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Compound Angles
In Fig. 2.5 a), A,P B, Q are four points on a circle in a horicaontal plane. ∠AQB=θ, ∠PAQ=π/2.a) Express sin∠ABQ in terms of AB, AQ and θ. Hence, find PQ in terms of AB and θ.b) Using the result of a), show that PQ=[√(AP2+BP2+2AP*BPcosθ)]/sinθ.c) Firthermore, V is a point vertically above P. Let ∠VAP=α,... 顯示更多 In Fig. 2.5 a), A,P B, Q are four points on a circle in a horicaontal plane. ∠AQB=θ, ∠PAQ=π/2. a) Express sin∠ABQ in terms of AB, AQ and θ. Hence, find PQ in terms of AB and θ. b) Using the result of a), show that PQ=[√(AP2+BP2+2AP*BPcosθ)]/sinθ. c) Firthermore, V is a point vertically above P. Let ∠VAP=α, ∠VBP=β and ∠VQP=φ. Using the result of b), show that cot2φ=(cot2α+cot2β+2cotβcotαcosθ)/sin2θ. d) If α=π/4, β=π/3 and φ=π/6, find θ, where θ<π/2. 1.png (33.15 KB)
最佳解答:
a) sin ∠ABQ / sin θ= AQ / AB sin ∠ABQ = AQ sinθ/ AB ∠ABQ = ∠APQ (∠s in the same segment ) AQ = PQ sin ∠APQ AQ = PQ ( AQ sinθ/ AB ) 1 = PQ sinθ/ AB PQ = AB / sinθ b) ∠ APB = 180* - θ cos ( 180* - θ) = ( AP2 + PB2 – AB2 ) / 2 ( AP )( PB ) - 2 ( AP )( PB ) cosθ= AP2 + PB2 – AB2 AB2 = AP2 + PB2 + 2 ( AP )( PB ) cosθ Then, PQ = √{AP2 + PB2 + 2 ( AP )( PB ) cosθ} / sinθ c) PQ = VP / tan φ, AP = VP / tan α, BP = VP / tan β Then, VP cot φ = √{VP2 cot2α + VP2 cot2 β+ 2 ( VP cot α)( VP cot β ) cosθ} / sinθ VP2 cot2φ = {VP2 cot2α + VP2 cot2 β+ 2 ( VP cot α)( VP cot β ) cosθ}/ sin2θ cot2φ = { cot2α + cot2 β+ 2 ( cot α)( cot β ) cosθ} / sin2θ d) cot2( pi / 6 )=cot2( pi / 4 )+cot2 ( pi / 3 )+ 2( cot pi / 4 )( cot pi / 3 )( cosθ) / sin2θ 3 = (1 + 1 / 3 + 2 cosθ/√3) / sin2θ 3( 1 – cos2θ) = 4 / 3 + 2 cosθ/√3 3 – 3 cos2θ= 4 / 3 + 2 cosθ/√3 9√3 cos2θ+ 6 cosθ- 5√3 = 0 By quadratic formula, θ= 0.955 rad ( cor. to 3 s.f. )
其他解答:
Compound Angles
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發問:In Fig. 2.5 a), A,P B, Q are four points on a circle in a horicaontal plane. ∠AQB=θ, ∠PAQ=π/2.a) Express sin∠ABQ in terms of AB, AQ and θ. Hence, find PQ in terms of AB and θ.b) Using the result of a), show that PQ=[√(AP2+BP2+2AP*BPcosθ)]/sinθ.c) Firthermore, V is a point vertically above P. Let ∠VAP=α,... 顯示更多 In Fig. 2.5 a), A,P B, Q are four points on a circle in a horicaontal plane. ∠AQB=θ, ∠PAQ=π/2. a) Express sin∠ABQ in terms of AB, AQ and θ. Hence, find PQ in terms of AB and θ. b) Using the result of a), show that PQ=[√(AP2+BP2+2AP*BPcosθ)]/sinθ. c) Firthermore, V is a point vertically above P. Let ∠VAP=α, ∠VBP=β and ∠VQP=φ. Using the result of b), show that cot2φ=(cot2α+cot2β+2cotβcotαcosθ)/sin2θ. d) If α=π/4, β=π/3 and φ=π/6, find θ, where θ<π/2. 1.png (33.15 KB)
最佳解答:
a) sin ∠ABQ / sin θ= AQ / AB sin ∠ABQ = AQ sinθ/ AB ∠ABQ = ∠APQ (∠s in the same segment ) AQ = PQ sin ∠APQ AQ = PQ ( AQ sinθ/ AB ) 1 = PQ sinθ/ AB PQ = AB / sinθ b) ∠ APB = 180* - θ cos ( 180* - θ) = ( AP2 + PB2 – AB2 ) / 2 ( AP )( PB ) - 2 ( AP )( PB ) cosθ= AP2 + PB2 – AB2 AB2 = AP2 + PB2 + 2 ( AP )( PB ) cosθ Then, PQ = √{AP2 + PB2 + 2 ( AP )( PB ) cosθ} / sinθ c) PQ = VP / tan φ, AP = VP / tan α, BP = VP / tan β Then, VP cot φ = √{VP2 cot2α + VP2 cot2 β+ 2 ( VP cot α)( VP cot β ) cosθ} / sinθ VP2 cot2φ = {VP2 cot2α + VP2 cot2 β+ 2 ( VP cot α)( VP cot β ) cosθ}/ sin2θ cot2φ = { cot2α + cot2 β+ 2 ( cot α)( cot β ) cosθ} / sin2θ d) cot2( pi / 6 )=cot2( pi / 4 )+cot2 ( pi / 3 )+ 2( cot pi / 4 )( cot pi / 3 )( cosθ) / sin2θ 3 = (1 + 1 / 3 + 2 cosθ/√3) / sin2θ 3( 1 – cos2θ) = 4 / 3 + 2 cosθ/√3 3 – 3 cos2θ= 4 / 3 + 2 cosθ/√3 9√3 cos2θ+ 6 cosθ- 5√3 = 0 By quadratic formula, θ= 0.955 rad ( cor. to 3 s.f. )
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