SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=3, BC=2, ∠ACB=90°. Find the value of sin(BA).
问题：如图所示，AB=3，BC=2，∠ACB=90°。求sin(BA)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(BC),2)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Sin(MeasureOfAngle(BAC)))
Answer: 2/3
Theorem sequence:
cosine_theorem(1,ACB)
cosine_theorem(1,CBA)
Problem: As shown in the diagram, AB=3, OB=2, ∠AOB=α°, ∠OBA=90°. Find tan(a).
问题：如图所示，AB=3，OB=2，∠AOB=α°，∠OBA=90°。求tan(a)。
Construction (CDL):
Shape(AO,OB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(OB),2)
Equal(MeasureOfAngle(AOB),α)
Equal(MeasureOfAngle(OBA),90)
Goal: Value(Tan(α))
Answer: 3/2
Theorem sequence:
cosine_theorem(1,OBA)
cosine_theorem(1,BAO)
Problem: As shown in the diagram, AB=4, AC=2, ∠BCA=90°. Find the value of sin(AB).
问题：如图所示，AB=4，AC=2，∠BCA=90°。求sin(AB)的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AC),2)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(Sin(MeasureOfAngle(ABC)))
Answer: 1/2
Theorem sequence:
cosine_theorem(1,BCA)
cosine_theorem(1,CAB)
Problem: As shown in the diagram, AC=5, BC=12, ∠ACB=90°. Find the value of cos(BA).
问题：如图所示，AC=5，BC=12，∠ACB=90°。求cos(BA)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),12)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Cos(MeasureOfAngle(BAC)))
Answer: 5/13
Theorem sequence:
cosine_theorem(1,ACB)
cosine_theorem(1,CBA)
Problem: As shown in the diagram, ∠BCA=48°, O is the center of ⊙O. Find the measure of ∠BOA.
问题：如图所示，∠BCA=48°，O是圆O的圆心。求∠BOA的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(BO,OA,OAB)
Shape(CB,OBC)
Shape(CA,AO,OB,BC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),48)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BOA))
Answer: 96
Theorem sequence:
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
Problem: As shown in the diagram, Div(LengthOfLine(OM)=LengthOfLine(OP)), ∠OMP=90°, ∠POM=α°. Find cos(a).
问题：如图所示，Div(LengthOfLine(OM)=LengthOfLine(OP))，∠OMP=90°，∠POM=α°。求cos(a)。
Construction (CDL):
Shape(PO,OM,MP)
Conditions (CDL):
Equal(Div(LengthOfLine(OM),LengthOfLine(OP)),4/5)
Equal(MeasureOfAngle(OMP),90)
Equal(MeasureOfAngle(POM),α)
Goal: Value(Cos(α))
Answer: 4/5
Theorem sequence:
sine_theorem(1,POM)
sine_theorem(1,OMP)
cosine_theorem(1,POM)
Problem: As shown in the diagram, AC=4, ∠ACB=90°, Tan(MeasureOfAngle(BAC))=1/2. Find the length of line BC.
问题：如图所示，AC=4，∠ACB=90°，Tan(MeasureOfAngle(BAC))=1/2。求直线BC的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(MeasureOfAngle(ACB),90)
Equal(Tan(MeasureOfAngle(BAC)),1/2)
Goal: Value(LengthOfLine(BC))
Answer: 2
Theorem sequence:
cosine_theorem(1,ACB)
cosine_theorem(1,CBA)
Problem: As shown in the diagram, ∠BOA=80°, the center of circle O is O. Find the measure of ∠BCA.
问题：如图所示，∠BOA=80°，⊙O的圆心为O。求∠BCA的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(BD,DA,OAB)
Shape(CD,DO,OB,OBC)
Shape(OD,DB,BO)
Shape(CA,AD,DC)
Collinear(CDB)
Collinear(ADO)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BOA),80)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 40
Theorem sequence:
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
Problem: As shown in the diagram, AC=3, CD=5/2, ∠BCA=90°, CD is the median of △ CAB. Find the value of tan(AB).
