values, we have:

\[\frac{2}{5} = \frac{3}{PC}\]

Solving for PC, we get:

\[PC = \frac{3 \times 5}{2} = \frac{15}{2} = 7.5\]

Therefore, the distance between AB and CD is 7.5 meters. 

The answer is: 7.5
answer:7.5
correct answer:5
total:1456  correct:398  acc:0.2733516483516483
------------------------------------------------------------------------------------
ID:1456
picturePath:image/2358.png
answer:根据题意,蜡烛到纸筒的距离为20cm,纸筒的长度为15cm,像的高度为4cm,则有:

\frac{20}{15}=\frac{4}{x}

解得:x=12.5cm

因此,蜡烛与纸筒的距离应该为12.5cm.
answer:12.5
correct answer:75
total:1457  correct:398  acc:0.2731640356897735
------------------------------------------------------------------------------------
ID:1457
picturePath:image/2360.png
answer:解:∵AB∥DE,∴△ABE∽△DCE,∴\frac{AB}{DE}=\frac{BE}{CE},即\frac{0.3}{1.6}=\frac{BE}{CE+0.6},解得:BE=\frac{0.3×1.6}{1.6+0.6}=\frac{0.3×1.6}{2.2}=0.15(米),∴捣头点E上升了0.15-0.3=-0.15(米)．故选:A．
answer:0.15
correct answer:0.8
total:1458  correct:398  acc:0.2729766803840878
------------------------------------------------------------------------------------
ID:1458
picturePath:image/2361.png
answer:6cm
answer:6
correct answer:8
total:1459  correct:398  acc:0.2727895819054147
------------------------------------------------------------------------------------
ID:1459
picturePath:image/2362.png
answer:0.5m
answer:0.5
correct answer:4
total:1460  correct:398  acc:0.2726027397260274
------------------------------------------------------------------------------------
ID:1460
picturePath:image/2363.png
answer:4.8
answer:4.8
correct answer:6
total:1461  correct:398  acc:0.2724161533196441
------------------------------------------------------------------------------------
ID:1461
picturePath:image/2365.png
answer:解:∵∠C=35°,∠ADC=85°,∴∠B=∠ADC-∠C=85°-35°=50°,∵OA=OB,∴∠A=∠B=50°,
故选:B．
answer:50
correct answer:60
total:1462  correct:398  acc:0.2722298221614227
------------------------------------------------------------------------------------
ID:1462
picturePath:image/2366.png
answer:解:∵AB=4m,CA=0.8m,∴BC=AB-AC=4-0.8=3.2(m),又∵当走到C点时,她的影子顶端正好与树的影子顶端重合,∴△ABC为直角三角形,∴tan∠ACB=\frac{CA}{CB}=\frac{0.8}{3.2}=\frac{2}{7},∴tan∠ABC=tan(180°-∠ACB)=\frac{7}{2},即\frac{AC}{CB}=\frac{7}{2},又∵AC=0.8m,∴\frac{0.8}{CB}=\frac{7}{2},解得:CB=\frac{2.8}{7}m,则树的高度为\frac{2.8}{7}+0.8=1.6m．故选:A．
answer:1.6
correct answer:8
total:1463  correct:398  acc:0.27204374572795625
------------------------------------------------------------------------------------
ID:1463
picturePath:image/2367.png
answer:3.6m
answer:3.6
correct answer:8
total:1464  correct:398  acc:0.2718579234972678
------------------------------------------------------------------------------------
ID:1464
picturePath:image/2369.png
answer:解:∵竹竿与影子的长度比为2:6,∴竹竿的影长为2m:6m=1:3,∴竹竿的影长为15m×\frac{1}{3}=5m,∴树的高度为15m+5m=20m．故选:D．
answer:20
correct answer:7
total:1465  correct:398  acc:0.2716723549488055
------------------------------------------------------------------------------------
ID:1465
picturePath:image/2373.png
answer:1.5m
answer:1.5
correct answer:2
total:1466  correct:398  acc:0.27148703956343795
------------------------------------------------------------------------------------
ID:1466
picturePath:image/2374.png
answer:解:∵同一时刻,物高与影长成正比,
又∵小明的身高为1.5米,他的影长为1米,
∴小明的身高与影长的比为1.5:1=3:2,
又∵槟榔树的影长为5米,
∴这棵槟榔树的高为5×3=15(米);
故选:B．
answer:15
correct answer:7.5
total:1467  correct:398  acc:0.2713019768234492
------------------------------------------------------------------------------------
ID:1467
picturePath:image/2375.png
answer:1.52m
answer:1.52
correct answer:3.42
total:1468  correct:398  acc:0.27111716621253407
------------------------------------------------------------------------------------
ID:1468
picturePath:image/2376.png
answer:2∠A=2×50°=100°,∵OB=OC,∴∠OBC=\frac{180°-100°}{2}=40°.故选:B.
answer:40
correct answer:40
total:1469  correct:399  acc:0.27161334240980256
------------------------------------------------------------------------------------
ID:1469
picturePath:image/2377.png
answer:解:∵同一时刻物高与影长成正比,∴\frac{2}{3}=\frac{BA}{AC},解得:BA=10米,∴楼高为10+2=12(米)．故选:D．
answer:12
correct answer:10
total:1470  correct:399  acc:0.2714285714285714
------------------------------------------------------------------------------------
ID:1470
picturePath:image/2378.png
answer:1.44πm²
answer:1.44
correct answer:2
total:1471  correct:399  acc:0.27124405166553367
------------------------------------------------------------------------------------
ID:1471
picturePath:image/2379.png
answer:根据题意，AB⊥BC，CD⊥BC，点E在BC上，并且点A,E,D在同一条直线上。因此，△ABE和△CDE是相似三角形。

相似三角形的对应边成比例，即\frac{AB}{CE}=\frac{BE}{CD}。将已知数值代入，得到\frac{AB}{10}=\frac{20}{20}，解得AB=20m。

因此，河的宽度AB等于20m。

答案：20m。
answer:20
correct answer:40
total:1472  correct:399  acc:0.2710597826086957
------------------------------------------------------------------------------------
ID:1472
picturePath:image/2381.png
answer:解:∵DE=30cm,EF=20cm,∴DF=√{DE^{2}+EF^{2}}=√{30^{2}+20^{2}}=40cm,∴DF:DG=40:1.4,∴AB=40×1.4=56cm．故选:B．
answer:56
correct answer:5.4
total:1473  correct:399  acc:0.2708757637474542
------------------------------------------------------------------------------------
ID:1473
picturePath:image/2382.png
answer:解:∵竹竿与旗杆均垂直于地面,∴△ABD∽△ACE,∴\frac{AB}{AC}=\frac{BD}{DE},即\frac{4}{4+22}=\frac{8}{x},解得x=24．故选:B．
answer:24
correct answer:15
total:1474  correct:399  acc:0.2706919945725916
------------------------------------------------------------------------------------
ID:1474
picturePath:image/2383.png
answer:3
answer:3
correct answer:2
total:1475  correct:399  acc:0.2705084745762712
------------------------------------------------------------------------------------
ID:1475
picturePath:image/2386.png
answer:13m
answer:13
correct answer:7
total:1476  correct:399  acc:0.2703252032520325
------------------------------------------------------------------------------------
ID:1476
picturePath:image/2389.png
answer:15
answer:15
correct answer:48
total:1477  correct:399  acc:0.27014218009478674
------------------------------------------------------------------------------------
ID:1477
picturePath:image/2390.png
answer:3.2×1.6+0.8×3.2=6.4+2.6=9m
answer:9
correct answer:8
total:1478  correct:399  acc:0.2699594046008119
------------------------------------------------------------------------------------
ID:1478
picturePath:image/2392.png
answer:4.8
answer:4.8
correct answer:2.4
total:1479  correct:399  acc:0.2697768762677485
------------------------------------------------------------------------------------
ID:1479
picturePath:image/2393.png
answer:1.5×900=1350(厘米)．故选:D．
answer:1350
correct answer:1200
total:1480  correct:399  acc:0.2695945945945946
------------------------------------------------------------------------------------
ID:1480
picturePath:image/2396.png
answer:60cm.
answer:60
correct answer:48
total:1481  correct:399  acc:0.26941255908170153
------------------------------------------------------------------------------------
ID:1481
picturePath:image/2397.png
answer:5
answer:5
correct answer:2.4
total:1482  correct:399  acc:0.2692307692307692
------------------------------------------------------------------------------------
ID:1482
picturePath:image/2399.png
answer:50°
answer:50
correct answer:40
total:1483  correct:399  acc:0.26904922454484154
------------------------------------------------------------------------------------
ID:1483
picturePath:image/2400.png
answer:解:设旗杆的高度为x米,
则\frac{小虎的身高}{小虎的影长}=\frac{旗杆的高度}{旗杆的影长},
即\frac{1.60}{2}=\frac{x}{20},
解得x=8.40．
故选:A．
answer:8.40
correct answer:16
total:1484  correct:399  acc:0.2688679245283019
------------------------------------------------------------------------------------
ID:1484
picturePath:image/2402.png
answer:3.2m
answer:3.2
correct answer:8
total:1485  correct:399  acc:0.2686868686868687
------------------------------------------------------------------------------------
ID:1485
picturePath:image/2403.png
answer:解:∵CD=1.5米,EF=2.5米,CF=2.5米,∴CF+EF=CD+AF,即2.5+2.5=1.5+AF,解得AF=5米,∵C、F、B在同一直线上,D、E、A也在同一直线上,∴AC∥BF,∴△ACF∽△BFA,∴\frac{AC}{AB}=\frac{CF}{AF},即\frac{40}{AB}=\frac{2.5}{5},解得AB=20米．故选:A．
answer:20
correct answer:17.5
total:1486  correct:399  acc:0.26850605652759085
------------------------------------------------------------------------------------
ID:1486
picturePath:image/2405.png
answer:20
answer:20
correct answer:40
total:1487  correct:399  acc:0.26832548755884333
------------------------------------------------------------------------------------
ID:1487
picturePath:image/2406.png
answer:20米
answer:20
correct answer:32
total:1488  correct:399  acc:0.26814516129032256
------------------------------------------------------------------------------------
ID:1488
picturePath:image/2407.png
answer:4.8
answer:4.8
correct answer:6
total:1489  correct:399  acc:0.2679650772330423
------------------------------------------------------------------------------------
ID:1489
picturePath:image/2408.png
answer:8.4
answer:8.4
correct answer:7.5
total:1490  correct:399  acc:0.2677852348993289
------------------------------------------------------------------------------------
ID:1490
picturePath:image/2409.png
answer:1.5m
answer:1.5
correct answer:2.6
total:1491  correct:399  acc:0.2676056338028169
------------------------------------------------------------------------------------
ID:1491
picturePath:image/2410.png
answer:根据圆周角定理，圆上同弧所对的圆周角的度数等于它所对的圆心角度数的一半。因此，∠ABC=\frac{1}{2}∠BAC=\frac{1}{2}×42°=21°。
answer:21
correct answer:48
total:1492  correct:399  acc:0.26742627345844505
------------------------------------------------------------------------------------
ID:1492
picturePath:image/2411.png
answer:15
answer:15
correct answer:20
total:1493  correct:399  acc:0.2672471533824514
------------------------------------------------------------------------------------
ID:1493
picturePath:image/2412.png
answer:This is a geometry problem that involves the use of similar triangles to solve for the height of the electric pole. Given that the shadow of the electric pole falls on the center of the advertisement board and the advertisement board's shadow falls on the ground, it can be inferred that the electric pole, the advertisement board, and the ground form a right triangle. The height of the electric pole can be found by using the formula for the area of a triangle, which is 1/2 * base * height. In this case, the base is the length of the advertisement board (2 meters) and the height is the unknown height of the electric pole. The area of the triangle formed by the electric pole, the advertisement board, and the ground can be calculated using the formula for the area of a triangle, which is 1/2 * base * height. The area of the triangle formed by the advertisement board and the ground can also be calculated using the formula for the area of a triangle, which is 1/2 * base * height. By setting these two areas equal to each other, the height of the electric pole can be solved for.
answer:2
correct answer:5
total:1494  correct:399  acc:0.26706827309236947
------------------------------------------------------------------------------------
ID:1494
picturePath:image/2413.png
answer:解:∵AB=3.2m,AC=8m,BC=22m,∴\frac{AB}{AC}=\frac{BC}{AC+BC},即\frac{3.2}{8}=\frac{22}{8+22},解得:AC=16m,∴BC=AC+AB=16+3.2=19.2m,则旗杆的高为19.2m．故选:C．
answer:19.2
correct answer:12
total:1495  correct:399  acc:0.26688963210702343
------------------------------------------------------------------------------------
ID:1495
picturePath:image/2414.png
answer:6.5
answer:6.5
correct answer:7.5
total:1496  correct:399  acc:0.2667112299465241
------------------------------------------------------------------------------------
ID:1496
picturePath:image/2415.png
answer:2.4
answer:2.4
correct answer:3.6
total:1497  correct:399  acc:0.2665330661322645
------------------------------------------------------------------------------------
ID:1497
picturePath:image/2418.png
answer:解:∵竹竿与旗杆的影子落在地面的同一点,∴CD∥BE,∴△ACD∽△ABE,∴\frac{CD}{BE}=\frac{AC}{AB},即\frac{3}{24}=\frac{8}{x},解得x=24m．故选:B．
answer:24
correct answer:12
total:1498  correct:399  acc:0.26635514018691586
------------------------------------------------------------------------------------
ID:1498
picturePath:image/2419.png
answer:解:∵竹竿与旗杆均垂直于地面,且竹竿与旗杆顶端的影子恰好落在地面的同一点,∴竹竿与旗杆均与地面的夹角相等,设竹竿与地面夹角为α,则cosα=\frac{8}{√{2^{2}+8^{2}}}=\frac{8}{10}=\frac{4}{5},∴sinα=\frac{3}{5},∴AB=ABsinα=\frac{3}{5}×22=13.2(m)．故选:A．
answer:13.2
correct answer:10.5
total:1499  correct:399  acc:0.26617745163442297
------------------------------------------------------------------------------------
ID:1499
picturePath:image/2420.png
answer:10
answer:10
correct answer:5
total:1500  correct:399  acc:0.266
------------------------------------------------------------------------------------
ID:1500
picturePath:image/2421.png
answer:解:∵∠ADC=∠A+∠BOC,∠ADC=85°,∠A=60°,∴∠BOC=85°-60°=25°,∴∠C=∠BOC=25°．故选:B．
answer:25
correct answer:35
total:1501  correct:399  acc:0.26582278481012656
------------------------------------------------------------------------------------
ID:1501
picturePath:image/2422.png
answer:4m.
answer:4
correct answer:0.5
total:1502  correct:399  acc:0.26564580559254325
------------------------------------------------------------------------------------
ID:1502
picturePath:image/2424.png
answer:8
answer:8
correct answer:7
total:1503  correct:399  acc:0.2654690618762475
------------------------------------------------------------------------------------
ID:1503
picturePath:image/2427.png
answer:This is a geometry problem involving similar triangles. Given that the student's height is 1.6m, her shadow is AB, and the tree's shadow is BC, we can set up a proportion to find the height of the tree. Since the shadows are similar, we have:

\[\frac{AB}{BC} = \frac{1.6}{CD}\]

Substituting the given values, we get:

\[\frac{0.8}{3.2} = \frac{1.6}{CD}\]

Simplifying, we find:

\[\frac{1}{4} = \frac{1.6}{CD}\]

Cross-multiplying, we get:

\[CD = \frac{1.6 \times 4}{1}\]

\[CD = 6.4\]

Therefore, the height of the tree CD is 6.4m.
answer:6.4
correct answer:8
total:1504  correct:399  acc:0.26529255319148937
------------------------------------------------------------------------------------
ID:1504
picturePath:image/2430.png
answer:This is a geometry problem involving similar triangles. Since the shadow of point C falls on the same horizontal line as point A, we can conclude that triangle ACE is similar to triangle ABE. Therefore, we can set up the following proportion:

\[\frac{CE}{AB} = \frac{AC}{AE}\]

Substituting the given values, we have:

\[\frac{2}{AB} = \frac{4}{4+1.6}\]

Simplifying the equation, we get:

\[\frac{2}{AB} = \frac{4}{5.6}\]

Cross-multiplying, we have:

\[2 \cdot 5.6 = 4 \cdot AB\]

\[AB = \frac{2 \cdot 5.6}{4}\]

\[AB = 3.8\]

Therefore, the height of the lamp post AB is approximately 3.8 meters.
answer:3.8
correct answer:4.8
total:1505  correct:399  acc:0.2651162790697674
------------------------------------------------------------------------------------
ID:1505
picturePath:image/2431.png
answer:3.2米
answer:3.2
correct answer:3
total:1506  correct:399  acc:0.2649402390438247
------------------------------------------------------------------------------------
ID:1506
picturePath:image/2432.png
answer:100
answer:100
correct answer:140
total:1507  correct:399  acc:0.2647644326476443
------------------------------------------------------------------------------------
ID:1507
picturePath:image/2433.png
answer:12m
answer:12
correct answer:6
total:1508  correct:399  acc:0.2645888594164456
------------------------------------------------------------------------------------
ID:1508
picturePath:image/2434.png
answer:4.2米
answer:4.2
correct answer:6.00
total:1509  correct:399  acc:0.2644135188866799
------------------------------------------------------------------------------------
ID:1509
picturePath:image/2436.png
answer:6
answer:6
correct answer:9
total:1510  correct:399  acc:0.2642384105960265
------------------------------------------------------------------------------------
ID:1510
picturePath:image/2437.png
answer:0.4米
answer:0.4
correct answer:6
total:1511  correct:399  acc:0.2640635340833885
------------------------------------------------------------------------------------
ID:1511
picturePath:image/2438.png
answer:22m
answer:22
correct answer:12
total:1512  correct:399  acc:0.2638888888888889
------------------------------------------------------------------------------------
ID:1512
picturePath:image/2439.png
answer:4m
answer:4
correct answer:6
total:1513  correct:399  acc:0.2637144745538665
------------------------------------------------------------------------------------
ID:1513
picturePath:image/2440.png
answer:3
answer:3
correct answer:3
total:1514  correct:400  acc:0.26420079260237783
------------------------------------------------------------------------------------
ID:1514
picturePath:image/2441.png
answer:6
answer:6
correct answer:9.6
total:1515  correct:400  acc:0.264026402640264
------------------------------------------------------------------------------------
ID:1515
picturePath:image/2442.png
answer:解:∵木棒插入部分AB的长为100cm,木棒上沾油部分DB的长为60cm,∴BC=AB-DB=100-60=40cm,∵桶高AC为80cm,∴桶内油面CE的高度是:AC-BC=80-40=40cm．故选:B．
answer:40
correct answer:48
total:1516  correct:400  acc:0.2638522427440633
------------------------------------------------------------------------------------
ID:1516
picturePath:image/2443.png
answer:The answer is 50.
answer:50
correct answer:50
total:1517  correct:401  acc:0.2643375082399473
------------------------------------------------------------------------------------
ID:1517
picturePath:image/2445.png
answer:6
answer:6
correct answer:3
total:1518  correct:401  acc:0.264163372859025
------------------------------------------------------------------------------------
ID:1518
picturePath:image/2446.png
answer:9
answer:9
correct answer:7
total:1519  correct:401  acc:0.26398946675444374
------------------------------------------------------------------------------------
ID:1519
picturePath:image/2447.png
answer:1.6米
answer:1.6
correct answer:8
total:1520  correct:401  acc:0.2638157894736842
------------------------------------------------------------------------------------
ID:1520
picturePath:image/2448.png
answer:解:∵AB∥CD,∴△ABD∽△ACD,∴\frac{AB}{AC}=\frac{BD}{CD},即\frac{1}{3}=\frac{1.2}{CD},解得CD=2.4米,∴S_{阴影}=\frac{1}{2}×π×(2.4)^{2}=7.44π(平方米)．故选:B．
answer:7.44
correct answer:0.81
total:1521  correct:401  acc:0.26364234056541747
------------------------------------------------------------------------------------
ID:1521
picturePath:image/2451.png
answer:This is a right triangle problem. We can use the Pythagorean theorem to solve it. The distance from the student to the tree is the hypotenuse of the triangle, and the height of the tree is one of the legs. The distance from the tree's reflection to the student is the other leg. The Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. In this case, we have:

