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設 ED = h m, ∠ACD = θ 在 ΔEDC 中： (h m)/CD = tan60° (h m)/CD = √3 CD = h/√3 m 在 ΔEDB 中：(h m)/BD = tan45° (h m)/BD = 1 BD = h m 在 ΔEDA 中： (h m)/AD = tan30° (h m)/AD = 1/√3 AD = h√3m 在 ΔDBC中： cos∠BCD = (BC2 + CD2 - BD2) / (2 × BC × CD) cosθ = [502 + (h/√3)2 - h2) / [2 × 50 × (h/3)] ...... [1] 在 ΔDAC中： cos∠ACD = (AC2 + CD2 - AD2) / (2 × AC × CD) cosθ = [1502 + (h/√3)2 - (h√32) / [2 × 150 × (h/3)] ...... [2] [1] = [2] : [502 + (h/√3)2 - h2] / [2 × 50 × (h/3)] = [1502 + (h/√3)2 - (h√32) / [2 × 150 × (h/3)] 2500 + (h2/3) - h2 = [22500 + (h2/3) - 3h2] / 3 7500 + h2 - 3h2 = 22500 + (h2/3) - 3h2 (2/3)h2 = 15000 h2 = 22500 h = 150 山的高度 = 150m
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圖片參考：https://s.yimg.com/lo/api/res/1.2/1dC3o06zlFbhCBfCkpf6FA--/YXBwaWQ9dHdhbnN3ZXJzO3E9ODU-/http://oi62.tinypic.com/imkx1x.jpg http://oi62.tinypic.com/imkx1x.jpg
最佳解答:
設 ED = h m, ∠ACD = θ 在 ΔEDC 中： (h m)/CD = tan60° (h m)/CD = √3 CD = h/√3 m 在 ΔEDB 中：(h m)/BD = tan45° (h m)/BD = 1 BD = h m 在 ΔEDA 中： (h m)/AD = tan30° (h m)/AD = 1/√3 AD = h√3m 在 ΔDBC中： cos∠BCD = (BC2 + CD2 - BD2) / (2 × BC × CD) cosθ = [502 + (h/√3)2 - h2) / [2 × 50 × (h/3)] ...... [1] 在 ΔDAC中： cos∠ACD = (AC2 + CD2 - AD2) / (2 × AC × CD) cosθ = [1502 + (h/√3)2 - (h√32) / [2 × 150 × (h/3)] ...... [2] [1] = [2] : [502 + (h/√3)2 - h2] / [2 × 50 × (h/3)] = [1502 + (h/√3)2 - (h√32) / [2 × 150 × (h/3)] 2500 + (h2/3) - h2 = [22500 + (h2/3) - 3h2] / 3 7500 + h2 - 3h2 = 22500 + (h2/3) - 3h2 (2/3)h2 = 15000 h2 = 22500 h = 150 山的高度 = 150m
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