问题：如图所示，AC=3，CD=5/2，∠BCA=90°，CD是三角形CAB的中线。求tan(AB)的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(AD,DC,CA)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(CD),5/2)
Equal(MeasureOfAngle(BCA),90)
IsMedianOfTriangle(CD,CAB)
Goal: Value(Tan(MeasureOfAngle(ABC)))
Answer: 3/4
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
right_triangle_property_length_of_median(1,BCA,D)
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AB=4, AC=4, BC=6. Find the value of cos(AB).
问题：如图所示，AB=4，AC=4，BC=6。求cos(AB)的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(BC),6)
Goal: Value(Cos(MeasureOfAngle(ABC)))
Answer: 3/4
Theorem sequence:
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AC=15, BC=8, ∠ACB=90°. Find the value of sin(BA).
问题：如图所示，AC=15，BC=8，∠ACB=90°。求sin(BA)的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),15)
Equal(LengthOfLine(BC),8)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Sin(MeasureOfAngle(BAC)))
Answer: 8/17
Theorem sequence:
cosine_theorem(1,ACB)
cosine_theorem(1,CBA)
Problem: As shown in the diagram, AC=3, OB=7, OC=3, ∠ACB=90°. Find the value of tan(OB).
问题：如图所示，AC=3，OB=7，OC=3，∠ACB=90°。求tan(OB)的值。
Construction (CDL):
Shape(AO,OC,CA)
Shape(AC,CB,BA)
Collinear(OCB)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(OB),7)
Equal(LengthOfLine(OC),3)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Tan(MeasureOfAngle(OBA)))
Answer: 3/4
Theorem sequence:
line_addition(1,OC,CB)
cosine_theorem(1,CBA)
cosine_theorem(1,BAC)
Problem: As shown in the diagram, AB=3, BC=2, ∠BCA=90°. Find the value of cos(AB).
问题：如图所示，AB=3，BC=2，∠BCA=90°。求cos(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(BC),2)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(Cos(MeasureOfAngle(ABC)))
Answer: 2/3
Theorem sequence:
cosine_theorem(1,BCA)
cosine_theorem(1,CAB)
Problem: As shown in the diagram, AC=6, CD=5, CD is the median of △ CAB, triangle BCA is a right triangle. Find the value of tan(AB).
问题：如图所示，AC=6，CD=5，CD是△CAB的中线，∠BCA是直角△BCA的直角。求tan(AB)的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(DB,BC,CD)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(CD),5)
IsMedianOfTriangle(CD,CAB)
RightTriangle(BCA)
Goal: Value(Tan(MeasureOfAngle(ABC)))
Answer: 3/4
Theorem sequence:
right_triangle_property_length_of_median(1,BCA,D)
cosine_theorem(1,CAB)
cosine_theorem(1,BCA)
Problem: As shown in the diagram, BC=2, Sin(MeasureOfAngle(BAC))=1/3, AC and CB are the legs of the right triangle ACB. Find the length of line AB.
问题：如图所示，BC=2，Sin(MeasureOfAngle(BAC))=1/3，三角形ACB为直角三角形。求直线AB的长度。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(BC),2)
Equal(Sin(MeasureOfAngle(BAC)),1/3)
RightTriangle(ACB)
Goal: Value(LengthOfLine(AB))
Answer: 6
Theorem sequence:
cosine_theorem(1,ACB)
cosine_theorem(1,CBA)
Problem: As shown in the diagram, ∠ECD=80°, O is the center of ⊙O, CD⊥OD, OE is perpendicular to CE. Find the measure of ∠APB.