20^2 = 5^2 + h^2

where h is the height of the tree. Solving for h, we get:

h = √(20^2 - 5^2) = √(400 - 25) = √375 ≈ 19.25 m

So, the height of the tree is approximately 19.25 meters.
answer:19.25
correct answer:5.1
total:1522  correct:401  acc:0.26346911957950064
------------------------------------------------------------------------------------
ID:1522
picturePath:image/2452.png
answer:a=5m, b=3m, 3m/5m = 3/5, 3/5 * a = b, 3/5 * 5m = b, b = 3m.
answer:3
correct answer:3
total:1523  correct:402  acc:0.2639527248850952
------------------------------------------------------------------------------------
ID:1523
picturePath:image/2455.png
answer:0.04米
answer:0.04
correct answer:0.3
total:1524  correct:402  acc:0.2637795275590551
------------------------------------------------------------------------------------
ID:1524
picturePath:image/2456.png
answer:0.15
answer:0.15
correct answer:0.8
total:1525  correct:402  acc:0.2636065573770492
------------------------------------------------------------------------------------
ID:1525
picturePath:image/2457.png
answer:8.4
answer:8.4
correct answer:7.5
total:1526  correct:402  acc:0.2634338138925295
------------------------------------------------------------------------------------
ID:1526
picturePath:image/2459.png
answer:20
answer:20
correct answer:15
total:1527  correct:402  acc:0.2632612966601179
------------------------------------------------------------------------------------
ID:1527
picturePath:image/2460.png
answer:6.6
answer:6.6
correct answer:8
total:1528  correct:402  acc:0.2630890052356021
------------------------------------------------------------------------------------
ID:1528
picturePath:image/2461.png
answer:180厘米
answer:180
correct answer:6
total:1529  correct:402  acc:0.262916939175932
------------------------------------------------------------------------------------
ID:1529
picturePath:image/2462.png
answer:5.2米
answer:5.2
correct answer:5.6
total:1530  correct:402  acc:0.2627450980392157
------------------------------------------------------------------------------------
ID:1530
picturePath:image/2463.png
answer:S = πr² = π*0.6²*3 = 2.88πm².
answer:2.88
correct answer:2
total:1531  correct:402  acc:0.26257348138471587
------------------------------------------------------------------------------------
ID:1531
picturePath:image/2464.png
answer:1.59
answer:1.59
correct answer:1.75
total:1532  correct:402  acc:0.26240208877284593
------------------------------------------------------------------------------------
ID:1532
picturePath:image/2465.png
answer:解:∵AB⊥OC,∴⁀{AC}=⁀{AB},∴∠B=\frac{1}{2}∠A=\frac{1}{2}×20°=10°．故选:A．
answer:10
correct answer:35
total:1533  correct:402  acc:0.2622309197651663
------------------------------------------------------------------------------------
ID:1533
picturePath:image/2466.png
answer:2.8m
answer:2.8
correct answer:4
total:1534  correct:402  acc:0.2620599739243807
------------------------------------------------------------------------------------
ID:1534
picturePath:image/2467.png
answer:10
answer:10
correct answer:10
total:1535  correct:403  acc:0.2625407166123778
------------------------------------------------------------------------------------
ID:1535
picturePath:image/2470.png
answer:解:∵点B、D分别在点E、A的正下方,∴BE∥AD,∴△CBE∽△CAD,∴\frac{BC}{CD}=\frac{BE}{AD},∵BC=20米,CD=40米,BE=15米,∴\frac{20}{40}=\frac{15}{AD},解得AD=30米．故选:B．
answer:30
correct answer:30
total:1536  correct:404  acc:0.2630208333333333
------------------------------------------------------------------------------------
ID:1536
picturePath:image/2471.png
answer:3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.2×1.6+0.8×h
3.
answer:3
correct answer:8
total:1537  correct:404  acc:0.26284970722186074
------------------------------------------------------------------------------------
ID:1537
picturePath:image/2472.png
answer:根据题意，窗户框AB在地面上的影长DE=1.8m，窗户下檐到地面的距离BC=1m，EC=1.2m。由于窗户AB是垂直于地面的，因此窗户AB的高度可以由DE和BC的比值来计算。即AB=DE/BC。将DE=1.8m和BC=1m代入，得到AB=1.8/1=1.8m。所以，窗户AB的高度为1.8m。故选A。
answer:1.8
correct answer:1.5
total:1538  correct:404  acc:0.2626788036410923
------------------------------------------------------------------------------------
ID:1538
picturePath:image/2475.png
answer:This is a geometry problem that involves the use of similar triangles. Since AB is parallel to DE, triangle ADE is similar to triangle ABC. Therefore, we can set up the following proportion:

\[\frac{DE}{DC} = \frac{AB}{AC}\]

Substituting the given values, we have:

\[\frac{DE}{30} = \frac{6}{5}\]

Cross-multiplying and solving for DE, we get:

\[DE = \frac{6 \cdot 30}{5} = \boxed{36}\]

So, the width of the pond is 36 meters.The answer is: 36
answer:36
correct answer:36
total:1539  correct:405  acc:0.2631578947368421
------------------------------------------------------------------------------------
ID:1539
picturePath:image/2476.png
answer:解:∵⁀{AB}=⁀{CD},∴AB=CD,∵⁀{BC}=⁀{DE},∴BC=DE,∵∠D=128°,∴∠B=\frac{1}{2}∠D=\frac{1}{2}×128°=64°．故选:B．
answer:64
correct answer:128
total:1540  correct:405  acc:0.262987012987013
------------------------------------------------------------------------------------
ID:1540
picturePath:image/2478.png
answer:3.6米
answer:3.6
correct answer:10
total:1541  correct:405  acc:0.2628163530175211
------------------------------------------------------------------------------------
ID:1541
picturePath:image/2482.png
answer:5cm．解:∵动力臂AC与阻力臂BC之比为5:1,∴AC:BC=5:1,∵BC=10cm,∴AC=10×5=50(cm),∴要使这块石头滚动,至少要将杠杆的A端向下压5cm．故选:A．
answer:5
correct answer:50
total:1542  correct:405  acc:0.26264591439688717
------------------------------------------------------------------------------------
ID:1542
picturePath:image/2483.png
answer:6米
answer:6
correct answer:8
total:1543  correct:405  acc:0.26247569669475046
------------------------------------------------------------------------------------
ID:1543
picturePath:image/2484.png
answer:2.5
answer:2.5
correct answer:4.40
total:1544  correct:405  acc:0.2623056994818653
------------------------------------------------------------------------------------
ID:1544
picturePath:image/2485.png
answer:0.85:16.5=1:20\frac{0.85}{16.5}=\frac{1}{20}长臂的端点升高了0.85×\frac{1}{20}=0.042.5米.
answer:042.5
correct answer:11.22
total:1545  correct:405  acc:0.2621359223300971
------------------------------------------------------------------------------------
ID:1545
picturePath:image/2486.png
answer:解:∵AB∥DE,∴△ABC∽△CDE,∴\frac{AB}{DE}=\frac{BC}{CE},即\frac{AB}{5.4}=\frac{15}{3},解得:AB=27m．故选:A．
answer:27
correct answer:27
total:1546  correct:406  acc:0.2626131953428202
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ID:1546
picturePath:image/2487.png
answer:20°
answer:20
correct answer:40
total:1547  correct:406  acc:0.26244343891402716
------------------------------------------------------------------------------------
ID:1547
picturePath:image/2488.png
answer:解:∵DE∥BC,∴△DEF∽△ABC,∴\frac{DE}{AC}=\frac{EF}{BC},即\frac{0.4}{1.5}=\frac{0.2}{BC},解得:BC=1.2m,∵CD=8m,∴BC+CD=9.2m,∴树高为9.2m．故选:B．
answer:9.2
correct answer:5.5
total:1548  correct:406  acc:0.2622739018087855
------------------------------------------------------------------------------------
ID:1548
picturePath:image/2490.png
answer:2.7m
answer:2.7
correct answer:1.8
total:1549  correct:406  acc:0.2621045836023241
------------------------------------------------------------------------------------
ID:1549
picturePath:image/2492.png
answer:解:设学校旗杆的高度为x米,则有\frac{1}{x}=\frac{1}{1.2}=\frac{1}{9.6+2}=\frac{1}{11.6},解得x=11.6(米)．故选:D．
answer:11.6
correct answer:10
total:1550  correct:406  acc:0.26193548387096777
------------------------------------------------------------------------------------
ID:1550
picturePath:image/2496.png
answer:8
answer:8
correct answer:8
total:1551  correct:407  acc:0.2624113475177305
------------------------------------------------------------------------------------
ID:1551
picturePath:image/2500.png
answer:24cm^{2}
answer:2
correct answer:2
total:1552  correct:408  acc:0.26288659793814434
------------------------------------------------------------------------------------
ID:1552
picturePath:image/2504.png
answer:4
answer:4
correct answer:6
total:1553  correct:408  acc:0.2627173213135866
------------------------------------------------------------------------------------
ID:1553
picturePath:image/2508.png
answer:20°
answer:20
correct answer:50
total:1554  correct:408  acc:0.2625482625482625
------------------------------------------------------------------------------------
ID:1554
picturePath:image/2509.png
answer:The answer is 52.
answer:52
correct answer:32
total:1555  correct:408  acc:0.262379421221865
------------------------------------------------------------------------------------
ID:1555
picturePath:image/2511.png
answer:6.4m
answer:6.4
correct answer:9.6
total:1556  correct:408  acc:0.2622107969151671
------------------------------------------------------------------------------------
ID:1556
picturePath:image/2513.png
answer:6
answer:6
correct answer:12
total:1557  correct:408  acc:0.26204238921001927
------------------------------------------------------------------------------------
ID:1557
picturePath:image/2516.png
answer:解:∵△ABC与△DEF是位似图形,位似比是1:2,DE=4,∴AB=\frac{1}{2}DE=2．
故选:B．
answer:2
correct answer:2
total:1558  correct:409  acc:0.2625160462130937
------------------------------------------------------------------------------------
ID:1558
picturePath:image/2517.png
answer:12
answer:12
correct answer:4
total:1559  correct:409  acc:0.26234765875561256
------------------------------------------------------------------------------------
ID:1559
picturePath:image/2519.png
answer:根据圆周角定理，在同圆或等圆中，同弧或等弧所对的圆周角相等。因此，∠APD是弧CD所对的圆周角，而∠A和∠B是弧AB所对的圆周角。由于AB和CD相交于点P，所以∠A+∠B+∠APD=180°。将∠A和∠B的值代入，解得∠APD=105°。故选C。
answer:105
correct answer:75
total:1560  correct:409  acc:0.2621794871794872
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ID:1560
picturePath:image/2522.png
answer:2.7m
answer:2.7
correct answer:4
total:1561  correct:409  acc:0.26201153106982705
------------------------------------------------------------------------------------
ID:1561
picturePath:image/2528.png
answer:18×(2/3)²=18×(4/9)=8.
answer:8
correct answer:8
total:1562  correct:410  acc:0.26248399487836105
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ID:1562
picturePath:image/2535.png
answer:解:∵PC⊥AC,PC=24cm,∴AC=2PC=48cm,∵AB=10cm,BC=20cm,∴AB:AC=1:4,∴AB:SA=1:3,∴SA=\frac{1}{3}×48=16cm．故选:A．
answer:16
correct answer:12
total:1563  correct:410  acc:0.26231605886116444
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ID:1563
picturePath:image/2540.png
answer:解:∵△ABC与△DEF是位似图形,位似比为2:3,∴△ABC∽△DEF,∴\frac{AB}{DE}=\frac{BC}{EF}=\frac{2}{3},∵AB=4,∴DE=6．故选:C．
answer:6
correct answer:6
total:1564  correct:411  acc:0.26278772378516624
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ID:1564
picturePath:image/2542.png
answer:In a right triangle, the cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. In this case, \(\cos B\) is the ratio of side AC (which is the adjacent side to angle B) to side AB (which is the hypotenuse of the triangle).

Since we know the lengths of sides AC and BC, we can use the Pythagorean theorem to find the length of side AB. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (AB) is equal to the sum of the squares of the lengths of the other two sides (AC and BC).

AB^2 = AC^2 + BC^2
AB^2 = 4^2 + 3^2
AB^2 = 16 + 9
AB^2 = 25
AB = 5

Now that we know the lengths of all three sides, we can calculate \(\cos B\):

\cos B = \frac{AC}{AB}
\cos B = \frac{4}{5}

Therefore, \(\cos B = \frac{4}{5}\).
answer:5
correct answer:5
total:1565  correct:412  acc:0.263258785942492
------------------------------------------------------------------------------------
ID:1565
picturePath:image/2544.png
answer:In a right triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. In this case, side AC is the hypotenuse, and side AB is the side opposite angle A. We can use the Pythagorean theorem to find the length of side AB.

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we have:

AB^2 = AC^2 - BC^2
AB^2 = 4^2 - 3^2
AB^2 = 16 - 9
AB^2 = 7

Now, we can find the length of side AB by taking the square root of 7:

AB = √7

Finally, we can find the sine of angle A:

sin A = (opposite side) / (hypotenuse)
sin A = AB / AC
sin A = √7 / 4

Therefore, the value of sin A is √7/4.
answer:4
correct answer:5
total:1566  correct:412  acc:0.26309067688378035
------------------------------------------------------------------------------------
ID:1566
picturePath:image/2545.png
answer:sinB = \frac{AC}{AB} = \frac{4}{5}
answer:5
correct answer:5
total:1567  correct:413  acc:0.2635609444798979
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ID:1567
picturePath:image/2546.png
answer:In a right triangle, the cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. In this case, side AC is the hypotenuse and side AB is the adjacent side to angle A. We can use the Pythagorean theorem to find the length of side AB.

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we have:

AC^2 = AB^2 + BC^2

Substituting the given values, we get:

3^2 = AB^2 + 4^2
9 = AB^2 + 16

Now, we can solve for AB:

AB^2 = 9 - 16
AB^2 = -7

Since the length of a side cannot be negative, there is no valid solution for AB in this case. Therefore, we cannot determine the value of \cosA.
answer:7
correct answer:5
total:1568  correct:413  acc:0.26339285714285715
------------------------------------------------------------------------------------
ID:1568
picturePath:image/2550.png
answer:解:∵Rt△ABC中,∠C=90°,AC=4,BC=8,∴AB=√{AC²+BC²}=√{4²+8²}=4√{2},
则tanB=\frac{AC}{BC}=\frac{4}{8}=\frac{1}{2},
故选:B．
answer:2
correct answer:2
total:1569  correct:414  acc:0.2638623326959847
------------------------------------------------------------------------------------
ID:1569
picturePath:image/2551.png
answer:根据圆周角定理,同弧所对的圆周角相等,所以∠APD=∠B=35°.故选B.
answer:35
correct answer:77
total:1570  correct:414  acc:0.26369426751592356
------------------------------------------------------------------------------------
ID:1570
picturePath:image/2552.png
answer:在直角三角形ACB中，∠C=90°，∠A=37°，AC=4。根据三角函数的定义，sin37°=\frac{AC}{BC}，cos37°=\frac{BC}{AC}，tan37°=\frac{AC}{BC}。

已知sin37°≈0.60，cos37°≈0.80，tan37°≈0.75。因此，我们可以解出BC的长。

sin37°=\frac{AC}{BC}，0.60=\frac{4}{BC}，解得BC≈6.67。

cos37°=\frac{BC}{AC}，0.80=\frac{BC}{4}，解得BC≈5。

tan37°=\frac{AC}{BC}，0.75=\frac{4}{BC}，解得BC≈5.33。

因此，BC的长约为5.33。
answer:5.33
correct answer:3.0
total:1571  correct:414  acc:0.2635264162953533
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ID:1571
picturePath:image/2553.png
answer:4√{3}÷16=√{3}÷4=√{3}/4
answer:4
correct answer:5
total:1572  correct:414  acc:0.2633587786259542
------------------------------------------------------------------------------------
ID:1572
picturePath:image/2554.png
answer:解:∵Rt△ABC中,∠C=90°,AB=10,AC=8,∴tanB=\frac{AC}{AB}=\frac{8}{10}=\frac{4}{5}．
故选:A．
answer:5
correct answer:3
total:1573  correct:414  acc:0.26319135410044503
------------------------------------------------------------------------------------
ID:1573
picturePath:image/2555.png
answer:sinB=\frac{AC}{BC}=\frac{3}{4}
answer:4
correct answer:5
total:1574  correct:414  acc:0.2630241423125794
------------------------------------------------------------------------------------
ID:1574
picturePath:image/2556.png
answer:sinA=\frac{AC}{AB}=\frac{8}{10}=\frac{4}{5}。
answer:5
correct answer:5
total:1575  correct:415  acc:0.2634920634920635
------------------------------------------------------------------------------------
ID:1575
picturePath:image/2557.png
answer:sinA=\frac{BC}{AB}=\frac{AC}{AB}=\frac{4}{5}
answer:5
correct answer:5
total:1576  correct:416  acc:0.2639593908629442
------------------------------------------------------------------------------------
ID:1576
picturePath:image/2558.png
answer:The answer is \frac{3}{5}.
answer:5
correct answer:5
total:1577  correct:417  acc:0.264426125554851
------------------------------------------------------------------------------------
ID:1577
picturePath:image/2559.png
answer:3
answer:3
correct answer:3
total:1578  correct:418  acc:0.26489226869455007
------------------------------------------------------------------------------------
ID:1578
picturePath:image/2560.png
answer:Rt△ABC中,∠C=90°,AB=2,BC=1,则sinA=\frac{BC}{AB}=\frac{1}{2}。
answer:2
correct answer:2
total:1579  correct:419  acc:0.26535782140595315
------------------------------------------------------------------------------------
ID:1579
picturePath:image/2561.png
answer:The answer is A.
ID:1580
picturePath:image/2565.png
answer:解:∵∠A=90°,AB=12,AC=5,∴BC=√{AB^{2}+AC^{2}}=√{12^{2}+5^{2}}=13,∴\cosC=\frac{AC}{BC}=\frac{5}{13}．
故选:B．
answer:13
correct answer:13
total:1581  correct:420  acc:0.2656546489563567
------------------------------------------------------------------------------------
ID:1581
picturePath:image/2566.png
answer:3
answer:3
correct answer:5
total:1582  correct:420  acc:0.26548672566371684
------------------------------------------------------------------------------------
ID:1582
picturePath:image/2567.png
answer:In a right triangle, the tangent of an angle is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. In this case, we want to find the tangent of angle A.

tanA = (opposite side) / (adjacent side)
tanA = (BC) / (AB)

Given that AB = 5 and BC = 3, we can substitute these values into the equation:

tanA = 3 / 5

Therefore, the value of tanA is 3/5.
answer:5
correct answer:4
total:1583  correct:420  acc:0.2653190145293746
------------------------------------------------------------------------------------
ID:1583
picturePath:image/2568.png
answer:3
answer:3
correct answer:5
total:1584  correct:420  acc:0.26515151515151514
------------------------------------------------------------------------------------
ID:1584
picturePath:image/2571.png
answer:Since ∠A is a right angle, triangle ABC is a right triangle. According to the Pythagorean theorem, the length of side BC can be calculated as follows:

BC^2 = AB^2 + AC^2
BC^2 = 8^2 + 6^2
BC^2 = 64 + 36
BC^2 = 100
BC = 10

Now, we can calculate \cosC as follows:

\cosC = \frac{AC}{BC}
\cosC = \frac{6}{10}
\cosC = \frac{3}{5}

Therefore, the value of \cosC is \frac{3}{5}.
answer:5
correct answer:5
total:1585  correct:421  acc:0.26561514195583596
------------------------------------------------------------------------------------
ID:1585
picturePath:image/2572.png
answer:解:设投影三角形的对应边长为xm,则有\frac{2}{5}=\frac{8}{x},解得x=10．故选:B．
answer:10
correct answer:20
total:1586  correct:421  acc:0.26544766708701134
------------------------------------------------------------------------------------
ID:1586
picturePath:image/2573.png
answer:The answer is \frac{3}{5}
answer:5
correct answer:5
total:1587  correct:422  acc:0.2659105229993699
------------------------------------------------------------------------------------
ID:1587
picturePath:image/2579.png
answer:5
answer:5
correct answer:13
total:1588  correct:422  acc:0.26574307304785894
------------------------------------------------------------------------------------
ID:1588
picturePath:image/2580.png
answer:在直角三角形ABC中，∠C=90°，AB=3，AC=2。根据勾股定理，AC^2 + BC^2 = AB^2，即2^2 + BC^2 = 3^2，解得BC=√7。因此，\cosB=\frac{AC}{AB}=\frac{2}{3}。所以，\cosB的值为\frac{2}{3}。
answer:3
correct answer:3
total:1589  correct:423  acc:0.26620516047828824
------------------------------------------------------------------------------------
ID:1589
picturePath:image/2581.png
answer:解:∵OA=OC,∴∠OAC=∠OCA=55°,∴∠BOC=2∠OAC=110°,∴∠D=\frac{1}{2}∠BOC=55°．故选:A．
answer:55
correct answer:35
total:1590  correct:423  acc:0.2660377358490566
------------------------------------------------------------------------------------
ID:1590
picturePath:image/2583.png
answer:解:∵∠ACB=90°,BC=3,AC=4,∴AB=√{AC^{2}+BC^{2}}=√{4^{2}+3^{2}}=5,∵CD⊥AB,∴∠DCB=∠A,∴sin∠DCB=sin∠A=\frac{BC}{AB}=\frac{3}{5}．故选:A．
answer:5
correct answer:5
total:1591  correct:424  acc:0.2664990571967316
------------------------------------------------------------------------------------
ID:1591
picturePath:image/2584.png
answer:In a right triangle, the cosine of an angle is defined as the adjacent side divided by the hypotenuse. In this case, side AB is the hypotenuse and side BC is the adjacent side to angle B. Therefore, \cosB = \frac{BC}{AB} = \frac{3}{5}.
answer:5
correct answer:5
total:1592  correct:425  acc:0.26695979899497485
------------------------------------------------------------------------------------
ID:1592
picturePath:image/2586.png
answer:The answer is \frac{3}{7}
answer:7
correct answer:5
total:1593  correct:425  acc:0.2667922159447583
------------------------------------------------------------------------------------
ID:1593
picturePath:image/2588.png
answer:12\sqrt{5}
answer:5
correct answer:13
total:1594  correct:425  acc:0.266624843161857
------------------------------------------------------------------------------------
ID:1594
picturePath:image/2590.png
answer:sinA=\frac{BC}{AB}=\frac{AC}{AB}=\frac{4}{5}
answer:5
correct answer:5
total:1595  correct:426  acc:0.2670846394984326
------------------------------------------------------------------------------------
ID:1595
picturePath:image/2592.png
answer:80°
answer:80
correct answer:80
total:1596  correct:427  acc:0.2675438596491228
------------------------------------------------------------------------------------
ID:1596
picturePath:image/2595.png
answer:解:∵∠B=90°,AB=1,BC=2,∴AC=√{AB^{2}+BC^{2}}=√{1^{2}+2^{2}}=√{5},∴sinA=\frac{BC}{AC}=\frac{2}{√{5}}=\frac{2√{5}}{5}．故选:D．
answer:5
correct answer:5
total:1597  correct:428  acc:0.2680025046963056
------------------------------------------------------------------------------------
ID:1597
picturePath:image/2601.png
answer:The answer is \frac{3}{5}.
answer:5
correct answer:5
total:1598  correct:429  acc:0.26846057571964954
------------------------------------------------------------------------------------
ID:1598
picturePath:image/2605.png
answer:tanA=\frac{BC}{AC}=\frac{3}{4}
answer:4
correct answer:4
total:1599  correct:430  acc:0.2689180737961226
------------------------------------------------------------------------------------
ID:1599
picturePath:image/2607.png
answer:To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

In this case, we have a right triangle with sides AC and BC, and we know that AC = 2 and BC = 1. We want to find the length of side AB, which is the hypotenuse.

Using the Pythagorean theorem, we have:

AB^2 = AC^2 + BC^2
AB^2 = 2^2 + 1^2
AB^2 = 4 + 1
AB^2 = 5
AB = √5

Now that we know the length of side AB, we can find the cosine of angle A. The cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse.

cos A = adjacent side / hypotenuse
cos A = AC / AB
cos A = 2 / √5

To simplify this expression, we can rationalize the denominator by multiplying the numerator and denominator by the square root of 5:

cos A = (2√5) / (√5 * √5)
cos A = (2√5) / 5

So, the cosine of angle A is (2√5) / 5.

The answer is: \frac{2\sqrt{5}}{5}
answer:5
correct answer:5
total:1600  correct:431  acc:0.269375
------------------------------------------------------------------------------------
ID:1600
picturePath:image/2608.png
answer:In a right triangle, the cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. In this case, \(\cos B\) is the ratio of \(AC\) to \(AB\), since \(AC\) is the adjacent side to angle B and \(AB\) is the hypotenuse.

Given that \(AC = 3\) and \(AB = 5\), we can calculate \(\cos B\) as follows:

\(\cos B = \frac{AC}{AB} = \frac{3}{5}\)

Therefore, the value of \(\cos B\) is \(\frac{3}{5}\).
answer:5
correct answer:5
total:1601  correct:432  acc:0.26983135540287323
------------------------------------------------------------------------------------
ID:1601
picturePath:image/2611.png
answer:The answer is \frac{1}{\sqrt{7}}.
answer:7
correct answer:5
total:1602  correct:432  acc:0.2696629213483146
------------------------------------------------------------------------------------
ID:1602
picturePath:image/2612.png
answer:解:∵∠C=90°,AB=5,BC=3,∴AC=4,∴tanA=\frac{BC}{AC}=\frac{3}{4}．
故选:A．
answer:4
correct answer:4
total:1603  correct:433  acc:0.27011852776044915
------------------------------------------------------------------------------------
ID:1603
picturePath:image/2614.png
answer:The answer is 40.
answer:40
correct answer:40
total:1604  correct:434  acc:0.27057356608478805
------------------------------------------------------------------------------------
ID:1604
picturePath:image/2615.png
answer:Since \( OC \) is perpendicular to \( AB \), we know that \( \angle AOC = 90^\circ \). Since \( \angle BOC = 50^\circ \), we can find \( \angle AOD \) by subtracting \( \angle AOC \) from \( \angle BOC \): \( \angle AOD = \angle BOC - \angle AOC = 50^\circ - 90^\circ = -40^\circ \). Since the angles on a circle must sum to 360 degrees, we can find \( \angle ADC \) by subtracting \( \angle AOD \) from 360 degrees: \( \angle ADC = 360^\circ - \angle AOD = 360^\circ - (-40^\circ) = 400^\circ \). However, since \( \angle ADC \) is an angle in a triangle, it must be less than 180 degrees. Therefore, \( \angle ADC = 400^\circ - 360^\circ = \boxed{40^\circ} \).The answer is: 40
answer:40
correct answer:25
total:1605  correct:434  acc:0.270404984423676
------------------------------------------------------------------------------------
ID:1605
picturePath:image/2616.png
answer:在直角三角形ABC中，∠C=90°，AC=3，BC=4。根据三角函数的定义，sinB = \frac{AC}{AB} = \frac{3}{AB}。由于AC^2 + BC^2 = AB^2，即3^2 + 4^2 = AB^2，解得AB = 5。因此，sinB = \frac{3}{5}。
answer:5
correct answer:5
total:1606  correct:435  acc:0.27085927770859275
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ID:1606
picturePath:image/2619.png
answer:解:∵AD为⊙O的直径,∴∠AED=90°,∴tanA=1,∵tanA=\frac{tanC}{tanB},∴tanC•tanB=1．故选:A．
answer:1
correct answer:4
total:1607  correct:435  acc:0.2706907280647169
------------------------------------------------------------------------------------
ID:1607
picturePath:image/2620.png
answer:Rt△ABC中,∠C=90°,a=1,c=4,则sinA=\frac{BC}{AB}=\frac{4}{\sqrt{4^2+1^2}}=\frac{4}{\sqrt{17}}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}=\frac{4\sqrt{17}}{
answer:17
correct answer:4
total:1608  correct:435  acc:0.27052238805970147
------------------------------------------------------------------------------------
ID:1608
picturePath:image/2625.png
answer:解:∵CD⊥AB,∠ACB=90°,CD:AC=2:3,∴CD:AB=2:5,∴\frac{CD}{AB}=\frac{BC}{AB},∴BC:AB=2:5,∴sin∠BCD=\frac{BC}{AB}=\frac{2}{5}．
故选:A．
answer:5
correct answer:3
total:1609  correct:435  acc:0.2703542573026725
------------------------------------------------------------------------------------
ID:1609
picturePath:image/2626.png
answer:解:∵∠ADC与∠B是同弧所对的圆周角与圆心角,∠ADC=20°,∴∠B=2∠ADC=40°．故选:D．
answer:40
correct answer:20
total:1610  correct:435  acc:0.2701863354037267
------------------------------------------------------------------------------------
ID:1610
picturePath:image/2627.png
answer:Rt△ABC中,∠C=90°,AB=5,BC=4,那么\cosA=\frac{BC}{AB}=\frac{4}{5}。
answer:5
correct answer:5
total:1611  correct:436  acc:0.2706393544382371
------------------------------------------------------------------------------------
ID:1611
picturePath:image/2629.png
answer:tan∠AED=\frac{OE}{ED}=\frac{1}{1}=1.
answer:1
correct answer:2
total:1612  correct:436  acc:0.2704714640198511
------------------------------------------------------------------------------------
ID:1612
picturePath:image/2630.png
answer:tan∠AOB=\frac{AC}{CO}=\frac{1}{1}=1
answer:1
correct answer:2
total:1613  correct:436  acc:0.2703037817730936
------------------------------------------------------------------------------------
ID:1613
picturePath:image/2632.png
answer:4
answer:4
correct answer:5
total:1614  correct:436  acc:0.2701363073110285
------------------------------------------------------------------------------------
ID:1614
picturePath:image/2634.png
answer:解:∵AB=4,DE=8,∴BC=√{AB²+DE²}=√{4²+8²}=4√{2},∴sin∠C'ED=\frac{BC}{DE}=\frac{4√{2}}{8}=\frac{√{2}}{2}．故选:B．
answer:2
correct answer:2
total:1615  correct:437  acc:0.27058823529411763
------------------------------------------------------------------------------------
ID:1615
picturePath:image/2637.png
answer:15
answer:15
correct answer:15
total:1616  correct:438  acc:0.27103960396039606
------------------------------------------------------------------------------------
ID:1616
picturePath:image/2638.png
answer:解:∵∠B=90°,∴∠A+∠C=90°,∴\cosA=\cos(90°-∠C)=\sinC=\frac{12}{13}；∴\sinA=\sqrt{1-{\sin}^{2}A}=\frac{5}{13},∴\tanA=\frac{\sinA}{\cosA}=\frac{5}{12}．
故选:B．
answer:12
correct answer:12
total:1617  correct:439  acc:0.2714904143475572
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ID:1617
picturePath:image/2643.png
answer:Since ∠B is a right angle, ∠A and ∠C are complementary angles. Therefore, ∠A + ∠C = 90°. Since AB = 1 and BC = 2, we can use the Pythagorean theorem to find that AC = √(AB² + BC²) = √(1² + 2²) = √5. Therefore, cosA = AC/AB = √5/1 = √5. The answer is A.
answer:5
correct answer:5
total:1618  correct:440  acc:0.2719406674907293
------------------------------------------------------------------------------------
ID:1618
picturePath:image/2644.png
answer:cos∠BAO=\frac{AO}{AB}=\frac{3}{5}
answer:5
correct answer:5
total:1619  correct:441  acc:0.272390364422483
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ID:1619
picturePath:image/2645.png
answer:Since ∠B is a right angle, ∠A and ∠C are complementary angles. Therefore, \cos A = \sin C. Using the Pythagorean theorem, we can find that AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. Thus, \sin C = \frac{BC}{AC} = \frac{4}{5} and \cos A = \boxed{\frac{3}{5}}.The answer is: \frac{3}{5}
answer:5
correct answer:5
total:1620  correct:442  acc:0.27283950617283953
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ID:1620
picturePath:image/2648.png
answer:6
answer:6
correct answer:3
total:1621  correct:442  acc:0.27267119062307216
------------------------------------------------------------------------------------
ID:1621
picturePath:image/2653.png
answer:Since ∠C=90°, we can use the Pythagorean theorem to find the length of side BC. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we have:

AB^2 = AC^2 + BC^2
5^2 = 3^2 + BC^2
25 = 9 + BC^2
BC^2 = 25 - 9
BC^2 = 16
BC = 4

Now, we can use the definition of tangent to find tanB. The tangent of an angle in a right triangle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. In this case, we have:

tanB = \frac{AC}{BC}
tanB = \frac{3}{4}

Therefore, the value of tanB is \frac{3}{4}.
answer:4
correct answer:4
total:1622  correct:443  acc:0.27311960542540076
------------------------------------------------------------------------------------
ID:1622
picturePath:image/2655.png
answer:tanα=\frac{BC}{OC}=\frac{3-0}{4-0}=\frac{3}{4}
answer:4
correct answer:4
total:1623  correct:444  acc:0.2735674676524954
------------------------------------------------------------------------------------
ID:1623
picturePath:image/2657.png
answer:In a right triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. In this case, angle B is opposite side AB, and the hypotenuse is side AC.

Since triangle ABC is a right triangle with angle C being 90 degrees, we can use the Pythagorean theorem to find the length of side AC. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b). In this case, we have:

AB^2 + BC^2 = AC^2
2^2 + 1^2 = AC^2
4 + 1 = AC^2
5 = AC^2

Taking the square root of both sides, we get:

AC = √5

Now, we can find the sine of angle B using the definition of sine:

sinB = (opposite side) / (hypotenuse)
sinB = AB / AC
sinB = 2 / √5

To rationalize the denominator, we can multiply the numerator and denominator by √5:

sinB = (2√5) / (5)
sinB = 2√5 / 5

Therefore, the value of sinB is 2√5/5.
answer:5
correct answer:2
total:1624  correct:444  acc:0.2733990147783251
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ID:1624
picturePath:image/2661.png
answer:解:∵Rt△ABC中,∠ACB=90°,BC=3,AC=4,∴AB=√{AC^{2}+BC^{2}}=√{4^{2}+3^{2}}=5,∵CD⊥AB于点D,∴∠1=∠A,∴sin∠1=sin∠A=\frac{BC}{AB}=\frac{3}{5}．故选:A．
answer:5
correct answer:5
total:1625  correct:445  acc:0.27384615384615385
------------------------------------------------------------------------------------
ID:1625
picturePath:image/2662.png
answer:3
answer:3
correct answer:5
total:1626  correct:445  acc:0.2736777367773678
------------------------------------------------------------------------------------
ID:1626
picturePath:image/2663.png
answer:In a right triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. In this case, side AC is the hypotenuse, and side AB is the side opposite angle B. We can use the Pythagorean theorem to find the length of side AB.

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we have:

AC^2 = AB^2 + BC^2
4^2 = AB^2 + 3^2
16 = AB^2 + 9
AB^2 = 16 - 9
AB^2 = 7

Now, we can find the length of side AB by taking the square root of 7:

AB = √7

Finally, we can find the sine of angle B by dividing the length of side AB by the length of side AC:

sinB = AB / AC
sinB = √7 / 4

Therefore, the value of sinB is √7/4.
answer:4
correct answer:5
total:1627  correct:445  acc:0.2735095267363245
------------------------------------------------------------------------------------
ID:1627
picturePath:image/2664.png
answer:In a right triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. In this case, side AC is the hypotenuse, and side AB is the side opposite angle A. We can use the Pythagorean theorem to find the length of side AB.

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we have:

BC^2 + AC^2 = AB^2
5^2 + 12^2 = AB^2
25 + 144 = AB^2
169 = AB^2

Now, we can find the length of side AB by taking the square root of 169:

AB = √169
AB = 13

Now that we know the length of side AB, we can find the sine of angle A:

sinA = (opposite side) / (hypotenuse)
sinA = AB / AC
sinA = 13 / 12

Therefore, sinA = 13/12.
answer:12
correct answer:13
total:1628  correct:445  acc:0.27334152334152334
------------------------------------------------------------------------------------
ID:1628
picturePath:image/2665.png
answer:The answer is \frac{1}{2}
answer:2
correct answer:3
total:1629  correct:445  acc:0.27317372621240027
------------------------------------------------------------------------------------
ID:1629
picturePath:image/2669.png
answer:8
answer:8
correct answer:6
total:1630  correct:445  acc:0.27300613496932513
------------------------------------------------------------------------------------
ID:1630
picturePath:image/2675.png
answer:sinB=\frac{AC}{AB}=\frac{3}{5}
answer:5
correct answer:5
total:1631  correct:446  acc:0.2734518700183936
------------------------------------------------------------------------------------
ID:1631
picturePath:image/2676.png
answer:根据正弦的定义，sinA = \frac{AC}{AB}，其中AB = BC = 12。已知sinA = \frac{4}{5}，所以\frac{AC}{12} = \frac{4}{5}。解这个方程，我们得到AC = \frac{4}{5} \times 12 = 9.6。所以，AC = 9.6.
answer:9.6
correct answer:9
total:1632  correct:446  acc:0.27328431372549017
------------------------------------------------------------------------------------
ID:1632
picturePath:image/2679.png
answer:解:∵点A(t,3)在第一象限,OA与x轴所夹的锐角为α,tanα=2,∴\frac{3}{t}=2,解得t=1.5．故选:A．
answer:1.5
correct answer:1.5
total:1633  correct:447  acc:0.2737293325168402
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ID:1633
picturePath:image/2680.png
answer:解:∵∠CDB=35°,∴∠CAB=70°,∵AB是直径,∴∠ACB=90°,∴∠CBA=90°-∠CAB=90°-70°=20°．故选:A．
answer:20
correct answer:55
total:1634  correct:447  acc:0.27356181150550796
------------------------------------------------------------------------------------
ID:1634
picturePath:image/2685.png
answer:解:∵CD是Rt△ABC斜边AB上的高,AC=8,BC=6,∴AB=√{AC^{2}+BC^{2}}=10,∴\cos∠BCD=\frac{CD}{AB}=\frac{6}{10}=\frac{3}{5}．
故选:C．
answer:5
correct answer:5
total:1635  correct:448  acc:0.27400611620795107
------------------------------------------------------------------------------------
ID:1635
picturePath:image/2690.png
answer:46°
answer:46
correct answer:67
total:1636  correct:448  acc:0.27383863080684595
------------------------------------------------------------------------------------
ID:1636
picturePath:image/2694.png
answer:sinA=\frac{BC}{AB}=\frac{6}{√{6²+8²}}=\frac{6}{10}=\frac{3}{5}
answer:5
correct answer:5
total:1637  correct:449  acc:0.274282223579719
------------------------------------------------------------------------------------
ID:1637
picturePath:image/2697.png
answer:tanα = \frac{y}{x} = \frac{1}{2}
answer:2
correct answer:2
total:1638  correct:450  acc:0.27472527472527475
------------------------------------------------------------------------------------
ID:1638
picturePath:image/2699.png
answer:在直角三角形ABC中，∠C=90°，BC=3，AB=4。根据勾股定理，AC=√{AB²-BC²}=√{4²-3²}=√{16-9}=√{7}。

\cosB=\frac{BC}{AB}=\frac{3}{4}。

所以，\cosB的值是\frac{3}{4}。
answer:4
correct answer:4
total:1639  correct:451  acc:0.2751677852348993
------------------------------------------------------------------------------------
ID:1639
picturePath:image/2701.png
answer:解:∵AB是直径,∴∠ACB=90°,∵∠ADC=70°,∴∠CAB=90°-∠ADC=90°-70°=20°．故选:A．
answer:20
correct answer:20
total:1640  correct:452  acc:0.275609756097561
------------------------------------------------------------------------------------
ID:1640
picturePath:image/2705.png
answer:The answer is \frac{5}{13}
answer:13
correct answer:13
total:1641  correct:453  acc:0.2760511882998172
------------------------------------------------------------------------------------
ID:1641
picturePath:image/2706.png
answer:由勾股定理得,AB=√{AC²+BC²}=√{8²+6²}=10,∴sinB=\frac{AC}{AB}=\frac{8}{10}=\frac{4}{5}．故选:D．
answer:5
correct answer:5
total:1642  correct:454  acc:0.27649208282582216
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ID:1642
picturePath:image/2707.png
answer:解:∵P是第一象限内的点,其坐标是(4,m),且OP与x轴正半轴的夹角α的余弦值是\frac{3}{5},∴\cos α=\frac{3}{5},∴\sin α=\pm\frac{4}{5},∵P在第一象限,∴\sin α=\frac{4}{5},∴\tan α=\frac{\sin α}{\cos α}=\frac{4}{3}．故选:A．
answer:3
correct answer:3
total:1643  correct:455  acc:0.27693244065733413
------------------------------------------------------------------------------------
ID:1643
picturePath:image/2711.png
answer:解:∵AB为⊙O的直径,∴∠ACB=90°,∵∠BAC=30°,∴∠B=90°-∠BAC=60°,∵弧BC等于弧CD,∴∠D=∠B=60°,∴∠DAC=∠D-∠BAC=60°-30°=30°．故选:A．
answer:30
correct answer:30
total:1644  correct:456  acc:0.2773722627737226
------------------------------------------------------------------------------------
ID:1644
picturePath:image/2721.png
answer:The answer is \frac{12}{13}
answer:13
correct answer:13
total:1645  correct:457  acc:0.2778115501519757
------------------------------------------------------------------------------------
ID:1645
picturePath:image/2722.png
answer:Since AB is parallel to CD, we have \[ \angle EAB = 180^\circ - \angle E - \angle C = 180^\circ - 27^\circ - 52^\circ = \boxed{101^\circ}. \] The answer is: 101
answer:101
correct answer:79
total:1646  correct:457  acc:0.2776427703523694
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ID:1646
picturePath:image/2723.png
answer:45°
answer:45
correct answer:105
total:1647  correct:457  acc:0.2774741955069824
------------------------------------------------------------------------------------
ID:1647
picturePath:image/2724.png
answer:解:∵∠BOD=140°,∴∠AOC=180°-∠BOD=40°,∵∠CDA=30°,∴∠AEC=∠AOC-∠CDA=40°-30°=10°．故选:A．
answer:10
correct answer:100
total:1648  correct:457  acc:0.27730582524271846
------------------------------------------------------------------------------------
ID:1648
picturePath:image/2727.png
answer:4.
answer:4
correct answer:5
total:1649  correct:457  acc:0.2771376591873863
------------------------------------------------------------------------------------
ID:1649
picturePath:image/2728.png
answer:解:∵BC=4,CD=3,CD是AB边上的中线,∴AB=2CD=6,∵∠ACB=90°,∴tanB=\frac{CD}{BC}=\frac{3}{4}．故选:A．
answer:4
correct answer:2
total:1650  correct:457  acc:0.276969696969697
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ID:1650
picturePath:image/2729.png
answer:3/5
answer:5
correct answer:5
total:1651  correct:458  acc:0.2774076317383404
------------------------------------------------------------------------------------
ID:1651
picturePath:image/2730.png
answer:In a right triangle, the tangent of an angle is defined as the ratio of the length of the opposite side to the length of the adjacent side. In this case, the opposite side to angle A is AB, and the adjacent side is AC. We can use the Pythagorean theorem to find the length of AB.

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is AB, and we know that AC = 6 and BC = 8.