问题：如图所示，∠ECD=80°，O是圆O的圆心，CD垂直于OD，OE垂直于CE。求∠APB的大小。
Construction (CDL):
Shape(AD,DC,OCA)
Shape(CE,EB,OBC)
Shape(CD,DO,OE,EC)
Shape(PA,PAP)
Shape(BP,OPB)
Shape(AP,PB,BE,EO,OD,DA)
Collinear(ADO)
Collinear(BEO)
Cocircular(O,APBC)
Conditions (CDL):
Equal(MeasureOfAngle(ECD),80)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CD,OD)
PerpendicularBetweenLine(OE,CE)
Goal: Value(MeasureOfAngle(APB))
Answer: 50
Theorem sequence:
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OBA,P)
quadrilateral_property_angle_sum(1,CDOE)
Problem: As shown in the diagram, AB=5, AC=4, ∠ACB=90°. Find the value of sin(CB).
问题：如图所示，AB=5，AC=4，∠ACB=90°。求sin(CB)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),4)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Sin(MeasureOfAngle(CBA)))
Answer: 4/5
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, ∠BOA=90°, O is the center of circle O. Find the value of sin(BP).
问题：如图所示，∠BOA=90°，⊙O的圆心为O。求sin(BP)的值。
Construction (CDL):
Shape(BP,PA,AO,OB)
Shape(PB,OBP)
Shape(AP,OPA)
Shape(BO,OA,OAB)
Cocircular(O,BPA)
Conditions (CDL):
Equal(MeasureOfAngle(BOA),90)
IsCentreOfCircle(O,O)
Goal: Value(Sin(MeasureOfAngle(BPA)))
Answer: sqrt(2)/2
Theorem sequence:
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,P)
Problem: As shown in the diagram, Cos(MeasureOfAngle(ABC))=2/3, AB=6, ∠BCA=90°. Find the length of line BC.
问题：如图所示，Cos(MeasureOfAngle(ABC))=2/3，AB=6，∠BCA=90°。求直线BC的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(Cos(MeasureOfAngle(ABC)),2/3)
Equal(LengthOfLine(AB),6)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(LengthOfLine(BC))
Answer: 4
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,BCA)
Problem: As shown in the diagram, AC=2, BC=3, ∠ACB=90°. Find the value of tan(CB).
问题：如图所示，AC=2，BC=3，∠ACB=90°。求tan(CB)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),2)
Equal(LengthOfLine(BC),3)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Tan(MeasureOfAngle(CBA)))
Answer: 2/3
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
cosine_theorem(1,BAC)
Problem: As shown in the diagram, OA=OB, OA=OC, OC=OB, ∠ABO=40°, the center of circle O is O. Find the measure of ∠BCA.
问题：如图所示，OA=OB，OA=OC，OC=OB，∠ABO=40°，O是圆O的圆心。求∠BCA的大小。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OB,BC,CO)
Shape(AO,OC,CA)
Shape(AC,OCA)
Shape(BA,OAB)
Shape(CB,OBC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(OA),LengthOfLine(OB))
Equal(LengthOfLine(OA),LengthOfLine(OC))
Equal(LengthOfLine(OC),LengthOfLine(OB))
Equal(MeasureOfAngle(ABO),40)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 50
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,OAB)
triangle_property_angle_sum(1,OAB)
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
Problem: As shown in the diagram, OA=4, OP=OB, OP=OQ, OQ=OB, B bisects segment OA, Q bisects segment PA. Find the value of cos(BQ).
问题：如图所示，OA=4，OP=OB，OP=OQ，OQ=OB，B平分线段OA，Q是线段PA的中点。求cos(BQ)的值。
Construction (CDL):
Shape(PO,OQ,QP)
Shape(BA,AQ,OBQ)
Shape(QO,OB,BQ)
Shape(QB,OBQ)
Shape(PQ,OQP)
Shape(BO,OP,OPB)
Collinear(OBA)
Collinear(PQA)
Cocircular(O,BQP)
Conditions (CDL):
Equal(LengthOfLine(OA),4)
Equal(LengthOfLine(OP),LengthOfLine(OB))
Equal(LengthOfLine(OP),LengthOfLine(OQ))
Equal(LengthOfLine(OQ),LengthOfLine(OB))
IsMidpointOfLine(B,OA)
IsMidpointOfLine(Q,PA)
Goal: Value(Cos(MeasureOfAngle(BQO)))
Answer: 1/4
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,QB,APO)
midsegment_of_triangle_property_length(1,QB,APO)
cosine_theorem(1,QOB)
Problem: As shown in the diagram, Cos(MeasureOfAngle(CAB))=3/5, AB=10, ∠BCA=90°. Find the length of line BC.