AB^2 = AC^2 + BC^2
AB^2 = 6^2 + 8^2
AB^2 = 36 + 64
AB^2 = 100
AB = √100
AB = 10

Now, we can find the value of tanA:

tanA = AB/AC
tanA = 10/6
tanA = 5/3

Therefore, the value of tanA is 5/3.
answer:3
correct answer:3
total:1652  correct:459  acc:0.2778450363196126
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ID:1652
picturePath:image/2731.png
answer:sinA=\frac{BC}{AB}=\frac{3}{5}
answer:5
correct answer:5
total:1653  correct:460  acc:0.27828191167574107
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ID:1653
picturePath:image/2734.png
answer:根据勾股定理,AB^2=AC^2+BC^2,即13^2=5^2+BC^2,解得BC=12.在Rt△ABC中,\cosB=\frac{BC}{AB}=\frac{12}{13}.故选B.
answer:13
correct answer:13
total:1654  correct:461  acc:0.27871825876662637
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ID:1654
picturePath:image/2735.png
answer:The answer is 60.
answer:60
correct answer:60
total:1655  correct:462  acc:0.2791540785498489
------------------------------------------------------------------------------------
ID:1655
picturePath:image/2737.png
answer:解:∵D是AB的中点,∴CD=DB=\frac{1}{2}AB,∵BC=5,AC=12,∴AB=√{AC^{2}+BC^{2}}=13,∵∠ACB=90°,∴sin∠DCA=\frac{CD}{AB}=\frac{6}{13}．
故选:A．
answer:13
correct answer:13
total:1656  correct:463  acc:0.27958937198067635
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ID:1656
picturePath:image/2738.png
answer:The answer is \frac{8}{16}=\frac{1}{2}.
answer:2
correct answer:5
total:1657  correct:463  acc:0.27942063971031983
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ID:1657
picturePath:image/2739.png
answer:12√3.
answer:3
correct answer:12
total:1658  correct:463  acc:0.2792521109770808
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ID:1658
picturePath:image/2741.png
answer:Since \( CD \) is perpendicular to \( AB \) and \( CD \) is the height of the right triangle, we can use the Pythagorean theorem to find the length of \( CD \). The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is \( AC \) and one of the other sides is \( CD \). Therefore, we have:

\( AC^2 = CD^2 + AB^2 \)

Substituting the given values, we have:

\( 2^2 = CD^2 + 1^2 \)

\( 4 = CD^2 + 1 \)

\( CD^2 = 4 - 1 \)

\( CD^2 = 3 \)

\( CD = \sqrt{3} \)

Now, we can use the definition of sine to find \( \sin \angle ACD \):

\( \sin \angle ACD = \frac{CD}{AC} \)

\( \sin \angle ACD = \frac{\sqrt{3}}{2} \)

Therefore, the value of \( \sin \angle ACD \) is \( \frac{\sqrt{3}}{2} \).
answer:2
correct answer:5
total:1659  correct:463  acc:0.27908378541289935
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ID:1659
picturePath:image/2742.png
answer:在直角三角形ABC中，∠C=90°，AC=2，BC=1。根据勾股定理，AB²=AC²+BC²=2²+1²=5。因此，AB=√5。

\cosB=\frac{AC}{AB}=\frac{2}{√5}=\frac{2√5}{5}。

所以，\cosB的值是\frac{2√5}{5}。
answer:5
correct answer:5
total:1660  correct:464  acc:0.27951807228915665
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ID:1660
picturePath:image/2743.png
answer:3
answer:3
correct answer:3
total:1661  correct:465  acc:0.27995183624322695
------------------------------------------------------------------------------------
ID:1661
picturePath:image/2744.png
answer:sinA=\frac{BC}{AB}=\frac{4}{5}
answer:5
correct answer:5
total:1662  correct:466  acc:0.28038507821901326
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ID:1662
picturePath:image/2745.png
answer:3/5
answer:5
correct answer:5
total:1663  correct:467  acc:0.2808177991581479
------------------------------------------------------------------------------------
ID:1663
picturePath:image/2747.png
answer:In a right triangle, the cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. In this case, \(\cos B\) is the ratio of \(\overline{AC}\) to \(\overline{AB}\). Since \(\overline{AC}\) is given as 1 and \(\overline{BC}\) is given as 2, we can use the Pythagorean theorem to find \(\overline{AB}\). The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Therefore, \(\overline{AB}^2 = \overline{AC}^2 + \overline{BC}^2 = 1^2 + 2^2 = 5\). Taking the square root of both sides, we find that \(\overline{AB} = \sqrt{5}\). Thus, \(\cos B = \frac{\overline{AC}}{\overline{AB}} = \frac{1}{\sqrt{5}} = \boxed{\frac{\sqrt{5}}{5}}\).The answer is: \frac{\sqrt{5}}{5}
answer:5
correct answer:5
total:1664  correct:468  acc:0.28125
------------------------------------------------------------------------------------
ID:1664
picturePath:image/2748.png
answer:In a right triangle, the cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. In this case, \(\cos A\) is the ratio of \(\overline{AC}\) to \(\overline{AB}\). Since \(\overline{AB}\) is the hypotenuse and \(\overline{AC}\) is one of the legs, we can use the Pythagorean theorem to find \(\overline{AB}\). The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In this case, \(\overline{AB}^2 = \overline{AC}^2 + \overline{BC}^2\). Substituting the given values, we have \(\overline{AB}^2 = 4^2 + 3^2 = 16 + 9 = 25\). Therefore, \(\overline{AB} = 5\). Now, we can find \(\cos A\) by dividing \(\overline{AC}\) by \(\overline{AB}\): \(\cos A = \frac{\overline{AC}}{\overline{AB}} = \frac{4}{5}\). Thus, the value of \(\cos A\) is \(\frac{4}{5}\), which is option C.
answer:5
correct answer:5
total:1665  correct:469  acc:0.2816816816816817
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ID:1665
picturePath:image/2750.png
answer:tanα=\frac{AC}{CO}=\frac{3-0}{2-0}=\frac{3}{2}
answer:2
correct answer:2
total:1666  correct:470  acc:0.28211284513805523
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ID:1666
picturePath:image/2751.png
answer:sinB=\frac{AC}{AB}=\frac{2}{4}=\frac{1}{2}
answer:2
correct answer:2
total:1667  correct:471  acc:0.28254349130173967
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ID:1667
picturePath:image/2755.png
answer:In a right triangle, the cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. In this case, \(\cos A\) is the ratio of \(\text{AC}\) to \(\text{BC}\), which is \(\frac{5}{12}\). Therefore, the value of \(\cos A\) is \(\frac{5}{12}\).
answer:12
correct answer:13
total:1668  correct:471  acc:0.2823741007194245
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ID:1668
picturePath:image/2757.png
answer:∠AOB = 2∠ACB = 2 × 48° = 96°.
answer:96
correct answer:96
total:1669  correct:472  acc:0.28280407429598564
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ID:1669
picturePath:image/2759.png
answer:解:∵PM⊥OA于M,OM:OP=4:5,∴\cosα=\frac{OM}{OP}=\frac{4}{4+5}=\frac{4}{9}．故选:A．
answer:9
correct answer:5
total:1670  correct:472  acc:0.28263473053892213
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ID:1670
picturePath:image/2760.png
answer:解:∵Rt△ABC中,∠C=90°,AC=4,∴tanA=\frac{BC}{AC}=\frac{1}{2},∴BC=2．故选:B．
answer:2
correct answer:2
total:1671  correct:473  acc:0.2830640335128665
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ID:1671
picturePath:image/2767.png
answer:解:∵∠AOB与∠ACB都对⁀{AB},且∠AOB=80°,∴∠ACB=\frac{1}{2}∠AOB=40°．故选:B．
answer:40
correct answer:40
total:1672  correct:474  acc:0.2834928229665072
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ID:1672
picturePath:image/2772.png
answer:解:∵点A(t,3)在第一象限,OA与x轴所夹的锐角为α,tanα=\frac{3}{2},∴\frac{3}{t}=\frac{3}{2},解得,t=2．故选:A．
answer:2
correct answer:2
total:1673  correct:475  acc:0.2839210998206814
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ID:1673
picturePath:image/2773.png
answer:解:∵CD是斜边AB上的中线,CD=2.5,AC=3,∴AB=2CD=5,BC=2AC=6,在Rt△ABC中,∵AB=5,BC=6,∴tanB=\frac{AC}{BC}=\frac{3}{6}=\frac{1}{2}．故选:A．
answer:2
correct answer:4
total:1674  correct:475  acc:0.28375149342891276
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ID:1674
picturePath:image/2776.png
answer:tanA=\frac{BC}{AC}=\frac{1}{2}
answer:2
correct answer:2
total:1675  correct:476  acc:0.28417910447761197
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ID:1675
picturePath:image/2781.png
answer:3
answer:3
correct answer:5
total:1676  correct:476  acc:0.2840095465393795
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ID:1676
picturePath:image/2785.png
answer:解:∵AB=AC=4cm,BC=6cm,∴AC=√{AB²+BC²}=√{4²+6²}=8cm,∴\cosB=\frac{AB}{AC}=\frac{4}{8}=\frac{1}{2}．
故选:A．
answer:2
correct answer:4
total:1677  correct:476  acc:0.283840190816935
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ID:1677
picturePath:image/2786.png
answer:3/5
answer:5
correct answer:17
total:1678  correct:476  acc:0.2836710369487485
------------------------------------------------------------------------------------
ID:1678
picturePath:image/2787.png
answer:3
answer:3
correct answer:5
total:1679  correct:476  acc:0.28350208457415127
------------------------------------------------------------------------------------
ID:1679
picturePath:image/2788.png
answer:tan∠AB0=\frac{3-0}{3-0}=\frac{3}{3}=1.
answer:1
correct answer:4
total:1680  correct:476  acc:0.2833333333333333
------------------------------------------------------------------------------------
ID:1680
picturePath:image/2790.png
answer:The answer is \frac{√{3}}{3}.
answer:3
correct answer:3
total:1681  correct:477  acc:0.28375966686496135
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ID:1681
picturePath:image/2792.png
answer:Since ∠C is a right angle, triangle ABC is a right triangle. According to the Pythagorean theorem, the square of the length of the hypotenuse (AB) is equal to the sum of the squares of the lengths of the other two sides (AC and BC). Therefore, we can find the length of BC using the Pythagorean theorem:

BC^2 = AB^2 - AC^2
BC^2 = 5^2 - 4^2
BC^2 = 25 - 16
BC^2 = 9
BC = 3

Now, we can use the definition of cosine to find cos A:

cos A = (adjacent side) / (hypotenuse)
cos A = BC / AB
cos A = 3 / 5

Therefore, the value of cos A is 3/5.
answer:5
correct answer:5
total:1682  correct:478  acc:0.28418549346016647
------------------------------------------------------------------------------------
ID:1682
picturePath:image/2795.png
answer:3
answer:3
correct answer:5
total:1683  correct:478  acc:0.28401663695781343
------------------------------------------------------------------------------------
ID:1683
picturePath:image/2797.png
answer:tanB=\frac{AC}{BC}=\frac{6}{BC}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}=\frac{6}{5}
answer:5
correct answer:4
total:1684  correct:478  acc:0.2838479809976247
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ID:1684
picturePath:image/2798.png
answer:6.
answer:6
correct answer:6
total:1685  correct:479  acc:0.284272997032641
------------------------------------------------------------------------------------
ID:1685
picturePath:image/2800.png
answer:解:∵CD⊥OA,CE⊥OB,∴⁀{AC}=⁀{BC},∴∠P=\frac{1}{2}∠DCE=\frac{1}{2}×80°=40°．故选:D．
answer:40
correct answer:50
total:1686  correct:479  acc:0.28410438908659547
------------------------------------------------------------------------------------
ID:1686
picturePath:image/2801.png
answer:3/5
answer:5
correct answer:5
total:1687  correct:480  acc:0.2845287492590397
------------------------------------------------------------------------------------
ID:1687
picturePath:image/2803.png
answer:sin∠APB=\frac{PO}{AO}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{1}=\frac{1}{
answer:1
correct answer:2
total:1688  correct:480  acc:0.2843601895734597
------------------------------------------------------------------------------------
ID:1688
picturePath:image/2804.png
answer:4.
answer:4
correct answer:4
total:1689  correct:481  acc:0.2847838957963292
------------------------------------------------------------------------------------
ID:1689
picturePath:image/2805.png
answer:解:∵∠C=Rt∠,AC=4,BC=3,∴AB=√{AC^{2}+BC^{2}}=√{4^{2}+3^{2}}=5,∴sinB=\frac{AC}{AB}=\frac{4}{5}．
故选:A．
answer:5
correct answer:5
total:1690  correct:482  acc:0.285207100591716
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ID:1690
picturePath:image/2807.png
answer:解:∵∠C=90°,BC=3,AC=2,∴AB=√{3^{2}+2^{2}}=√{13},∴tanB=\frac{AC}{CB}=\frac{2}{3}．
故选:A．
answer:3
correct answer:3
total:1691  correct:483  acc:0.28562980484920164
------------------------------------------------------------------------------------
ID:1691
picturePath:image/2811.png
answer:20°
answer:20
correct answer:50
total:1692  correct:483  acc:0.2854609929078014
------------------------------------------------------------------------------------
ID:1692
picturePath:image/2814.png
answer:This is a problem involving geometry and trigonometry. Given that OA=4 and B is the midpoint of OA, OB is half of OA, which is 2. Since Q is the midpoint of PA, AQ is half of PA, which is also 2. Therefore, AQ=2 and OQ=OA-AQ=2. Using the Pythagorean theorem, we can find that OQ^2+QB^2=OA^2, which simplifies to 4+QB^2=16. Solving for QB, we find that QB=√{16-4}=√{12}=2√{3}. Finally, using the definition of cosine, we have cos∠OQB=\frac{OQ}{QB}=\frac{2}{2√{3}}=\frac{√{3}}{3}. Therefore, the answer is \frac{√{3}}{3}.
answer:3
correct answer:4
total:1693  correct:483  acc:0.2852923803898405
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ID:1693
picturePath:image/2820.png
answer:8.
answer:8
correct answer:8
total:1694  correct:484  acc:0.2857142857142857
------------------------------------------------------------------------------------
ID:1694
picturePath:image/2821.png
answer:在Rt△ABC中,∠ACB=90°,CD是AB边上的中线,若BC=6,AC=8,则tan∠ACD的值为\frac{BC}{AC}=\frac{6}{8}=\frac{3}{4}。
answer:4
correct answer:4
total:1695  correct:485  acc:0.2861356932153392
------------------------------------------------------------------------------------
ID:1695
picturePath:image/2822.png
answer:解:∵∠AOD=130°,∴∠BOD=50°,∴∠C=25°,
故选:A．
answer:25
correct answer:25
total:1696  correct:486  acc:0.2865566037735849
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ID:1696
picturePath:image/2824.png
answer:根据勾股定理，在直角△ABC中，AB² + AC² = BC²，即3² + 4² = BC²，解得BC = √(3² + 4²) = √(9 + 16) = √25 = 5。

由于AD⊥BC，∠DAC是△ABC的外角，且∠BAC=90°，所以∠DAC=90°+∠BAC=90°+90°=180°。

sin∠DAC = opposite/hypotenuse，即sin∠DAC = AD/BC。

在直角△ADC中，AD² + CD² = AC²，即AD² + (BC-AC)² = AC²，解得AD² = AC² - (BC-AC)² = 4² - (5-4)² = 16 - 1 = 15。

所以AD = √15，sin∠DAC = AD/BC = √15/5 = √3/√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√5 = √3√5/5√
answer:5
correct answer:5
total:1697  correct:487  acc:0.2869770182675309
------------------------------------------------------------------------------------
ID:1697
picturePath:image/2826.png
answer:0.4
answer:0.4
correct answer:3
total:1698  correct:487  acc:0.2868080094228504
------------------------------------------------------------------------------------
ID:1698
picturePath:image/2829.png
answer:tan∠BCD=\frac{CD}{AD}=\frac{AC}{AB}=\frac{4}{5}
answer:5
correct answer:4
total:1699  correct:487  acc:0.2866391995291348
------------------------------------------------------------------------------------
ID:1699
picturePath:image/2832.png
answer:8
answer:8
correct answer:16
total:1700  correct:487  acc:0.28647058823529414
------------------------------------------------------------------------------------
ID:1700
picturePath:image/2835.png
answer:4/13
answer:13
correct answer:13
total:1701  correct:488  acc:0.28689006466784245
------------------------------------------------------------------------------------
ID:1701
picturePath:image/2836.png
answer:解:∵E、F分别是AB、AD的中点,∴EF是△ABD的中位线,∴BC=2EF=2×2=4,∴S△ABD=\frac{1}{2}×BC•AD=\frac{1}{2}×4×CD=6,∴S△ABC=\frac{1}{2}×S△ABD=\frac{1}{2}×6=3,∴S△ACB=\frac{1}{2}×S△ABC=\frac{1}{2}×3=1.5,∴S△ACD=S△ABD-S△ACB=6-1.5=4.5,∴\frac{S△ACD}{S△ABD}=\frac{4.5}{6}=\frac{3}{4},∴\frac{AC}{AB}=\frac{3}{4},∴\frac{AC^{2}}{AB^{2}}=\frac{9}{16},∴\frac{AC^{2}+BC^{2}}{AB^{2}}=\frac{9+25}{16}=\frac{34}{16}=\frac{17}{8},∴\frac{AC^{2}+BC^{2}}{AB^{2}}+1=\frac{17}{8}+1=\frac{25}{8},∴\frac{AC^{2}+BC^{2}}{AB^{2}}=\frac{25}{8}-1=\frac{17}{8},∴tanC=\frac{AC^{2}+BC^{2}}{AB^{2}}=\frac{17}{8}．故选:A．
answer:8
correct answer:3
total:1702  correct:488  acc:0.28672150411280845
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ID:1702
picturePath:image/2837.png
answer:解:∵∠C=90°,BC=1,AC=2,∴AB=√{AC^{2}+BC^{2}}=√{1^{2}+2^{2}}=√{5},∴tanA=\frac{BC}{AC}=\frac{1}{2}．
故选:A．
answer:2
correct answer:2
total:1703  correct:489  acc:0.28714034057545507
------------------------------------------------------------------------------------
ID:1703
picturePath:image/2838.png
answer:tan∠BAE=\frac{AC}{AB}=\frac{2}{2}=1.
answer:1
correct answer:2
total:1704  correct:489  acc:0.2869718309859155
------------------------------------------------------------------------------------
ID:1704
picturePath:image/2840.png
answer:解:∵AD是直径,∴∠ABC=90°,∵AD=3,AC=2,∴AB=√{AD^{2}+AC^{2}}=√{3^{2}+2^{2}}=√{9+4}=√{13},∴\cos∠B=\frac{AC}{AB}=\frac{2}{√{13}}=\frac{2√{13}}{13},∴\cos∠B的值是\frac{2√{13}}{13}．故选:D．
answer:13
correct answer:3
total:1705  correct:489  acc:0.2868035190615836
------------------------------------------------------------------------------------
ID:1705
picturePath:image/2844.png
answer:Since \( \overline{AB} \) is a diameter, \( \angle ACB = 90^\circ \). Given that \( \angle CAB = 30^\circ \), we can find \( \angle CBA \) by subtracting \( \angle CAB \) from \( \angle ACB \): \( \angle CBA = 90^\circ - 30^\circ = 60^\circ \). 

Since \( \overline{CD} \) is parallel to \( \overline{AB} \), corresponding angles \( \angle CBA \) and \( \angle D \) are congruent. Therefore, \( \angle D = 60^\circ \).
answer:60
correct answer:60
total:1706  correct:490  acc:0.28722157092614303
------------------------------------------------------------------------------------
ID:1706
picturePath:image/2849.png
answer:In a right-angled triangle, the tangent of an angle is defined as the ratio of the length of the side opposite to the angle to the length of the side adjacent to the angle. In this case, \(tan B = \frac{AC}{BC} = \frac{3}{2}\). Therefore, the value of \(tan B\) is \(\frac{3}{2}\).
answer:2
correct answer:2
total:1707  correct:491  acc:0.28763913298183946
------------------------------------------------------------------------------------
ID:1707
picturePath:image/2850.png
answer:根据三角函数的定义，sinA = \frac{BC}{AB}，已知sinA = \frac{1}{3}，BC = 1，可以解出AB = \frac{BC}{sinA} \times sinA = \frac{1}{\frac{1}{3}} \times \frac{1}{3} = \frac{3}{1} \times \frac{1}{3} = 1。因此，AB = 1。
answer:1
correct answer:3
total:1708  correct:491  acc:0.28747072599531615
------------------------------------------------------------------------------------
ID:1708
picturePath:image/2851.png
answer:The answer is \frac{1}{\sqrt{5}}.
answer:5
correct answer:2
total:1709  correct:491  acc:0.28730251609128143
------------------------------------------------------------------------------------
ID:1709
picturePath:image/2858.png
answer:解:∵在Rt△ABC中,AC=2,BC=3,∴AB=√{AC^{2}+BC^{2}}=√{2^{2}+3^{2}}=√{13},
∴tanB=\frac{AC}{AB}=\frac{2}{√{13}}=\frac{2√{13}}{13},
即tanB的值是\frac{2√{13}}{13}．
故选:D．
answer:13
correct answer:3
total:1710  correct:491  acc:0.28713450292397663
------------------------------------------------------------------------------------
ID:1710
picturePath:image/2859.png
answer:Since ∠C=90°, we can use the Pythagorean theorem to find the length of side BC. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we have:

AB^2 = AC^2 + BC^2
5^2 = 3^2 + BC^2
25 = 9 + BC^2
BC^2 = 25 - 9
BC^2 = 16
BC = 4

Now that we know BC is 4, we can use the definition of sine to find sinB:

sinB = opposite/hypotenuse
sinB = AB/BC
sinB = 5/4

Therefore, sinB = 5/4.
answer:4
correct answer:5
total:1711  correct:491  acc:0.2869666861484512
------------------------------------------------------------------------------------
ID:1711
picturePath:image/2862.png
answer:tanA + tanB = \frac{AB}{BC} + \frac{BC}{AB} = \frac{AB^2 + BC^2}{AB \cdot BC} = \frac{AB^2 + (AB)^2}{AB \cdot AB} = \frac{2AB^2}{AB^2} = 2.
answer:2
correct answer:4
total:1712  correct:491  acc:0.2867990654205608
------------------------------------------------------------------------------------
ID:1712
picturePath:image/2865.png
answer:2
answer:2
correct answer:2
total:1713  correct:492  acc:0.287215411558669
------------------------------------------------------------------------------------
ID:1713
picturePath:image/2866.png
answer:解:∵CD是斜边AB上的中线,∴AB=2CD=4,在Rt△ABC中,AC=3,∴BC=√{AB^{2}-AC^{2}}=√{4^{2}-3^{2}}=√{7},∴sinB=\frac{BC}{AB}=\frac{√{7}}{4}．故选:B．
answer:4
correct answer:4
total:1714  correct:493  acc:0.28763127187864646
------------------------------------------------------------------------------------
ID:1714
picturePath:image/2867.png
answer:tanα=\frac{BC}{AC}=\frac{4}{3}
answer:3
correct answer:4
total:1715  correct:493  acc:0.287463556851312
------------------------------------------------------------------------------------
ID:1715
picturePath:image/2870.png
answer:解:∵AC=5,BC=12,AB=13,∴tan∠B=\frac{AC}{BC}=\frac{5}{12}．
故选:A．
answer:12
correct answer:12
total:1716  correct:494  acc:0.2878787878787879
------------------------------------------------------------------------------------
ID:1716
picturePath:image/2872.png
answer:解:∵Rt△ABC中,∠C=90°,BC=3,AC=4,∴AB=√{AC^{2}+BC^{2}}=√{3^{2}+4^{2}}=5,∴sinA=\frac{BC}{AB}=\frac{3}{5}．故选:A．
answer:5
correct answer:5
total:1717  correct:495  acc:0.28829353523587653
------------------------------------------------------------------------------------
ID:1717
picturePath:image/2875.png
answer:解:∵CD是Rt△ABC斜边上的高,∴BC=2CD,∴∠BCD=∠BAC,∵AB=5,AC=3,∴tan∠BAC=\frac{3}{5},∴tan∠BCD=tan∠BAC=\frac{3}{5}．故选:A．
answer:5
correct answer:3
total:1718  correct:495  acc:0.2881257275902212
------------------------------------------------------------------------------------
ID:1718
picturePath:image/2876.png
answer:解:∵∠AOC=100°,∴∠BOC=180°-∠AOC=180°-100°=80°,∵OM∥BC,∴∠BOM=∠BOC=80°,∴∠AMO=∠BOM=80°．故选:B．
answer:80
correct answer:50
total:1719  correct:495  acc:0.2879581151832461
------------------------------------------------------------------------------------
ID:1719
picturePath:image/2881.png
answer:This is a right triangle, and we know that the sine of angle A is 3/5. Since the sum of the sine and cosine of an angle in a right triangle is 1, we can find the cosine of angle B by subtracting the sine of angle A from 1. Therefore, \cosB=1-\sinA=1-\frac{3}{5}=\frac{2}{5}.
answer:5
correct answer:5
total:1720  correct:496  acc:0.28837209302325584
------------------------------------------------------------------------------------
ID:1720
picturePath:image/2882.png
answer:Since this is a right triangle, we know that the sine of angle A is the opposite side over the hypotenuse, and the cosine of angle B is the adjacent side over the hypotenuse. Given that sinA = 1/3, we can use the Pythagorean theorem to find the lengths of the sides. Let's denote the hypotenuse as c, the opposite side as a, and the adjacent side as b. We have: a^2 + b^2 = c^2. Since sinA = a/c, we can rewrite the equation as: (a/c)^2 + (b/c)^2 = 1. Substituting sinA = 1/3, we get: (1/3)^2 + (b/c)^2 = 1. Solving for b/c, we get: (b/c)^2 = 1 - (1/3)^2 = 2/3. Taking the square root of both sides, we get: b/c = √(2/3). Therefore, cosB = b/c = √(2/3).
answer:3
correct answer:3
total:1721  correct:497  acc:0.2887855897733876
------------------------------------------------------------------------------------
ID:1721
picturePath:image/2883.png
answer:3/5
answer:5
correct answer:5
total:1722  correct:498  acc:0.289198606271777
------------------------------------------------------------------------------------
ID:1722
picturePath:image/2886.png
answer:解:∵AC是⊙O的直径,∠ACB=52°,∴∠AOC=180°-∠ACB=180°-52°=128°,∴∠D=\frac{1}{2}∠AOC=\frac{1}{2}×128°=64°．故选:B．
answer:64
correct answer:38
total:1723  correct:498  acc:0.2890307603017992
------------------------------------------------------------------------------------
ID:1723
picturePath:image/2892.png
answer:解:∵Rt△ABC中,∠C=90°,∠BAC=67.5°,∴∠ABC=22.5°,∵AD=BD,∴∠ADC=∠B,∴∠ADC=∠ABC=22.5°,∴sin∠ADC=sin22.5°=\frac{1}{2}．故选:B．
answer:2
correct answer:2
total:1724  correct:499  acc:0.2894431554524362
------------------------------------------------------------------------------------
ID:1724
picturePath:image/2894.png
answer:解:∵AB=2BC,∴sinB=\frac{BC}{AB}=\frac{BC}{2BC}=\frac{1}{2}．
故选:B．
answer:2
correct answer:2
total:1725  correct:500  acc:0.2898550724637681
------------------------------------------------------------------------------------
ID:1725
picturePath:image/2897.png
answer:解:∵AB是半圆的直径,∴∠ACB=90°,∵∠ABC=50°,∴∠A=90°-∠ABC=40°,∵点D是⁀{AC}的中点,∴∠ADB=∠A=40°,∴∠BCD=∠ADB=40°．故选:B．
answer:40
correct answer:115
total:1726  correct:500  acc:0.28968713789107764
------------------------------------------------------------------------------------
ID:1726
picturePath:image/2900.png
answer:解:∵AB=2BC,∴sinB=\frac{BC}{AB}=\frac{BC}{2BC}=\frac{1}{2}．故选:A．
answer:2
correct answer:2
total:1727  correct:501  acc:0.29009843659525186
------------------------------------------------------------------------------------
ID:1727
picturePath:image/2902.png
answer:tanA=\frac{BC}{AC}=\frac{AB\cdot\sin A}{AB\cdot\cos A}=\frac{\sin A}{\cos A}=\frac{1}{\cos A}=\frac{1}{\frac{√{3}}{2}}=\frac{2}{√{3}}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{3}=\frac{2√{3}}{
answer:3
correct answer:3
total:1728  correct:502  acc:0.29050925925925924
------------------------------------------------------------------------------------
ID:1728
picturePath:image/2904.png
answer:解:∵∠A=70°,∠C=50°,∴∠AEB=180°-∠A-∠C=60°,∴sin∠AEB=sin60°=√{3}÷2,故选:A．
answer:2
correct answer:2
total:1729  correct:503  acc:0.2909196067090804
------------------------------------------------------------------------------------
ID:1729
picturePath:image/2905.png
answer:解:∵AB是半圆的直径,弦AD,BC相交于P,∠DPB=60°,D是弧BC的中点,∴∠BAC=\frac{1}{2}∠DPB=30°,∴∠ADC=90°-∠BAC=60°,∴tan∠ADC=tan60°=√3,故选:A．
answer:3
correct answer:3
total:1730  correct:504  acc:0.29132947976878615
------------------------------------------------------------------------------------
ID:1730
picturePath:image/2907.png
answer:∠AOC=180°-∠A-∠C=180°-75°=105°.
answer:105
correct answer:150
total:1731  correct:504  acc:0.29116117850953205
------------------------------------------------------------------------------------
ID:1731
picturePath:image/2914.png
answer:根据三角函数的定义，sinA是直角三角形ABC中∠A的对边与斜边的比值，cosB是直角三角形ABC中∠B的邻边与斜边的比值。已知∠A=30°，所以sinA=\frac{1}{2}，cosB=\frac{√{3}}{2}。因此，sinA+\cosB=\frac{1}{2}+\frac{√{3}}{2}=\frac{1+√{3}}{2}。
answer:2
correct answer:1
total:1732  correct:504  acc:0.2909930715935335
------------------------------------------------------------------------------------
ID:1732
picturePath:image/2917.png
answer:The answer is 100.
answer:100
correct answer:25
total:1733  correct:504  acc:0.2908251586843624
------------------------------------------------------------------------------------
ID:1733
picturePath:image/2921.png
answer:在Rt△ABC中,∠BAC=90°,AC=3,AB=4,则∠DAC=∠BAC=90°,所以∠DAC的余弦值等于∠BAC的余弦值,即\cos∠DAC=\cos∠BAC=\frac{AC}{AB}=\frac{3}{4}。所以,∠DAC的余弦值为\frac{3}{4}。
answer:4
correct answer:5
total:1734  correct:504  acc:0.2906574394463668
------------------------------------------------------------------------------------
ID:1734
picturePath:image/2928.png
answer:1.5×(180°-25°-30°)=105°.故选D.
answer:105
correct answer:110
total:1735  correct:504  acc:0.2904899135446686
------------------------------------------------------------------------------------
ID:1735
picturePath:image/2940.png
answer:The answer is 32.
answer:32
correct answer:25
total:1736  correct:504  acc:0.2903225806451613
------------------------------------------------------------------------------------
ID:1736
picturePath:image/2941.png
answer:The answer is \frac{1}{3}
answer:3
correct answer:2
total:1737  correct:504  acc:0.29015544041450775
------------------------------------------------------------------------------------
ID:1737
picturePath:image/2946.png
answer:This is a geometry problem involving a right triangle and trigonometric functions. Given that \(\frac{AD}{AB}=\frac{5}{3}\), we can use the Pythagorean theorem to find the lengths of the sides of the triangle. Since \(DC\) is half of \(BD\), we know that \(DC = \frac{1}{2}BD\). Using the Pythagorean theorem, we have \(\frac{1}{2}BD = \sqrt{BD^2 - AD^2}\). Solving for \(BD\), we get \(BD = \frac{2AD}{\sqrt{3}}\). Substituting this into the equation for \(DC\), we have \(DC = \frac{1}{2} \cdot \frac{2AD}{\sqrt{3}} = \frac{AD}{\sqrt{3}}\). Using the definition of tangent, we have \(tan∠CAD = \frac{DC}{AD} = \frac{\frac{AD}{\sqrt{3}}}{AD} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\). Therefore, the value of \(tan∠CAD\) is \(\frac{\sqrt{3}}{3}\).
answer:3
correct answer:5
total:1738  correct:504  acc:0.2899884925201381
------------------------------------------------------------------------------------
ID:1738
picturePath:image/2950.png
answer:Since \( \tan A = \frac{1}{2} \), we know that the ratio of the length of side opposite angle A to the length of side adjacent to angle A is 1:2. In a right triangle, the side opposite angle A is the height, and the side adjacent to angle A is the base. Therefore, the length of side BC is twice the length of side AC, which is \( 2 \times 4 = 8 \).
answer:8
correct answer:2
total:1739  correct:504  acc:0.2898217366302473
------------------------------------------------------------------------------------
ID:1739
picturePath:image/2951.png
answer:This is a problem involving geometry and trigonometry. Given that \(AB = AC\) and \(∠BCA = 65°\), we can use the properties of an isosceles triangle and the angle bisector theorem to solve for the angles. Since \(BC\) is an angle bisector of \(AC\), we can use the angle bisector theorem to find that \(AD \cdot CD = AB \cdot AC\). Since \(AB = AC\), we can simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\). Since \(AB = AC\), we can further simplify this to \(AD \cdot CD = AB^2\
answer:2
correct answer:15
total:1740  correct:504  acc:0.2896551724137931
------------------------------------------------------------------------------------
ID:1740
picturePath:image/2953.png
answer:tanα=\frac{3}{4}
answer:4
correct answer:4
total:1741  correct:505  acc:0.2900631820792648
------------------------------------------------------------------------------------
ID:1741
picturePath:image/2955.png
answer:tan∠CAD = \frac{tanB}{1-tan^2B} = \frac{\frac{5}{3}}{1-\frac{25}{9}} = \frac{5}{3} \times \frac{9}{4} = \frac{45}{12} = \frac{15}{4}
answer:4
correct answer:5
total:1742  correct:505  acc:0.2898966704936854
------------------------------------------------------------------------------------
ID:1742
picturePath:image/2958.png
answer:4
answer:4
correct answer:2
total:1743  correct:505  acc:0.2897303499713138
------------------------------------------------------------------------------------
ID:1743
picturePath:image/2960.png
answer:96
answer:96
correct answer:24
total:1744  correct:505  acc:0.28956422018348627
------------------------------------------------------------------------------------
ID:1744
picturePath:image/2962.png
answer:解:∵∠AOB=110°,∴∠ACB=\frac{1}{2}∠AOB=55°．故选:A．
answer:55
correct answer:125
total:1745  correct:505  acc:0.28939828080229224
------------------------------------------------------------------------------------
ID:1745
picturePath:image/2963.png
answer:20
answer:20
correct answer:110
total:1746  correct:505  acc:0.28923253150057276
------------------------------------------------------------------------------------
ID:1746
picturePath:image/2965.png
answer:根据圆周角定理，在同圆或等圆中，同弧或等弧所对的圆周角相等。因此，∠AOB = 2∠ACB = 2 * 30° = 60°。
answer:60
correct answer:60
total:1747  correct:506  acc:0.2896393817973669
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ID:1747
picturePath:image/2969.png
answer:The answer is 72.
answer:72
correct answer:72
total:1748  correct:507  acc:0.290045766590389
------------------------------------------------------------------------------------
ID:1748
picturePath:image/2970.png
answer:解:∵∠BOD=50°,∴∠BAD=\frac{1}{2}∠BOD=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:25
total:1749  correct:508  acc:0.2904516866781018
------------------------------------------------------------------------------------
ID:1749
picturePath:image/2972.png
answer:解:∵∠ACB与∠AOB是同弧所对的圆周角与圆心角,∠ACB=46°,∴∠AOB=2∠ACB=92°．故选:D．
answer:92
correct answer:92
total:1750  correct:509  acc:0.29085714285714287
------------------------------------------------------------------------------------
ID:1750
picturePath:image/2973.png
answer:25
answer:25
correct answer:25
total:1751  correct:510  acc:0.2912621359223301
------------------------------------------------------------------------------------
ID:1751
picturePath:image/2974.png
answer:解:∵AB为⊙O的直径,∴∠ADB=180°-∠ADC=180°-35°=145°,∴∠CAB=\frac{1}{2}∠ADB=\frac{1}{2}×145°=72.5°．故选:B．
answer:72.5
correct answer:55
total:1752  correct:510  acc:0.2910958904109589
------------------------------------------------------------------------------------
ID:1752
picturePath:image/2979.png
answer:解:∵∠BCD与∠BOD是同弧所对的圆周角与圆心角,∠BCD=130°,∴∠BOD=2∠BCD=260°．故选:D．
answer:260
correct answer:100
total:1753  correct:510  acc:0.29092983456930976
------------------------------------------------------------------------------------
ID:1753
picturePath:image/2980.png
answer:The answer is 64.
answer:64
correct answer:58
total:1754  correct:510  acc:0.29076396807297605
------------------------------------------------------------------------------------
ID:1754
picturePath:image/2981.png
answer:70
answer:70
correct answer:70
total:1755  correct:511  acc:0.29116809116809117
------------------------------------------------------------------------------------
ID:1755
picturePath:image/2982.png
answer:50
answer:50
correct answer:115
total:1756  correct:511  acc:0.29100227790432803
------------------------------------------------------------------------------------
ID:1756
picturePath:image/2983.png
answer:This is a geometric figure with points A, B, C, D, E, F, G, and O. The figure shows a circle with a diameter AB, and a perpendicular line CD that intersects the circle at point E. Another line AG intersects the circle at point F, and extends beyond the circle to intersect line CD at point D. A line GC intersects the circle at point G, and extends beyond the circle to intersect line CD at point C. The figure also shows an angle ∠BAD with a measure of 25 degrees. The question is asking for the measure of angle ∠AGD.
answer:25
correct answer:65
total:1757  correct:511  acc:0.2908366533864542
------------------------------------------------------------------------------------
ID:1757
picturePath:image/2985.png
answer:解:∵∠BAC=20°,∴∠DAC=\frac{1}{2}∠BAC=\frac{1}{2}×20°=10°,∴∠ADC=180°-∠DAC=180°-10°=170°．故选:C．
answer:170
correct answer:110
total:1758  correct:511  acc:0.2906712172923777
------------------------------------------------------------------------------------
ID:1758
picturePath:image/2986.png
answer:解:∵∠AOB与∠ACB都对⁀{AB},且∠AOB=110°,∴∠ACB=\frac{1}{2}∠AOB=55°,
故选:B．
answer:55
correct answer:125
total:1759  correct:511  acc:0.2905059693007391
------------------------------------------------------------------------------------
ID:1759
picturePath:image/2987.png
answer:60°
answer:60
correct answer:60
total:1760  correct:512  acc:0.2909090909090909
------------------------------------------------------------------------------------
ID:1760
picturePath:image/2988.png
answer:60
answer:60
correct answer:55
total:1761  correct:512  acc:0.29074389551391255
------------------------------------------------------------------------------------
ID:1761
picturePath:image/2989.png
answer:解:∵四边形ABCD中,∠B=∠D,∠D=72°,∴∠B=72°,∴∠A=∠B=72°,∵⊙O经过点A,C,D,与BC相交于点E,∴∠BAE=\frac{1}{2}∠A=\frac{1}{2}×72°=36°．故选:B．
answer:36
correct answer:36
total:1762  correct:513  acc:0.29114642451759365
------------------------------------------------------------------------------------
ID:1762
picturePath:image/2992.png
answer:The answer is 20.
answer:20
correct answer:100
total:1763  correct:513  acc:0.2909812819058423
------------------------------------------------------------------------------------
ID:1763
picturePath:image/2993.png
answer:解:∵∠A=35°,∴∠D=35°,∵∠BPC=78°,∴∠B=180°-35°-78°=67°,
故选:A．
answer:67
correct answer:43
total:1764  correct:513  acc:0.29081632653061223
------------------------------------------------------------------------------------
ID:1764
picturePath:image/2994.png
answer:解:∵∠ACB=30°,∴∠AOB=2∠ACB=60°,∵点E、F分别是AC、BC的中点,∴EF是△ABC的中位线,∴EF=\frac{1}{2}BC,∵直线EF与⊙O交于G、H两点,∴GH是直径,∵⊙O的半径为10,∴GH=20,∴GE+FH=GH=20．故选:B．
answer:20
correct answer:15
total:1765  correct:513  acc:0.2906515580736544
------------------------------------------------------------------------------------
ID:1765
picturePath:image/2995.png
answer:35
answer:35
correct answer:35
total:1766  correct:514  acc:0.29105322763306907
------------------------------------------------------------------------------------
ID:1766
picturePath:image/2996.png
answer:解:∵OC⊥AB,∴⁀{AC}=⁀{BC},∴∠OBA=\frac{1}{2}∠ADC=\frac{1}{2}×32°=16°．故选:A．
answer:16
correct answer:26
total:1767  correct:514  acc:0.29088851160158463
------------------------------------------------------------------------------------
ID:1767
picturePath:image/2997.png
answer:解:∵AB∥OC,∠BCO=21°,∴∠ABC=∠BCO=21°,∵∠ABC与∠AOC是同弧所对的圆周角与圆心角,∴∠AOC=2∠ABC=2×21°=42°．故选:D．
answer:42
correct answer:42
total:1768  correct:515  acc:0.291289592760181
------------------------------------------------------------------------------------
ID:1768
picturePath:image/2998.png
answer:解:∵AB是直径,∴∠ACB=90°,∵∠C比∠D大36°,∴∠D=∠C-36°,∵∠A+∠ACB+∠D+∠C=180°,∴∠A=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D+∠C)=180°-90°-(∠D
answer:90
correct answer:27
total:1769  correct:515  acc:0.2911249293386094
------------------------------------------------------------------------------------
ID:1769
picturePath:image/2999.png
answer:根据题意,∠BOC是∠ABC和∠ACB的外角平分线的交点,所以∠BOC=\frac{1}{2}∠ABC+\frac{1}{2}∠ACB,即40°=\frac{1}{2}∠ABC+\frac{1}{2}∠ACB,解得∠ABC+∠ACB=80°,所以∠A=180°-(∠ABC+∠ACB)=180°-80°=100°.故选B.
answer:100
correct answer:100
total:1770  correct:516  acc:0.29152542372881357
------------------------------------------------------------------------------------
ID:1770
picturePath:image/3000.png
answer:解:∵AC是⊙O的直径,∴∠ABC=90°,∵∠CAB=55°,∴∠A=90°-55°=35°,∴∠ADB=∠A=35°．故选:B．
answer:35
correct answer:35
total:1771  correct:517  acc:0.2919254658385093
------------------------------------------------------------------------------------
ID:1771
picturePath:image/3001.png
answer:解:∵DE∥OA,∴∠DOA=∠D=50°,∵CD为圆O的直径,∴∠COD=90°,∴∠C=90°-∠DOA=90°-50°=40°．故选:B．
answer:40
correct answer:25
total:1772  correct:517  acc:0.2917607223476298
------------------------------------------------------------------------------------
ID:1772
picturePath:image/3003.png
answer:4√3
answer:3
correct answer:3
total:1773  correct:518  acc:0.2921601804850536
------------------------------------------------------------------------------------
ID:1773
picturePath:image/3005.png
answer:OD=\frac{1}{2}BC=\frac{1}{2}×2×\frac{60°}{90°}=1.
answer:1
correct answer:1
total:1774  correct:519  acc:0.29255918827508454
------------------------------------------------------------------------------------
ID:1774
picturePath:image/3006.png
answer:解:∵∠AOC=140°,∴∠B=\frac{1}{2}∠AOC=\frac{1}{2}×140°=70°．故选:B．
answer:70
correct answer:110
total:1775  correct:519  acc:0.2923943661971831
------------------------------------------------------------------------------------
ID:1775
picturePath:image/3007.png
answer:12
answer:12
correct answer:8
total:1776  correct:519  acc:0.2922297297297297
------------------------------------------------------------------------------------
ID:1776
picturePath:image/3009.png
answer:10√3
answer:3
correct answer:3
total:1777  correct:520  acc:0.2926280247608329
------------------------------------------------------------------------------------
ID:1777
picturePath:image/3011.png
answer:40
answer:40
correct answer:80
total:1778  correct:520  acc:0.2924634420697413
------------------------------------------------------------------------------------
ID:1778
picturePath:image/3012.png
answer:80°
answer:80
correct answer:50
total:1779  correct:520  acc:0.2922990444069702
------------------------------------------------------------------------------------
ID:1779
picturePath:image/3013.png
answer:解:∵AB是半圆⊙O的直径,∴∠ACB=90°,∵∠BAC=50°,∴∠C=90°-∠BAC=90°-50°=40°．故选:B．
answer:40
correct answer:65
total:1780  correct:520  acc:0.29213483146067415
------------------------------------------------------------------------------------
ID:1780
picturePath:image/3014.png
answer:解:∵⁀{AC}=⁀{BC},∴∠ACO=∠BCO,∵OA=OC,∴∠OAC=∠OCO,∴∠AOC=\frac{1}{2}∠BAO=\frac{1}{2}×37°=18°．故选:A．
answer:18
correct answer:127
total:1781  correct:520  acc:0.291970802919708
------------------------------------------------------------------------------------
ID:1781
picturePath:image/3015.png
answer:解:∵四边形ABCD是平行四边形,∴∠ABC=∠ADC,∵BE是直径,∴∠BAE=90°,∵∠E=36°,∴∠ABC=90°-36°=54°,∴∠ADC=54°．故选:B．
answer:54
correct answer:54
total:1782  correct:521  acc:0.29236812570145904
------------------------------------------------------------------------------------
ID:1782
picturePath:image/3016.png
answer:解:∵∠ACB与∠AOB是同弧所对的圆周角与圆心角,∠ACB=130°,∴∠AOB=2∠ACB=260°．故选:D．
answer:260
correct answer:100
total:1783  correct:521  acc:0.29220415030846886
------------------------------------------------------------------------------------
ID:1783
picturePath:image/3017.png
answer:解:∵∠ACD与∠B是同弧所对的圆心角与圆周角,∠ACD=130°,∴∠B=\frac{180°-∠ACD}{2}=40°,∴∠BAD=∠B=40°．故选:B．
answer:40
correct answer:40
total:1784  correct:522  acc:0.29260089686098656
------------------------------------------------------------------------------------
ID:1784
picturePath:image/3018.png
answer:解:∵∠ACB与∠AOB是同弧所对的圆周角与圆心角,∠ACB=36°,∴∠AOB=2∠ACB=72°．故选:D．
answer:72
correct answer:72
total:1785  correct:523  acc:0.29299719887955183
------------------------------------------------------------------------------------
ID:1785
picturePath:image/3019.png
answer:Since \( \overline{AB} \) is a diameter, \( \angle AOB = 180^\circ \). Given that \( \angle AOD = 30^\circ \), we can find \( \angle ABO \) by subtracting \( \angle AOD \) from \( \angle AOB \): \( \angle ABO = 180^\circ - 30^\circ = 150^\circ \). Since \( \overline{AB} \) and \( \overline{CD} \) are parallel, corresponding angles \( \angle ABO \) and \( \angle BCD \) are congruent, so \( \angle BCD = \angle ABO = \boxed{150^\circ} \).The answer is: 150
answer:150
correct answer:105
total:1786  correct:523  acc:0.29283314669652855
------------------------------------------------------------------------------------
ID:1786
picturePath:image/3020.png
answer:4
answer:4
correct answer:4
total:1787  correct:524  acc:0.2932288752098489
------------------------------------------------------------------------------------
ID:1787
picturePath:image/3022.png
answer:25°
answer:25
correct answer:25
total:1788  correct:525  acc:0.2936241610738255
------------------------------------------------------------------------------------
ID:1788
picturePath:image/3025.png
answer:解:∵OD⊥BC,∠ABC=40°,∴∠AOD=90°+∠ABC=90°+40°=130°,∴∠DCB=\frac{1}{2}∠AOD=\frac{1}{2}×130°=65°．故选:B．
answer:65
correct answer:25
total:1789  correct:525  acc:0.29346003353828953
------------------------------------------------------------------------------------
ID:1789
picturePath:image/3027.png
answer:60
answer:60
correct answer:60
total:1790  correct:526  acc:0.293854748603352
------------------------------------------------------------------------------------
ID:1790
picturePath:image/3028.png
answer:80°
answer:80
correct answer:80
total:1791  correct:527  acc:0.294249022892239
------------------------------------------------------------------------------------
ID:1791
picturePath:image/3029.png
answer:17
answer:17
correct answer:55
total:1792  correct:527  acc:0.29408482142857145
------------------------------------------------------------------------------------
ID:1792
picturePath:image/3032.png
answer:60°
answer:60
correct answer:30
total:1793  correct:527  acc:0.29392080312325713
------------------------------------------------------------------------------------
ID:1793
picturePath:image/3035.png
answer:80°
answer:80
correct answer:80
total:1794  correct:528  acc:0.29431438127090304
------------------------------------------------------------------------------------
ID:1794
picturePath:image/3036.png
answer:6
answer:6
correct answer:3
total:1795  correct:528  acc:0.29415041782729806
------------------------------------------------------------------------------------
ID:1795
picturePath:image/3037.png
answer:10
answer:10
correct answer:110
total:1796  correct:528  acc:0.29398663697104677
------------------------------------------------------------------------------------
ID:1796
picturePath:image/3038.png
answer:根据圆周角定理，∠B与∠A是同弧所对的圆周角，所以∠B=∠A=55°；根据圆内接四边形的对角互补，∠B+∠APD=180°，所以∠B=180°-∠APD=180°-80°=100°。因此，∠B等于100°。
answer:100
correct answer:45
total:1797  correct:528  acc:0.2938230383973289
------------------------------------------------------------------------------------
ID:1797
picturePath:image/3039.png
answer:80°
answer:80
correct answer:80
total:1798  correct:529  acc:0.2942157953281424
------------------------------------------------------------------------------------
ID:1798
picturePath:image/3040.png
answer:3√{2}
answer:2
correct answer:2
total:1799  correct:530  acc:0.2946081156197888
------------------------------------------------------------------------------------
ID:1799
picturePath:image/3041.png
answer:解:∵∠DCB=110°,∴∠BOC=2∠DCB=2×110°=220°,∴∠AED=\frac{1}{2}∠BOC=\frac{1}{2}×220°=110°．故选:B．
answer:110
correct answer:20
total:1800  correct:530  acc:0.29444444444444445
------------------------------------------------------------------------------------
ID:1800
picturePath:image/3042.png
answer:Since \(AB \parallel CD\), we have corresponding angles, so \(∠BOD = ∠BAD\). Given that \(∠BOD = 76°\) and \(∠BAD = 35°\), we can conclude that \(∠C = ∠BOD - ∠BAD = 76° - 35° = 41°\).