问题：如图所示，Cos(MeasureOfAngle(CAB))=3/5，AB=10，∠BCA=90°。求直线BC的长度。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(Cos(MeasureOfAngle(CAB)),3/5)
Equal(LengthOfLine(AB),10)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(LengthOfLine(BC))
Answer: 8
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, AC=8, BC=6, CD=5, ∠BCA=90°, D bisects segment AB. Find the value of tan(DC).
问题：如图所示，AC=8，BC=6，CD=5，∠BCA=90°，D是线段AB的中点。求tan(DC)的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(CD),5)
Equal(MeasureOfAngle(BCA),90)
IsMidpointOfLine(D,AB)
Goal: Value(Tan(MeasureOfAngle(DCA)))
Answer: 3/4
Theorem sequence:
line_addition(1,AD,DB)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
cosine_theorem(1,CAD)
Problem: As shown in the diagram, ∠DOA=130°, the center of circle O is O. Find the measure of ∠BCE.
问题：如图所示，∠DOA=130°，圆O的圆心为O。求∠BCE的大小。
Construction (CDL):
Shape(DO,OA,OAD)
Shape(BE,ED,ODB)
Shape(OD,DE,EO)
Shape(CE,EB,BC)
Shape(CB,OBC)
Shape(AO,OE,EC,OCA)
Collinear(AOEB)
Collinear(CED)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(MeasureOfAngle(DOA),130)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BCE))
Answer: 25
Theorem sequence:
adjacent_complementary_angle(1,EOD,DOA)
arc_property_center_angle(1,ODB,O)
arc_property_circumference_angle_external(1,ODB,C)
Problem: As shown in the diagram, AB=3, AC=4, ∠CAB=90°, AD⊥CD. Find the value of sin(CA).
问题：如图所示，AB=3，AC=4，∠CAB=90°，AD⊥CD。求sin(CA)的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(AB,BD,DA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),4)
Equal(MeasureOfAngle(CAB),90)
PerpendicularBetweenLine(AD,CD)
Goal: Value(Sin(MeasureOfAngle(CAD)))
Answer: 4/5
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
adjacent_complementary_angle(1,BDA,ADC)
triangle_property_angle_sum(1,ABD)
angle_addition(1,CAD,DAB)
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AB=5, AC=4, ∠BCA=90°. Find the value of tan(AB).
问题：如图所示，AB=5，AC=4，∠BCA=90°。求tan(AB)的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),4)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(Tan(MeasureOfAngle(ABC)))
Answer: 4/3
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AB=5, AC=4, ∠BCA=90°, CD⊥BD. Find the value of tan(BC).
问题：如图所示，AB=5，AC=4，∠BCA=90°，CD垂直于BD。求tan(BC)的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),4)
Equal(MeasureOfAngle(BCA),90)
PerpendicularBetweenLine(CD,BD)
Goal: Value(Tan(MeasureOfAngle(BCD)))
Answer: 3/4
Theorem sequence:
triangle_property_angle_sum(1,CDB)
triangle_property_angle_sum(1,BCA)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
cosine_theorem(1,ABC)
Problem: As shown in the diagram, AB=8, ∠BCA=90°, ∠CAB=15°. Find the value of AC*BC.
问题：如图所示，AB=8，∠BCA=90°，∠CAB=15°。求AC*BC的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(MeasureOfAngle(BCA),90)
Equal(MeasureOfAngle(CAB),15)
Goal: Value(Mul(LengthOfLine(AC),LengthOfLine(BC)))
Answer: (-2*sqrt(2)+2*sqrt(6))*(2*sqrt(2)+2*sqrt(6))
Theorem sequence:
triangle_property_angle_sum(1,CAB)
sine_theorem(1,ABC)
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=13, BC=5, ∠ACB=90°. Find the value of sin(BA).