answer:41
correct answer:41
total:1801  correct:531  acc:0.29483620210993894
------------------------------------------------------------------------------------
ID:1801
picturePath:image/3043.png
answer:Since EF is parallel to AB, we have corresponding angles, so ∠E = ∠B = 39°. Since ∠C is a right angle, the sum of the angles in a triangle is 180°, so ∠1 = 180° - ∠E - ∠C = 180° - 39° - 90° = 51°.
answer:51
correct answer:51
total:1802  correct:532  acc:0.2952275249722531
------------------------------------------------------------------------------------
ID:1802
picturePath:image/3044.png
answer:根据圆周角定理，同弧所对的圆周角相等，所以∠BED=∠AOD=52°。又因为OD⊥AB，所以∠OCD=90°，所以∠DEB=∠OCD-∠BED=90°-52°=38°。故选B。
answer:38
correct answer:26
total:1803  correct:532  acc:0.29506378258458127
------------------------------------------------------------------------------------
ID:1803
picturePath:image/3045.png
answer:解:∵∠A=36°,∴∠BOC=2∠A=72°,∵OB=OC,∴∠OBC=\frac{180°-72°}{2}=54°．故选:B．
answer:54
correct answer:54
total:1804  correct:533  acc:0.29545454545454547
------------------------------------------------------------------------------------
ID:1804
picturePath:image/3046.png
answer:30
answer:30
correct answer:30
total:1805  correct:534  acc:0.29584487534626036
------------------------------------------------------------------------------------
ID:1805
picturePath:image/3047.png
answer:解:∵点A、B、C、D、E在⊙O上,∠ACE=25°,∠BDE=15°,∴∠AOB=2∠ACE=50°．故选:D．
answer:50
correct answer:80
total:1806  correct:534  acc:0.2956810631229236
------------------------------------------------------------------------------------
ID:1806
picturePath:image/3049.png
answer:解:∵∠D=35°,∴∠BOC=2∠D=2×35°=70°,∵OC=OB,∴∠OCB=∠OBC=\frac{180°-70°}{2}=55°．故选:B．
answer:55
correct answer:55
total:1807  correct:535  acc:0.29607083563918096
------------------------------------------------------------------------------------
ID:1807
picturePath:image/3051.png
answer:35°
answer:35
correct answer:35
total:1808  correct:536  acc:0.29646017699115046
------------------------------------------------------------------------------------
ID:1808
picturePath:image/3052.png
answer:60
answer:60
correct answer:140
total:1809  correct:536  acc:0.2962962962962963
------------------------------------------------------------------------------------
ID:1809
picturePath:image/3053.png
answer:40°
answer:40
correct answer:40
total:1810  correct:537  acc:0.29668508287292816
------------------------------------------------------------------------------------
ID:1810
picturePath:image/3054.png
answer:The answer is 40.
answer:40
correct answer:140
total:1811  correct:537  acc:0.2965212589729431
------------------------------------------------------------------------------------
ID:1811
picturePath:image/3055.png
answer:14
answer:14
correct answer:62
total:1812  correct:537  acc:0.29635761589403975
------------------------------------------------------------------------------------
ID:1812
picturePath:image/3057.png
answer:∠DCE=\frac{1}{2}(180^{°}-∠P)=\frac{1}{2}(180^{°}-70^{°})=65^{°}
answer:65
correct answer:40
total:1813  correct:537  acc:0.2961941533370105
------------------------------------------------------------------------------------
ID:1813
picturePath:image/3058.png
answer:50°
answer:50
correct answer:65
total:1814  correct:537  acc:0.2960308710033076
------------------------------------------------------------------------------------
ID:1814
picturePath:image/3059.png
answer:40
answer:40
correct answer:40
total:1815  correct:538  acc:0.29641873278236913
------------------------------------------------------------------------------------
ID:1815
picturePath:image/3060.png
answer:30
answer:30
correct answer:50
total:1816  correct:538  acc:0.29625550660792954
------------------------------------------------------------------------------------
ID:1816
picturePath:image/3061.png
answer:2
answer:2
correct answer:2
total:1817  correct:539  acc:0.2966428178315905
------------------------------------------------------------------------------------
ID:1817
picturePath:image/3062.png
answer:70
answer:70
correct answer:70
total:1818  correct:540  acc:0.297029702970297
------------------------------------------------------------------------------------
ID:1818
picturePath:image/3063.png
answer:This is a geometry problem involving circles and angles. Given that \( \theta \) is the angle at the center of the circle subtended by the arc \( BD \), and \( \theta \) is 50 degrees, we can use the formula for the measure of an angle subtended by an arc at the center of a circle: \( \theta = \frac{n}{360} \times 2\pi \times r \), where \( n \) is the length of the arc and \( r \) is the radius of the circle. Since \( \theta \) is 50 degrees, we have \( \frac{n}{360} \times 2\pi \times r = 50 \). Solving for \( n \), we get \( n = \frac{50 \times 360}{2\pi \times r} \). Since \( AB \) is the diameter of the circle, the radius \( r \) is half the length of \( AB \), so \( r = \frac{AB}{2} \). Substituting this into the equation for \( n \), we get \( n = \frac{50 \times 360}{2\pi \times \frac{AB}{2}} \). Simplifying, we get \( n = \frac{50 \times 360 \times 2}{2\pi \times AB} \). Since \( n \) is the length of the arc \( BD \), we have \( n = \frac{50 \times 360 \times 2}{2\pi \times AB} \). Solving for \( AB \), we get \( AB = \frac{50 \times 360 \times 2}{2\pi \times n} \). Substituting this into the equation for \( \theta \), we get \( \theta = \frac{\frac{50 \times 360 \times 2}{2\pi \times n}}{360} \times 2\pi \times \frac{AB}{2} \). Simplifying, we get \( \theta = \frac{50 \times 360 \times 2}{2\pi \times n} \times \frac{AB}{2} \). Since \( \theta \) is 50 degrees, we have \( \frac{50 \times 360 \times 2}{2\pi \times n} \times \frac{AB}{2} = 50 \). Solving for \( AB \), we get \( AB = \frac{2\pi \times n}{50 \times 360 \times 2} \times 50 \). Substituting this into the equation for \( \theta \), we get \( \theta = \frac{\frac{2\pi \times n}{50 \times 360 \times 2} \times 50}{360} \times 2\pi \times \frac{AB}{2} \). Simplifying, we get \( \theta = \frac{2\pi \times n}{50 \times 360 \times 2} \times \frac{AB}{2} \). Since \( \theta \) is 50 degrees, we have \( \frac{2\pi \times n}{50 \times 360 \times 2} \times \frac{AB}{2} = 50 \). Solving for \( AB \), we get \( AB = \frac{50 \times 360 \times 2}{2\pi \times n} \times 50 \). Substituting this into the equation for \( \theta \), we get \( \theta = \frac{\frac{50 \times 360 \times 2}{2\pi \times n} \times 50}{360} \times 2\pi \times \frac{AB}{2} \). Simplifying, we get \( \theta = \frac{50 \times 360 \times 2}{2\pi \times n} \times \frac{AB}{2} \). Since \( \theta \) is 50 degrees, we have \( \frac{50 \times 360 \times 2}{2\pi \times n} \times \frac{AB}{2} = 50 \). Solving for \( AB \), we get \( AB = \frac{2\pi \times n}{50 \times 360 \times 2} \times 50 \). Substituting this into the equation for \( \theta \), we get \( \theta = \frac{\frac{2\pi \times n}{50 \times 360 \times 2} \times 50}{360} \times 2\pi \times \frac{AB}{2} \). Simplifying, we get \( \theta = \frac{2\pi \times n}{50 \times 360 \
answer:360
correct answer:65
total:1819  correct:540  acc:0.29686641011544807
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ID:1819
picturePath:image/3064.png
answer:The answer is 40.
answer:40
correct answer:40
total:1820  correct:541  acc:0.29725274725274725
------------------------------------------------------------------------------------
ID:1820
picturePath:image/3065.png
answer:70
answer:70
correct answer:50
total:1821  correct:541  acc:0.2970895112575508
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ID:1821
picturePath:image/3066.png
answer:解:∵AB为⊙O的直径,弦CD⊥AB,∴⁀{AC}=⁀{AD},∵∠CEA=28°,∴∠AED=∠CEA=28°,∴∠ABD=∠AED=28°．故选:B．
answer:28
correct answer:28
total:1822  correct:542  acc:0.29747530186608123
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ID:1822
picturePath:image/3067.png
answer:解:∵∠D=35°,∴∠BOC=2∠D=2×35°=70°,∴∠AOC=180°-70°=110°,∵OA=OC,∴∠OCA=\frac{180^{°}-110^{°}}{2}=35°．故选:A．
answer:35
correct answer:55
total:1823  correct:542  acc:0.2973121228743829
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ID:1823
picturePath:image/3068.png
answer:解:∵∠C=40°,∴∠AOB=2∠C=80°,∵OA=OB,∴∠OAB=∠OBA=\frac{180°-80°}{2}=50°,
故选:A．
answer:50
correct answer:50
total:1824  correct:543  acc:0.29769736842105265
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ID:1824
picturePath:image/3071.png
answer:20
answer:20
correct answer:50
total:1825  correct:543  acc:0.29753424657534244
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ID:1825
picturePath:image/3072.png
answer:解:∵∠ABC=70°,∠ACB=30°,∴∠BAC=180°-70°-30°=80°,∵D是⁀{BAC}的中点,∴∠ADB=\frac{1}{2}∠BAC=40°,∴∠DBC=∠ADB=40°,
故选:A．
answer:40
correct answer:50
total:1826  correct:543  acc:0.2973713033953998
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ID:1826
picturePath:image/3073.png
answer:The answer is 4.
answer:4
correct answer:4
total:1827  correct:544  acc:0.2977558839627805
------------------------------------------------------------------------------------
ID:1827
picturePath:image/3075.png
answer:The answer is D.
ID:1828
picturePath:image/3077.png
answer:Since \( \angle ABO \) is an inscribed angle, it is half the measure of its intercepted arc. Therefore, \( \angle ABO = \frac{1}{2} \cdot 50^{\circ} = 25^{\circ} \). Since \( \angle ABO \) and \( \angle ACB \) are alternate interior angles, they are congruent. Therefore, \( \angle ACB = 25^{\circ} \). The answer is: B.
answer:25
correct answer:40
total:1829  correct:544  acc:0.2974302897758338
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ID:1829
picturePath:image/3079.png
answer:The answer is 60.
answer:60
correct answer:60
total:1830  correct:545  acc:0.2978142076502732
------------------------------------------------------------------------------------
ID:1830
picturePath:image/3080.png
answer:解:∵AD是⊙O的直径,∴∠ACD=90°,∵∠D=∠B=25°,∴∠CAD=90°-∠D=90°-25°=65°．故选:A．
answer:65
correct answer:65
total:1831  correct:546  acc:0.298197706171491
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ID:1831
picturePath:image/3081.png
answer:解:∵∠AOC=130°,∴∠BOC=50°,∴∠D=\frac{1}{2}∠BOC=\frac{1}{2}×50°=25°．故选:B．
answer:25
correct answer:25
total:1832  correct:547  acc:0.29858078602620086
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ID:1832
picturePath:image/3082.png
answer:60
answer:60
correct answer:30
total:1833  correct:547  acc:0.298417894162575
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ID:1833
picturePath:image/3083.png
answer:根据题意,∠A=90°,∠B=56°,则∠BCD=90°-56°=34°.故选B.
answer:34
correct answer:62
total:1834  correct:547  acc:0.29825517993456924
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ID:1834
picturePath:image/3084.png
answer:18°
answer:18
correct answer:29
total:1835  correct:547  acc:0.29809264305177113
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ID:1835
picturePath:image/3085.png
answer:50°
answer:50
correct answer:25
total:1836  correct:547  acc:0.29793028322440085
------------------------------------------------------------------------------------
ID:1836
picturePath:image/3086.png
answer:∠1+∠2=∠A=50°
answer:50
correct answer:230
total:1837  correct:547  acc:0.29776810016330973
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ID:1837
picturePath:image/3087.png
answer:80°
answer:80
correct answer:80
total:1838  correct:548  acc:0.2981501632208923
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ID:1838
picturePath:image/3088.png
answer:解:∵⊙O是△ABC的外接圆,∠ACB=60°,∴∠AOB=2∠ACB=120°,∵AO=BO,∴∠ABO=\frac{1}{2}(180°-120°)=30°．故选:B．
answer:30
correct answer:30
total:1839  correct:549  acc:0.29853181076672103
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ID:1839
picturePath:image/3089.png
answer:解:∵AD是⊙O的直径,∴∠ACD=90°,∵∠ABC=40°,∴∠CAD=90°-∠ABC=90°-40°=50°．故选:B．
answer:50
correct answer:50
total:1840  correct:550  acc:0.29891304347826086
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ID:1840
picturePath:image/3090.png
answer:解:∵⊙O是△ABC的外接圆,∠AOB=80°,∴∠C=\frac{1}{2}∠AOB=\frac{1}{2}×80°=40°．故选:B．
answer:40
correct answer:40
total:1841  correct:551  acc:0.2992938620315046
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ID:1841
picturePath:image/3092.png
answer:解:∵∠A=68°,∴∠BOC=2∠A=2×68°=136°．故选:D．
answer:136
correct answer:136
total:1842  correct:552  acc:0.2996742671009772
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ID:1842
picturePath:image/3093.png
answer:The answer is 60.
answer:60
correct answer:60
total:1843  correct:553  acc:0.30005425935973956
------------------------------------------------------------------------------------
ID:1843
picturePath:image/3094.png
answer:This is a geometry problem involving circles and lines. Given that \( \angle BAC = 90^\circ \) and \( AC = 12 \), we can deduce that \( \triangle ABC \) is a right triangle. The diameter of the circle is \( AD \), which means that \( AE \) is also the radius of the circle. Since \( AE \) is perpendicular to \( BD \), the shortest distance from \( C \) to \( E \) would be when \( CE \) is parallel to \( BD \). In this case, \( CE \) would be the height of the right triangle \( \triangle ABD \), which can be found using the formula for the area of a triangle: \( \frac{1}{2} \times \text{base} \times \text{height} \). The base of the triangle is \( BD \), which is the hypotenuse of \( \triangle ABC \), and the height is \( CE \). Therefore, the area of \( \triangle ABD \) is \( \frac{1}{2} \times 12 \times CE \). Since \( AB = 10 \), the area of \( \triangle ABD \) is also \( \frac{1}{2} \times 10 \times CE \). Setting these two expressions equal to each other, we can solve for \( CE \): \( \frac{1}{2} \times 12 \times CE = \frac{1}{2} \times 10 \times CE \). Simplifying, we get \( 6CE = 5CE \), which implies that \( CE = 6 \). Therefore, the minimum value of \( CE \) is \( \boxed{6} \).The answer is: 6
answer:6
correct answer:8
total:1844  correct:553  acc:0.29989154013015185
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ID:1844
picturePath:image/3095.png
answer:解:∵∠ABC与∠AOC是同弧所对的圆周角与圆心角,∠ABC=70°,∴∠AOC=2∠ABC=140°．故选:D．
answer:140
correct answer:140
total:1845  correct:554  acc:0.3002710027100271
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ID:1845
picturePath:image/3097.png
answer:This is a geometric figure with labeled points A, B, C, and D. The figure is a triangle with a right angle at point A. The angle at point D is 90 degrees, and the angle at point C is 30 degrees. The angle at point B is not labeled, but it can be determined by the sum of the angles in a triangle, which is 180 degrees. Since the sum of the angles at points A, C, and D is 90 + 30 + 142 = 262 degrees, the angle at point B is 180 - 262 = -82 degrees. However, an angle cannot be negative, so there must be an error in the figure or the given information.
answer:82
correct answer:22
total:1846  correct:554  acc:0.3001083423618635
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ID:1846
picturePath:image/3098.png
answer:50°
answer:50
correct answer:50
total:1847  correct:555  acc:0.3004872766648619
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ID:1847
picturePath:image/3099.png
answer:70
answer:70
correct answer:55
total:1848  correct:555  acc:0.3003246753246753
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ID:1848
picturePath:image/3100.png
answer:The answer is 35.
answer:35
correct answer:35
total:1849  correct:556  acc:0.300703082747431
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ID:1849
picturePath:image/3101.png
answer:12.5°
answer:12.5
correct answer:25
total:1850  correct:556  acc:0.3005405405405405
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ID:1850
picturePath:image/3102.png
answer:解:∵∠D=35°,∴∠BOC=2∠D=2×35°=70°,∴∠AOC=180°-70°=110°,∵OA=OC,∴∠OAC=\frac{1}{2}(180°-110°)=35°．故选:A．
answer:35
correct answer:55
total:1851  correct:556  acc:0.3003781739600216
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ID:1851
picturePath:image/3103.png
answer:tan∠CBD=\frac{CD}{CB}=\frac{AB}{AC}=\frac{8}{8+10}=\frac{8}{18}=\frac{4}{9}$.
answer:9
correct answer:4
total:1852  correct:556  acc:0.3002159827213823
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ID:1852
picturePath:image/3106.png
answer:解:∵∠AOC=100°,∴∠BOC=180°-∠AOC=80°,∴∠D=\frac{1}{2}∠BOC=40°．故选:B．
answer:40
correct answer:40
total:1853  correct:557  acc:0.3005936319481921
------------------------------------------------------------------------------------
ID:1853
picturePath:image/3109.png
answer:3
answer:3
correct answer:7
total:1854  correct:557  acc:0.3004314994606257
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ID:1854
picturePath:image/3111.png
answer:根据圆周角定理，同弧所对的圆周角相等，因此∠BOC = 2∠A = 80°。又因为OB = OC，所以∠OCB = (180° - ∠BOC) / 2 = (180° - 80°) / 2 = 60°。故选B。
answer:60
correct answer:50
total:1855  correct:557  acc:0.30026954177897575
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ID:1855
picturePath:image/3114.png
answer:The answer is 30.
answer:30
correct answer:60
total:1856  correct:557  acc:0.30010775862068967
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ID:1856
picturePath:image/3115.png
answer:72°
answer:72
correct answer:72
total:1857  correct:558  acc:0.30048465266558966
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ID:1857
picturePath:image/3116.png
answer:根据题意,∠ABC=∠ADC=65°,∵OA=OB=OC,∴∠OAB=∠OBA,∠OBC=∠OCB,∠OCA=∠OAC,∴∠OAB+∠OBC+∠OCA=65°+65°=130°,∴∠DAO+∠DCO=180°-130°=50°,故选B.
answer:50
correct answer:165
total:1858  correct:558  acc:0.30032292787944026
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ID:1858
picturePath:image/3119.png
answer:70°
answer:70
correct answer:40
total:1859  correct:558  acc:0.300161377084454
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ID:1859
picturePath:image/3123.png
answer:由于△ABC是圆O的内接三角形，根据圆内接四边形的性质，内角和为360°，因此∠AOC的度数等于360°-∠ABC，即360°-70°=290°。
answer:290
correct answer:140
total:1860  correct:558  acc:0.3
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ID:1860
picturePath:image/3127.png
answer:解:∵∠AOB=100°,∴∠ACB=\frac{1}{2}∠AOB=50°．故选:B．
answer:50
correct answer:130
total:1861  correct:558  acc:0.2998387963460505
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ID:1861
picturePath:image/3128.png
answer:解:∵AB为⊙O直径,∴∠ACB=90°,∵∠ABC=55°,∴∠BAC=90°-55°=35°．故选:B．
answer:35
correct answer:35
total:1862  correct:559  acc:0.30021482277121375
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ID:1862
picturePath:image/3129.png
answer:8√3
answer:3
correct answer:3
total:1863  correct:560  acc:0.30059044551798175
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ID:1863
picturePath:image/3130.png
answer:30°
answer:30
correct answer:30
total:1864  correct:561  acc:0.3009656652360515
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ID:1864
picturePath:image/3133.png
answer:解:∵AB⊥CD,∴⁀{AC}=⁀{AD},∴∠ACD=\frac{1}{2}∠CDB=\frac{1}{2}×64°=32°．故选:B．
answer:32
correct answer:26
total:1865  correct:561  acc:0.30080428954423594
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ID:1865
picturePath:image/3135.png
answer:3
answer:3
correct answer:3
total:1866  correct:562  acc:0.3011789924973205
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ID:1866
picturePath:image/3137.png
answer:19
answer:19
correct answer:19
total:1867  correct:563  acc:0.3015532940546331
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ID:1867
picturePath:image/3138.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵∠ABC=70°,∴∠BAC=90°-∠ABC=20°,∴∠BDC=∠BAC=20°．故选:B．
answer:20
correct answer:20
total:1868  correct:564  acc:0.3019271948608137
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ID:1868
picturePath:image/3139.png
answer:解:∵∠B=30°,∠APD=80°,∴∠A=\frac{1}{2}∠APD=\frac{1}{2}×80°=40°．故选:A．
answer:40
correct answer:70
total:1869  correct:564  acc:0.3017656500802568
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ID:1869
picturePath:image/3140.png
answer:80°
answer:80
correct answer:80
total:1870  correct:565  acc:0.30213903743315507
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ID:1870
picturePath:image/3142.png
answer:30
answer:30
correct answer:30
total:1871  correct:566  acc:0.30251202565473007
------------------------------------------------------------------------------------
ID:1871
picturePath:image/3143.png
answer:3
answer:3
correct answer:2
total:1872  correct:566  acc:0.30235042735042733
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ID:1872
picturePath:image/3144.png
answer:50°
answer:50
correct answer:50
total:1873  correct:567  acc:0.3027229044313935
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ID:1873
picturePath:image/3145.png
answer:解:∵AB∥CD,∠BAD=36°,∴∠D=∠BAD=36°,∴∠AOC=2∠D=72°．故选:C．
answer:72
correct answer:72
total:1874  correct:568  acc:0.3030949839914621
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ID:1874
picturePath:image/3147.png
answer:100°
answer:100
correct answer:100
total:1875  correct:569  acc:0.30346666666666666
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ID:1875
picturePath:image/3148.png
answer:50
answer:50
correct answer:50
total:1876  correct:570  acc:0.30383795309168443
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ID:1876
picturePath:image/3150.png
answer:根据圆周角定理，同弧所对的圆周角的度数等于它所对的圆心角度数的一半。因此，∠AOC的度数等于∠D的一半，即20°。
answer:20
correct answer:100
total:1877  correct:570  acc:0.3036760788492275
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ID:1877
picturePath:image/3153.png
answer:∠AOC=2∠D=2×33°=66°.
answer:66
correct answer:114
total:1878  correct:570  acc:0.3035143769968051
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ID:1878
picturePath:image/3154.png
answer:The answer is 22.
answer:22
correct answer:88
total:1879  correct:570  acc:0.30335284725918044
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ID:1879
picturePath:image/3155.png
answer:解:∵AO∥BC,∠AOB=50°,∴∠ABC=∠AOB=50°,∴∠ACB=180°-50°=130°,∴∠OAC=\frac{1}{2}∠ACB=\frac{1}{2}×130°=65°．故选:A．
answer:65
correct answer:25
total:1880  correct:570  acc:0.30319148936170215
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ID:1880
picturePath:image/3157.png
answer:25
answer:25
correct answer:25
total:1881  correct:571  acc:0.30356193514088253
------------------------------------------------------------------------------------
ID:1881
picturePath:image/3162.png
answer:The answer is 15.
answer:15
correct answer:30
total:1882  correct:571  acc:0.30340063761955366
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ID:1882
picturePath:image/3164.png
answer:解:∵OA⊥BC,∠AOB=48°,∴⁀{AB}=⁀{AC},∴∠ADC=\frac{1}{2}∠AOB=\frac{1}{2}×48°=24°．故选:B．
answer:24
correct answer:24
total:1883  correct:572  acc:0.30377057886351566
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ID:1883
picturePath:image/3165.png
answer:解:∵DC=OE,∴∠CED=∠OEB,∵∠C=20°,∴∠CED=∠C=20°,∴∠OEB=20°,∵OA=OB,∴∠OAB=∠OBA=\frac{180°-20°}{2}=80°,∴∠EOB=180°-∠OEB-∠OAB=180°-20°-80°=80°．故选:A．
answer:80
correct answer:60
total:1884  correct:572  acc:0.3036093418259023
------------------------------------------------------------------------------------
ID:1884
picturePath:image/3167.png
answer:60°
answer:60
correct answer:60
total:1885  correct:573  acc:0.3039787798408488
------------------------------------------------------------------------------------
ID:1885
picturePath:image/3168.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵∠CAB=24°,∴∠ABC=90°-24°=66°,∴∠ADC=\frac{1}{2}∠ABC=\frac{1}{2}×66°=33°．故选:B．
answer:33
correct answer:66
total:1886  correct:573  acc:0.30381760339342523
------------------------------------------------------------------------------------
ID:1886
picturePath:image/3169.png
answer:100°
answer:100
correct answer:100
total:1887  correct:574  acc:0.3041865394806571
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ID:1887
picturePath:image/3170.png
answer:解:∵∠BOC=112°,∴∠AOC=\frac{1}{2}∠BOC=56°,∵AD∥OC,∴∠AOD=∠AOC=56°．故选:B．
answer:56
correct answer:44
total:1888  correct:574  acc:0.3040254237288136
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ID:1888
picturePath:image/3171.png
answer:50°
answer:50
correct answer:40
total:1889  correct:574  acc:0.3038644785600847
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ID:1889
picturePath:image/3172.png
answer:解:∵∠ACD与∠BAD是同弧所对的圆周角与圆心角,∴∠BAD=2∠ACD=40°．故选:B．
answer:40
correct answer:70
total:1890  correct:574  acc:0.3037037037037037
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ID:1890
picturePath:image/3173.png
answer:The answer is 48.
answer:48
correct answer:48
total:1891  correct:575  acc:0.30407191961924906
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ID:1891
picturePath:image/3176.png
answer:根据圆周角定理，同弧所对的圆周角相等，因此∠AOB=2∠BOC。已知∠BOC=40°，所以∠AOB=2×40°=80°。
answer:80
correct answer:80
total:1892  correct:576  acc:0.3044397463002114
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ID:1892
picturePath:image/3180.png
answer:5
answer:5
correct answer:5
total:1893  correct:577  acc:0.3048071843634443
------------------------------------------------------------------------------------
ID:1893
picturePath:image/3182.png
answer:16
answer:16
correct answer:5
total:1894  correct:577  acc:0.30464625131995776
------------------------------------------------------------------------------------
ID:1894
picturePath:image/3184.png
answer:解:∵∠A=35°,∴∠BOC=2∠A=2×35°=70°．故选:B．
answer:70
correct answer:70
total:1895  correct:578  acc:0.3050131926121372
------------------------------------------------------------------------------------
ID:1895
picturePath:image/3186.png
answer:100
answer:100
correct answer:100
total:1896  correct:579  acc:0.30537974683544306
------------------------------------------------------------------------------------
ID:1896
picturePath:image/3189.png
answer:The answer is \frac{2\sqrt{3}}{2}=\sqrt{3}
answer:3
correct answer:3
total:1897  correct:580  acc:0.3057459146020032
------------------------------------------------------------------------------------
ID:1897
picturePath:image/3190.png
answer:This is a geometry problem involving a circle. The expression \(\frac{1}{2}DB+OD\) represents the sum of the distances from point D to points B and O, respectively. Since point O is the center of the circle, the shortest distance from point D to point O is a radius of the circle. Therefore, the minimum value of \(\frac{1}{2}DB+OD\) is equal to the radius of the circle. Since the diameter AC is 4, the radius is half of that, which is 2. Therefore, the minimum value of \(\frac{1}{2}DB+OD\) is 2. The answer is A.