问题：如图所示，AB=13，BC=5，∠ACB=90°。求sin(BA)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(BC),5)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Sin(MeasureOfAngle(BAC)))
Answer: 5/13
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
cosine_theorem(1,ACB)
Problem: As shown in the diagram, BC=5, CD=3, EF=2, E bisects segment AB, F bisects segment AD. Find the value of tan(BC).
问题：如图所示，BC=5，CD=3，EF=2，E是线段AB的中点，F平分线段AD。求tan(BC)的值。
Construction (CDL):
Shape(AE,EF,FA)
Shape(FE,EB,BD,DF)
Shape(DB,BC,CD)
Collinear(AEB)
Collinear(AFD)
Conditions (CDL):
Equal(LengthOfLine(BC),5)
Equal(LengthOfLine(CD),3)
Equal(LengthOfLine(EF),2)
IsMidpointOfLine(E,AB)
IsMidpointOfLine(F,AD)
Goal: Value(Tan(MeasureOfAngle(BCD)))
Answer: 4/3
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,EF,ABD)
midsegment_of_triangle_property_length(1,EF,ABD)
cosine_theorem(1,CDB)
Problem: As shown in the diagram, AC=2, BC=1, ∠ACB=90°. Find the value of tan(BA).
问题：如图所示，AC=2，BC=1，∠ACB=90°。求tan(BA)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),2)
Equal(LengthOfLine(BC),1)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Tan(MeasureOfAngle(BAC)))
Answer: 1/2
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
cosine_theorem(1,ACB)
Problem: As shown in the diagram, BC=2, BD=BE, ABCD is a square. Find the value of tan(BA).
问题：如图所示，BC=2，BD=BE，四边形ABCD是正方形。求tan(BA)的值。
Construction (CDL):
Shape(AE,EB,BA)
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(EBC)
Conditions (CDL):
Equal(LengthOfLine(BC),2)
Equal(LengthOfLine(BD),LengthOfLine(BE))
Square(ABCD)
Goal: Value(Tan(MeasureOfAngle(BAE)))
Answer: sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,BCD)
right_triangle_property_pythagorean(1,BCD)
adjacent_complementary_angle(1,EBA,ABC)
right_triangle_judgment_angle(1,EBA)
right_triangle_property_pythagorean(1,EBA)
cosine_theorem(1,AEB)
Problem: As shown in the diagram, AC=2, AD=3, O is the center of ⊙O, the diameter of ⊙O is AD. Find the value of cos(EB).
问题：如图所示，AC=2，AD=3，⊙O的圆心为O，AD是⊙O的直径。求cos(EB)的值。
Construction (CDL):
Shape(CA,AE,EC)
Shape(CE,EB,BC)
Shape(EA,AO,OD,DE)
Shape(DO,OA,OAD)
Shape(BE,ED,ODB)
Shape(CB,OBC)
Shape(AC,OCA)
Collinear(CED)
Collinear(AOD)
Collinear(AEB)
Cocircular(O,DBCA)
Conditions (CDL):
Equal(LengthOfLine(AC),2)
Equal(LengthOfLine(AD),3)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AD,O)
Goal: Value(Cos(MeasureOfAngle(EBC)))
Answer: sqrt(5)/3
Theorem sequence:
diameter_of_circle_property_right_angle(1,DCA,O)
right_triangle_judgment_angle(1,DCA)
right_triangle_property_pythagorean(1,DCA)
cosine_theorem(1,DCA)
arc_property_circumference_angle_external(1,OCA,B)
arc_property_circumference_angle_external(1,OCA,D)
Problem: As shown in the diagram, ∠CAE=30°, O is the center of ⊙O, AB is the diameter of ⊙O. Find the measure of ∠ADE.
问题：如图所示，∠CAE=30°，O是圆O的圆心，AB是⊙O的直径。求∠ADE的大小。
Construction (CDL):
Shape(AO,OE,EC,CA)
Shape(CE,EB,BC)
Shape(AD,DE,EO,OA)
Shape(BE,ED,ODB)
Shape(CB,OBC)
Shape(AC,OCA)
Shape(DA,OAD)
Collinear(AOEB)
Collinear(CED)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(MeasureOfAngle(CAE),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
Goal: Value(MeasureOfAngle(ADE))
Answer: 60
Theorem sequence:
diameter_of_circle_property_right_angle(1,BCA,O)
triangle_property_angle_sum(1,CAB)
arc_property_circumference_angle_external(1,OCA,B)
arc_property_circumference_angle_external(1,OCA,D)
Problem: As shown in the diagram, AC=3, BC=2, ∠BCA=90°. Find the value of tan(AB).
问题：如图所示，AC=3，BC=2，∠BCA=90°。求tan(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(BC),2)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(Tan(MeasureOfAngle(ABC)))
Answer: 3/2
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
cosine_theorem(1,BCA)
Problem: As shown in the diagram, BC=1, ∠BCA=90°, Sin(MeasureOfAngle(CAB))=1/3. Find the length of line AB.
问题：如图所示，BC=1，∠BCA=90°，Sin(MeasureOfAngle(CAB))=1/3。求直线AB的长度。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(BC),1)
Equal(MeasureOfAngle(BCA),90)
Equal(Sin(MeasureOfAngle(CAB)),1/3)
Goal: Value(LengthOfLine(AB))
Answer: 3
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=2, BC=1, ∠ACB=90°. Find the value of sin(BA).
问题：如图所示，AB=2，BC=1，∠ACB=90°。求sin(BA)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(BC),1)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Sin(MeasureOfAngle(BAC)))
Answer: 1/2
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
cosine_theorem(1,ACB)
Problem: As shown in the diagram, AB=5, AC=3, ∠BCA=90°. Find the value of sin(AB).
问题：如图所示，AB=5，AC=3，∠BCA=90°。求sin(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),3)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(Sin(MeasureOfAngle(ABC)))
Answer: 3/5
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AC=4, ∠BDO=90°, the center of ⊙O is O, the diameter of ⊙O is AB, OD is perpendicular to CD. Find the length of line OD.
问题：如图所示，AC=4，∠BDO=90°，O是⊙O的圆心，AB是⊙O的直径，OD⊥CD。求直线OD的长度。
Construction (CDL):
Shape(DO,OB,BD)
Shape(AO,OD,DC,CA)
Shape(BO,OA,OAB)
Shape(CD,DB,OBC)
Shape(AC,OCA)
Collinear(AOB)
Collinear(CDB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(MeasureOfAngle(BDO),90)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
PerpendicularBetweenLine(OD,CD)
Goal: Value(LengthOfLine(OD))
Answer: 2
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OA,O)
line_addition(1,AO,OB)
diameter_of_circle_property_right_angle(1,BCA,O)
similar_triangle_judgment_aa(1,OBD,ABC)
similar_triangle_property_line_ratio(1,DOB,CAB)
similar_triangle_property_line_ratio(1,BDO,BCA)
Problem: As shown in the diagram, AC=3, CD=2, ∠BCA=90°, CD is the median of triangle CAB. Find the value of sin(DB).
问题：如图所示，AC=3，CD=2，∠BCA=90°，CD是三角形CAB的中线。求sin(DB)的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(BC,CD,DB)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(CD),2)
Equal(MeasureOfAngle(BCA),90)
IsMedianOfTriangle(CD,CAB)
Goal: Value(Sin(MeasureOfAngle(DBC)))
Answer: 3/4
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
right_triangle_property_length_of_median(1,BCA,D)
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AC=3, BC=4, ∠BCA=90°, AD⊥CD. Find the value of tan(DC).