answer:2
correct answer:3
total:1898  correct:580  acc:0.3055848261327713
------------------------------------------------------------------------------------
ID:1898
picturePath:image/3191.png
answer:50°
answer:50
correct answer:25
total:1899  correct:580  acc:0.3054239073196419
------------------------------------------------------------------------------------
ID:1899
picturePath:image/3193.png
answer:50°
answer:50
correct answer:40
total:1900  correct:580  acc:0.30526315789473685
------------------------------------------------------------------------------------
ID:1900
picturePath:image/3194.png
answer:40
answer:40
correct answer:40
total:1901  correct:581  acc:0.3056286165176223
------------------------------------------------------------------------------------
ID:1901
picturePath:image/3195.png
answer:20
answer:20
correct answer:20
total:1902  correct:582  acc:0.305993690851735
------------------------------------------------------------------------------------
ID:1902
picturePath:image/3198.png
answer:根据圆周角定理，同弧所对的圆周角相等，所以∠BOC=∠ABC=65°．故选B．
answer:65
correct answer:100
total:1903  correct:582  acc:0.30583289542827113
------------------------------------------------------------------------------------
ID:1903
picturePath:image/3199.png
answer:解:∵弦CD⊥AB,∴⁀{BC}=⁀{BD},∵∠OCD=32°,∴∠BOC=2∠OCD=64°,∴∠A=\frac{1}{2}∠BOC=32°．故选:B．
answer:32
correct answer:29
total:1904  correct:582  acc:0.30567226890756305
------------------------------------------------------------------------------------
ID:1904
picturePath:image/3201.png
answer:∠AOB = 2∠ACB = 2*33° = 66°.
answer:66
correct answer:66
total:1905  correct:583  acc:0.30603674540682413
------------------------------------------------------------------------------------
ID:1905
picturePath:image/3205.png
answer:解:∵AB是直径,∴∠ACB=90°,∵∠D=20°,∴∠B=90°-20°=70°,∴∠BAC=∠B=70°．故选:B．
answer:70
correct answer:70
total:1906  correct:584  acc:0.30640083945435465
------------------------------------------------------------------------------------
ID:1906
picturePath:image/3206.png
answer:The answer is 40°.
answer:40
correct answer:70
total:1907  correct:584  acc:0.3062401678028317
------------------------------------------------------------------------------------
ID:1907
picturePath:image/3208.png
answer:解:∵∠AOB=110°,∴∠ACB=\frac{1}{2}∠AOB=55°．故选:A．
answer:55
correct answer:125
total:1908  correct:584  acc:0.3060796645702306
------------------------------------------------------------------------------------
ID:1908
picturePath:image/3209.png
answer:根据圆周角定理，在同圆或等圆中，同弧或等弧所对的圆周角相等。因此，∠AOB等于2∠ACB。将∠ACB=30°代入，得到∠AOB=2×30°=60°。故选C。
answer:60
correct answer:60
total:1909  correct:585  acc:0.30644316396018856
------------------------------------------------------------------------------------
ID:1909
picturePath:image/3210.png
answer:解:∵AD是⊙O的直径,∴∠C=90°,∵∠CAD=56°,∴∠D=90°-56°=34°,∴∠B=∠D=34°．故选:B．
answer:34
correct answer:34
total:1910  correct:586  acc:0.3068062827225131
------------------------------------------------------------------------------------
ID:1910
picturePath:image/3211.png
answer:50°
answer:50
correct answer:130
total:1911  correct:586  acc:0.30664573521716376
------------------------------------------------------------------------------------
ID:1911
picturePath:image/3212.png
answer:由于△ABC是⊙O的内接三角形，根据圆周角定理，同弧所对的圆周角相等，因此∠AOC=2∠ABC。将∠ABC=70°代入，得到∠AOC=2×70°=140°。所以，∠AOC的大小是140°。
answer:140
correct answer:140
total:1912  correct:587  acc:0.3070083682008368
------------------------------------------------------------------------------------
ID:1912
picturePath:image/3213.png
answer:60
answer:60
correct answer:60
total:1913  correct:588  acc:0.30737062205959226
------------------------------------------------------------------------------------
ID:1913
picturePath:image/3214.png
answer:解:∵∠α=130°,∴∠A=\frac{1}{2}∠α=\frac{1}{2}×130°=65°．故选:A．
answer:65
correct answer:115
total:1914  correct:588  acc:0.3072100313479624
------------------------------------------------------------------------------------
ID:1914
picturePath:image/3215.png
answer:解:∵AB是直径,∴∠ACB=90°,∵CA=CD,∴∠CAD=∠DCD,∵∠ACD=40°,∴∠CAD=20°,∴∠B=90°-20°=70°．故选:B．
answer:70
correct answer:70
total:1915  correct:589  acc:0.30757180156657965
------------------------------------------------------------------------------------
ID:1915
picturePath:image/3216.png
answer:解:∵∠ACB是⊙O的弧AB所对的圆周角,∠ACB=30°,∴∠AOB=2∠ACB=60°,∵线段PA,PB的中点分别为D,E,∴DE是△AOB的中位线,∴DE=\frac{1}{2}AB,∵OA=OB=8cm,∴AB=2OA=16cm,∴DE=\frac{1}{2}×16=8(cm)．故选:B．
answer:8
correct answer:4
total:1916  correct:589  acc:0.30741127348643005
------------------------------------------------------------------------------------
ID:1916
picturePath:image/3217.png
answer:解:连接AC,∵OA=OC=100mm,∴AC=√{100²+100²}=200mm,∵AB=160mm,∴BC=AB-AC=160-200=-40mm,∴圆木被锯部分的最大高度为40mm．故选:A．
answer:40
correct answer:40
total:1917  correct:590  acc:0.3077725612936881
------------------------------------------------------------------------------------
ID:1917
picturePath:image/3223.png
answer:解:∵⁀{AB}的度数为60°,∴∠AOB=60°,∵AC是⊙O的直径,∴∠BOC=180°-∠AOB=180°-60°=120°,∴∠OBC=\frac{1}{2}∠BOC=\frac{1}{2}×120°=60°．故选:B．
answer:60
correct answer:30
total:1918  correct:590  acc:0.30761209593326383
------------------------------------------------------------------------------------
ID:1918
picturePath:image/3225.png
answer:4.8
answer:4.8
correct answer:3
total:1919  correct:590  acc:0.30745179781136006
------------------------------------------------------------------------------------
ID:1919
picturePath:image/3226.png
answer:解:设油的最大深度为xcm,则AB-2x=80-2x(cm),
由勾股定理得:
(82-x)²=(80-2x)²,
解得x=20cm,
故选:B．
answer:20
correct answer:32
total:1920  correct:590  acc:0.3072916666666667
------------------------------------------------------------------------------------
ID:1920
picturePath:image/3228.png
answer:2.4厘米/分钟
answer:2.4
correct answer:1.6
total:1921  correct:590  acc:0.3071317022384175
------------------------------------------------------------------------------------
ID:1921
picturePath:image/3229.png
answer:解:∵⁀{AC}=⁀{BD},∴∠AOC=∠DOB=40°,∴∠COD=180°-∠AOC-∠DOB=100°．故选:B．
answer:100
correct answer:40
total:1922  correct:590  acc:0.30697190426638915
------------------------------------------------------------------------------------
ID:1922
picturePath:image/3230.png
answer:130
answer:130
correct answer:100
total:1923  correct:590  acc:0.30681227249089965
------------------------------------------------------------------------------------
ID:1923
picturePath:image/3231.png
answer:20
answer:20
correct answer:20
total:1924  correct:591  acc:0.30717255717255715
------------------------------------------------------------------------------------
ID:1924
picturePath:image/3232.png
answer:解:∵DE∥AB,∴∠AOB=∠DEO=40°,∴∠BOC=2∠AOB=80°．故选:D．
answer:80
correct answer:110
total:1925  correct:591  acc:0.307012987012987
------------------------------------------------------------------------------------
ID:1925
picturePath:image/3233.png
answer:52°
answer:52
correct answer:52
total:1926  correct:592  acc:0.3073727933541018
------------------------------------------------------------------------------------
ID:1926
picturePath:image/3234.png
answer:The answer is A.
ID:1927
picturePath:image/3235.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵∠ABC=30°,∴∠A=90°-∠ABC=60°,∵OD⊥BC,∴∠ODC=90°,∴∠DCB=∠ODC-∠A=90°-60°=30°．故选:B．
answer:30
correct answer:30
total:1928  correct:593  acc:0.30757261410788383
------------------------------------------------------------------------------------
ID:1928
picturePath:image/3236.png
answer:5m
answer:5
correct answer:2
total:1929  correct:593  acc:0.30741316744427166
------------------------------------------------------------------------------------
ID:1929
picturePath:image/3238.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵∠CAB=25°,∴∠ABC=90°-25°=65°,∴∠ADC=∠ABC=65°．故选:B．
answer:65
correct answer:65
total:1930  correct:594  acc:0.3077720207253886
------------------------------------------------------------------------------------
ID:1930
picturePath:image/3239.png
answer:This is a geometry problem involving circles and angles. Given that \( \angle C = 90^\circ \) and \( \angle A = 25^\circ \), we can find \( \angle B \) by subtracting \( \angle A \) from \( 90^\circ \): \( \angle B = 90^\circ - 25^\circ = 65^\circ \). Since \( D \) is on the circumference of the circle with \( C \) as the center and \( BC \) as the radius, \( \angle BDC \) is a right angle, which is \( 90^\circ \). Therefore, \( \angle BDC = 90^\circ \). The sum of angles in a triangle is \( 180^\circ \), so \( \angle CDB = 180^\circ - \angle BDC - \angle B = 180^\circ - 90^\circ - 65^\circ = 25^\circ \), which is the same as \( \angle A \). This means that \( \angle CDB \) is congruent to \( \angle A \), and since \( \angle BDC \) is a right angle, \( \angle BDC \) is also congruent to \( \angle A \). Therefore, \( \angle BDC = \angle A = 25^\circ \).
answer:25
correct answer:50
total:1931  correct:594  acc:0.3076126359399275
------------------------------------------------------------------------------------
ID:1931
picturePath:image/3240.png
answer:40°
answer:40
correct answer:40
total:1932  correct:595  acc:0.3079710144927536
------------------------------------------------------------------------------------
ID:1932
picturePath:image/3243.png
answer:解:∵AB是半圆的直径,∴∠ACB=90°,∵∠BAC=20°,∴∠ABC=90°-20°=70°,∵D是⁀{AC}的中点,∴∠ADC=\frac{1}{2}∠ABC=\frac{1}{2}×70°=35°,∴∠DAC=∠ADC-∠BAC=35°-20°=15°．故选:A．
answer:15
correct answer:35
total:1933  correct:595  acc:0.30781169167097777
------------------------------------------------------------------------------------
ID:1933
picturePath:image/3245.png
answer:8πcm
answer:8
correct answer:8
total:1934  correct:596  acc:0.3081695966907963
------------------------------------------------------------------------------------
ID:1934
picturePath:image/3248.png
answer:解:∵圆O中,圆弧AB=圆弧AC,∴∠B=∠C=80°,∴∠A=180°-∠B-∠C=180°-80°-80°=20°．故选:A．
answer:20
correct answer:20
total:1935  correct:597  acc:0.3085271317829457
------------------------------------------------------------------------------------
ID:1935
picturePath:image/3250.png
answer:20°
answer:20
correct answer:60
total:1936  correct:597  acc:0.30836776859504134
------------------------------------------------------------------------------------
ID:1936
picturePath:image/3252.png
answer:解:∵⁀{AB}是半圆,∴∠AOB=180°,∵AD∥OC,∴⁀{AD}的度数等于⁀{AB}的度数,∵⁀{CD}=62°,∴⁀{AD}=62°,∴∠DAB=62°,∴∠BDC=\frac{1}{2}∠DAB=31°．故选:B．
answer:31
correct answer:56
total:1937  correct:597  acc:0.3082085699535364
------------------------------------------------------------------------------------
ID:1937
picturePath:image/3255.png
answer:A. 90° B. 120° C. 180° D. 300°
answer:300
correct answer:60
total:1938  correct:597  acc:0.3080495356037152
------------------------------------------------------------------------------------
ID:1938
picturePath:image/3260.png
answer:30
answer:30
correct answer:90
total:1939  correct:597  acc:0.3078906652913873
------------------------------------------------------------------------------------
ID:1939
picturePath:image/3263.png
answer:Since \( \overline{CD} \) is a diameter of the circle, \( \angle ACD \) is \( 90^\circ \). Since \( \overline{CD} \) is also parallel to \( \overline{AB} \), we have \( \angle ADC = \angle ABD = 56^\circ \). Since \( \overline{AB} \) is a diameter of the circle, \( \angle ABE = 90^\circ \). Therefore, \( \angle AEB = 360^\circ - 90^\circ - 56^\circ - 56^\circ = \boxed{108^\circ} \).The answer is: 108
answer:108
correct answer:34
total:1940  correct:597  acc:0.3077319587628866
------------------------------------------------------------------------------------
ID:1940
picturePath:image/3264.png
answer:解:∵∠C=45°,∴∠B=45°,∵∠AMD=75°,∴∠BAM=75°-45°=30°,∴∠D=∠BAM=30°,故选:B．
answer:30
correct answer:30
total:1941  correct:598  acc:0.30808861411643484
------------------------------------------------------------------------------------
ID:1941
picturePath:image/3265.png
answer:解:∵∠AOC=100°,∴∠B=\frac{1}{2}∠AOC=50°．故选:B．
answer:50
correct answer:130
total:1942  correct:598  acc:0.3079299691040165
------------------------------------------------------------------------------------
ID:1942
picturePath:image/3268.png
answer:50°
answer:50
correct answer:50
total:1943  correct:599  acc:0.3082861554297478
------------------------------------------------------------------------------------
ID:1943
picturePath:image/3271.png
answer:4√3
answer:3
correct answer:2
total:1944  correct:599  acc:0.3081275720164609
------------------------------------------------------------------------------------
ID:1944
picturePath:image/3272.png
answer:解:∵∠A=36°,∠C=27°,∴∠BOC=∠A+∠C=36°+27°=63°,∴∠B=\frac{1}{2}∠BOC=\frac{1}{2}×63°=31.5°．故选:B．
answer:31.5
correct answer:63
total:1945  correct:599  acc:0.3079691516709512
------------------------------------------------------------------------------------
ID:1945
picturePath:image/3273.png
answer:5
answer:5
correct answer:3
total:1946  correct:599  acc:0.3078108941418294
------------------------------------------------------------------------------------
ID:1946
picturePath:image/3274.png
answer:解:∵∠AOB与∠ACB都对⁀{AB},且∠AOB=64°,∴∠ACB=\frac{1}{2}∠AOB=32°,
故选:B．
answer:32
correct answer:32
total:1947  correct:600  acc:0.3081664098613251
------------------------------------------------------------------------------------
ID:1947
picturePath:image/3275.png
answer:The answer is 35.
answer:35
correct answer:35
total:1948  correct:601  acc:0.3085215605749487
------------------------------------------------------------------------------------
ID:1948
picturePath:image/3276.png
answer:解:∵∠AOB与∠ACB都对⁀{AB},且∠AOB=70°,∴∠ACB=\frac{1}{2}∠AOB=35°．故选:A．
answer:35
correct answer:35
total:1949  correct:602  acc:0.30887634684453563
------------------------------------------------------------------------------------
ID:1949
picturePath:image/3277.png
answer:解:∵BD是⊙O的直径,∴∠BCD=90°,∵∠CBD=30°,∴∠BDC=90°-∠CBD=60°,∴∠A=\frac{1}{2}∠BDC=\frac{1}{2}×60°=30°．故选:B．
answer:30
correct answer:60
total:1950  correct:602  acc:0.3087179487179487
------------------------------------------------------------------------------------
ID:1950
picturePath:image/3279.png
answer:50°
answer:50
correct answer:50
total:1951  correct:603  acc:0.3090722706304459
------------------------------------------------------------------------------------
ID:1951
picturePath:image/3280.png
answer:解:∵∠CAD=110°,∴∠AOB=2∠CAD=220°,∵AO=BO,∴∠B=\frac{1}{2}(360°-220°)=110°．故选:A．
answer:110
correct answer:70
total:1952  correct:603  acc:0.3089139344262295
------------------------------------------------------------------------------------
ID:1952
picturePath:image/3281.png
answer:60
answer:60
correct answer:150
total:1953  correct:603  acc:0.3087557603686636
------------------------------------------------------------------------------------
ID:1953
picturePath:image/3283.png
answer:解:∵OC⊥AB于点D,∴⁀{AC}=⁀{BC},∴∠BOC=2∠A=72°,∴∠P=\frac{1}{2}∠BOC=36°．故选:A．
answer:36
correct answer:27
total:1954  correct:603  acc:0.3085977482088025
------------------------------------------------------------------------------------
ID:1954
picturePath:image/3284.png
answer:解:∵∠AED=20°,∴∠BED=∠AED=20°,∵AB是直径,∴∠EDB=90°,∴∠BCD=∠BED=20°．故选:A．
answer:20
correct answer:110
total:1955  correct:603  acc:0.30843989769820973
------------------------------------------------------------------------------------
ID:1955
picturePath:image/3286.png
answer:Since AD is the diameter, it is perpendicular to the chord BC. Given that ∠AOB is 50°, and the sum of the angles in a triangle is 180°, we can find the measure of ∠ADC by subtracting ∠AOB from 180°. Therefore, ∠ADC = 180° - ∠AOB = 180° - 50° = 130°.
answer:130
correct answer:25
total:1956  correct:603  acc:0.30828220858895705
------------------------------------------------------------------------------------
ID:1956
picturePath:image/3287.png
answer:The answer is 40.
answer:40
correct answer:40
total:1957  correct:604  acc:0.3086356668369954
------------------------------------------------------------------------------------
ID:1957
picturePath:image/3288.png
answer:解:∵∠ADC=35°,∴∠AOC=2∠ADC=2×35°=70°,∴∠ACB=\frac{1}{2}∠AOC=\frac{1}{2}×70°=35°．故选:A．
answer:35
correct answer:55
total:1958  correct:604  acc:0.3084780388151175
------------------------------------------------------------------------------------
ID:1958
picturePath:image/3289.png
answer:解:过O点作OC⊥AB,垂足为C,∵AB=80cm,∴OC=√{100^{2}-80^{2}}=20cm,∴油的最大深度为20cm．故选:A．
answer:20
correct answer:20
total:1959  correct:605  acc:0.30883103624298114
------------------------------------------------------------------------------------
ID:1959
picturePath:image/3292.png
answer:解:∵⊙O是△ABC的外接圆,∠AOB=130°,∴∠ACB=\frac{1}{2}∠AOB=\frac{1}{2}×130°=65°．故选:B．
answer:65
correct answer:115
total:1960  correct:605  acc:0.3086734693877551
------------------------------------------------------------------------------------
ID:1960
picturePath:image/3293.png
answer:This is a geometry problem involving circles and angles. Given that AB is the diameter, we know that ∠ACB = 90°. Since AB = AC, triangle ABC is an isosceles triangle, so ∠ABC = ∠ACB = 45°. Therefore, ∠EBC = ∠ABC = 45°. The answer is A.
answer:45
correct answer:22.5
total:1961  correct:605  acc:0.30851606323304437
------------------------------------------------------------------------------------
ID:1961
picturePath:image/3295.png
answer:解:∵∠A=50°,∴∠COB=2∠A=100°,∵∠B=30°,∴∠BDC=∠COB-∠B=100°-30°=70°．故选:A．
answer:70
correct answer:70
total:1962  correct:606  acc:0.308868501529052
------------------------------------------------------------------------------------
ID:1962
picturePath:image/3297.png
answer:解:∵BC⊥OA,∴⁀{AB}=⁀{AC},∵∠ADC=26°,∴∠AOB=2∠ADC=52°．故选:B．
answer:52
correct answer:52
total:1963  correct:607  acc:0.3092205807437596
------------------------------------------------------------------------------------
ID:1963
picturePath:image/3301.png
answer:90°
answer:90
correct answer:45
total:1964  correct:607  acc:0.3090631364562118
------------------------------------------------------------------------------------
ID:1964
picturePath:image/3302.png
answer:根据圆周角定理，同弧所对的圆周角相等，所以∠BOC=∠ACO=30°．故选B．
answer:30
correct answer:60
total:1965  correct:607  acc:0.3089058524173028
------------------------------------------------------------------------------------
ID:1965
picturePath:image/3303.png
answer:解:∵直径AB垂直于弦CD,∴⁀{AC}=⁀{BD},∴∠E=∠F,∵∠E=∠CAB=36°,∴∠F=36°,∴∠BCD=∠F=36°．故选:A．
answer:36
correct answer:36
total:1966  correct:608  acc:0.30925737538148523
------------------------------------------------------------------------------------
ID:1966
picturePath:image/3304.png
answer:解:∵AB的垂直平分线与⊙A交于E、F两点,∴FA=FB,∵∠BFC=20°,∴∠BFA=∠BFC=20°,∴∠BAC=40°,∴∠DBC=\frac{1}{2}∠BAC=20°．故选:B．
answer:20
correct answer:30
total:1967  correct:608  acc:0.3091001525165226
------------------------------------------------------------------------------------
ID:1967
picturePath:image/3305.png
answer:解:∵OC∥BD,∴∠OCB=∠A=30°,∴∠CBD=∠OCB=30°．故选:A．
answer:30
correct answer:30
total:1968  correct:609  acc:0.3094512195121951
------------------------------------------------------------------------------------
ID:1968
picturePath:image/3306.png
answer:解:∵BE=DE,∴∠BEC=∠DEC,∵∠B=40°,∴∠BEC=∠DEC=40°,∴∠A=∠BEC+∠DEC=80°．故选:D．
answer:80
correct answer:40
total:1969  correct:609  acc:0.3092940578974099
------------------------------------------------------------------------------------
ID:1969
picturePath:image/3307.png
answer:解:∵∠B=24°,∴∠AOC=2∠B=48°,∵OD∥AC,∴∠AOD=∠B=24°,∴∠BCD=∠AOC-∠AOD=48°-24°=24°．故选:A．
answer:24
correct answer:57
total:1970  correct:609  acc:0.30913705583756346
------------------------------------------------------------------------------------
ID:1970
picturePath:image/3308.png
answer:The answer is 40.
answer:40
correct answer:65
total:1971  correct:609  acc:0.3089802130898021
------------------------------------------------------------------------------------
ID:1971
picturePath:image/3312.png
answer:解:∵AB为⊙O的直径,∴∠AOC=90°,∵∠ACO=50°,∴∠BOC=90°-50°=40°,∴∠B=\frac{1}{2}∠BOC=20°．故选:A．
answer:20
correct answer:40
total:1972  correct:609  acc:0.3088235294117647
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ID:1972
picturePath:image/3313.png
answer:根据圆内接五边形的性质，∠B+∠E=180°-∠CAD=180°-42°=138°。故选C。
answer:138
correct answer:222
total:1973  correct:609  acc:0.30866700456158136
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ID:1973
picturePath:image/3314.png
answer:25
answer:25
correct answer:25
total:1974  correct:610  acc:0.3090172239108409
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ID:1974
picturePath:image/3315.png
answer:解:∵∠BOC与∠BAC是同弧所对的圆心角与圆周角,∠BOC=100°,∴∠BAC=\frac{1}{2}∠BOC=50°．故选:B．
answer:50
correct answer:50
total:1975  correct:611  acc:0.30936708860759493
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ID:1975
picturePath:image/3316.png
answer:根据圆周角定理，同弧所对的圆周角相等，所以∠PDC=∠NDC。又∵∠PNA=60°，∴∠NDC=∠PNA=60°，∵M是AB的中点，∴CD是AB的垂直平分线，∴PC=PB，∴∠PDC=∠PBC=60°，∴∠PDC=60°．故选：B．
answer:60
correct answer:60
total:1976  correct:612  acc:0.3097165991902834
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ID:1976
picturePath:image/3317.png
answer:解:∵弦CD⊥AB于E,∴⁀{BC}=⁀{BD},∴∠AOC=2∠A=2×35°=70°．故选:B．
answer:70
correct answer:110
total:1977  correct:612  acc:0.3095599393019727
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ID:1977
picturePath:image/3318.png
answer:解:∵∠AOB=72°,∴∠AOC=72°,∵B为弧AC的中点,∴∠ABC=∠EBC=30°,∴∠OAC=∠AOB-∠ABC=72°-30°=42°,故选:B．
answer:42
correct answer:18
total:1978  correct:612  acc:0.30940343781597573
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ID:1978
picturePath:image/3320.png
answer:解:∵BC是⊙O的直径,∴∠BAC=90°,∵∠D=36°,∴∠BCA=90°-∠D=90°-36°=54°．故选:B．
answer:54
correct answer:54
total:1979  correct:613  acc:0.3097524002021223
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ID:1979
picturePath:image/3321.png
answer:The answer is 100°.
answer:100
correct answer:40
total:1980  correct:613  acc:0.3095959595959596
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ID:1980
picturePath:image/3323.png
answer:解:∵∠AOB和∠ACB是弧AB所对的角,∴∠AOB=2∠ACB,∵∠AOB=100°,∴∠ACB=50°,
故选:B．
answer:50
correct answer:50
total:1981  correct:614  acc:0.3099444724886421
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ID:1981
picturePath:image/3324.png
answer:72°
answer:72
correct answer:72
total:1982  correct:615  acc:0.31029263370333
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ID:1982
picturePath:image/3325.png
answer:解:∵AB是⊙O的直径,∴∠ACB=90°,∵∠CAB=40°,∴∠D=90°-∠CAB=90°-40°=50°．故选:B．
answer:50
correct answer:50
total:1983  correct:616  acc:0.3106404437720625
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ID:1983
picturePath:image/3327.png
answer:解:∵∠COB=68°,∴∠AOC=180°-∠COB=180°-68°=112°,∴∠BDC=\frac{1}{2}∠AOC=\frac{1}{2}×112°=56°．故选:B．
answer:56
correct answer:34
total:1984  correct:616  acc:0.31048387096774194
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ID:1984
picturePath:image/3328.png
answer:20
answer:20
correct answer:70
total:1985  correct:616  acc:0.3103274559193955
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ID:1985
picturePath:image/3329.png
answer:55°
answer:55
correct answer:35
total:1986  correct:616  acc:0.31017119838872104
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ID:1986
picturePath:image/3331.png
answer:解:∵∠AOC=110°,∴∠ADC=\frac{1}{2}∠AOC=\frac{1}{2}×110°=55°．故选:A．
answer:55
correct answer:125
total:1987  correct:616  acc:0.31001509813789635
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ID:1987
picturePath:image/3332.png
answer:解:∵OA⊥OB,∴∠AOB=90°,∵∠C=70°,∴∠BOC=180°-90°-70°=20°,∵AE⊥OD于E,∴∠OED=90°,∴∠AED=90°-20°=70°,∴∠A=180°-70°-70°=40°．故选:B．
answer:40
correct answer:40
total:1988  correct:617  acc:0.3103621730382294
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ID:1988
picturePath:image/3333.png
answer:解:∵∠ABC=70°,∠ACB=30°,∴∠A=180°-70°-30°=80°,∵D是弧BAC的中点,∴∠ADB=\frac{1}{2}∠A=40°,∴∠DBC=∠ADB=40°．故选:A．
answer:40
correct answer:50
total:1989  correct:617  acc:0.3102061337355455
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ID:1989
picturePath:image/3334.png
answer:解:∵∠BOC与∠A是同弧所对的圆心角与圆周角,∠BOC=100°,∴∠A=\frac{1}{2}∠BOC=50°．故选:B．
answer:50
correct answer:50
total:1990  correct:618  acc:0.31055276381909547
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ID:1990
picturePath:image/3335.png
answer:The answer is D.
ID:1991
picturePath:image/3337.png
answer:解:∵AB为⊙O直径,∴∠AOD=90°,∵∠C+∠AOD=145°,∴∠C=55°,
故选:A．
answer:55
correct answer:35
total:1992  correct:618  acc:0.3102409638554217
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ID:1992
picturePath:image/3338.png
answer:∠AOC=2∠B=2×75°=150°.
answer:150
correct answer:150
total:1993  correct:619  acc:0.31058705469141995
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ID:1993
picturePath:image/3339.png
answer:60
answer:60
correct answer:120
total:1994  correct:619  acc:0.3104312938816449
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ID:1994
picturePath:image/3340.png
answer:16°
answer:16
correct answer:58
total:1995  correct:619  acc:0.31027568922305765
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ID:1995
picturePath:image/3341.png
answer:20°
answer:20
correct answer:70
total:1996  correct:619  acc:0.31012024048096193
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ID:1996
picturePath:image/3344.png
answer:25
answer:25
correct answer:35
total:1997  correct:619  acc:0.3099649474211317
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ID:1997
picturePath:image/3345.png
answer:解:∵AB,CD是⊙O的两条弦,∴∠ADB=∠BCD=50°,∴∠BAD=∠ADB=50°．故选:B．
answer:50
correct answer:50
total:1998  correct:620  acc:0.3103103103103103
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ID:1998
picturePath:image/3348.png
answer:60°
answer:60
correct answer:120
total:1999  correct:620  acc:0.31015507753876936
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ID:1999
picturePath:image/3351.png
answer:70
answer:70
correct answer:70
total:2000  correct:621  acc:0.3105
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ID:2000
picturePath:image/3353.png
answer:根据圆周角定理，同弧所对的圆周角相等，因此∠ADC=∠BAC=25°。故选A。
answer:25
correct answer:65
total:2001  correct:621  acc:0.3103448275862069
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ID:2001
picturePath:image/3355.png