问题：如图所示，AC=3，BC=4，∠BCA=90°，AD⊥CD。求tan(DC)的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(BC),4)
Equal(MeasureOfAngle(BCA),90)
PerpendicularBetweenLine(AD,CD)
Goal: Value(Tan(MeasureOfAngle(DCA)))
Answer: tan(pi*(-acos(3/5)/pi+1/2))
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
triangle_property_angle_sum(1,CAB)
adjacent_complementary_angle(1,ADC,CDB)
triangle_property_angle_sum(1,CDB)
angle_addition(1,BCD,DCA)
cosine_theorem(1,ABC)
Problem: As shown in the diagram, AB=13, AC=5, BC=12. Find the value of tan(CB).
问题：如图所示，AB=13，AC=5，BC=12。求tan(CB)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),12)
Goal: Value(Tan(MeasureOfAngle(CBA)))
Answer: 5/12
Theorem sequence:
cosine_theorem(1,BAC)
Problem: As shown in the diagram, AC=4, BC=3, ∠BCA=90°. Find the value of sin(CA).
问题：如图所示，AC=4，BC=3，∠BCA=90°。求sin(CA)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(BC),3)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(Sin(MeasureOfAngle(CAB)))
Answer: 3/5
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
cosine_theorem(1,ABC)
Problem: As shown in the diagram, ∠AOC=100°, O is the center of circle O, MO∥BC. Find the measure of ∠AMO.
问题：如图所示，∠AOC=100°，O是圆O的圆心，MO平行于BC。求∠AMO的大小。
Construction (CDL):
Shape(AM,MO,OA)
Shape(OM,MB,BC,CO)
Shape(AO,OC,OCA)
Shape(CB,OBC)
Shape(BM,MA,OAB)
Collinear(AMB)
Cocircular(O,BCA)
Conditions (CDL):
Equal(MeasureOfAngle(AOC),100)
IsCentreOfCircle(O,O)
ParallelBetweenLine(MO,BC)
Goal: Value(MeasureOfAngle(AMO))
Answer: 50
Theorem sequence:
arc_property_center_angle(1,OCA,O)
arc_property_circumference_angle_external(1,OCA,B)
parallel_property_corresponding_angle(1,MO,BC,A)
Problem: As shown in the diagram, ∠BCA=90°, Sin(MeasureOfAngle(CAB))=3/5. Find the value of cos(AB).
问题：如图所示，∠BCA=90°，Sin(MeasureOfAngle(CAB))=3/5。求cos(AB)的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),90)
Equal(Sin(MeasureOfAngle(CAB)),3/5)
Goal: Value(Cos(MeasureOfAngle(ABC)))
Answer: cos(pi*(-asin(3/5)/pi+1/2))
Theorem sequence:
triangle_property_angle_sum(1,BCA)
Problem: As shown in the diagram, ∠BCA=90°, Sin(MeasureOfAngle(CAB))=1/3. Find the value of cos(AB).
问题：如图所示，∠BCA=90°，Sin(MeasureOfAngle(CAB))=1/3。求cos(AB)的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),90)
Equal(Sin(MeasureOfAngle(CAB)),1/3)
Goal: Value(Cos(MeasureOfAngle(ABC)))
Answer: cos(pi*(-asin(1/3)/pi+1/2))
Theorem sequence:
triangle_property_angle_sum(1,BCA)
Problem: As shown in the diagram, Cos(MeasureOfAngle(BAC))=4/5, ∠ACB=90°. Find the value of sin(CB).
问题：如图所示，Cos(MeasureOfAngle(BAC))=4/5，∠ACB=90°。求sin(CB)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(Cos(MeasureOfAngle(BAC)),4/5)
Equal(MeasureOfAngle(ACB),90)
Goal: Value(Sin(MeasureOfAngle(CBA)))
Answer: sin(pi*(-acos(4/5)/pi+1/2))
Theorem sequence:
triangle_property_angle_sum(1,ACB)