如图：在山脚下A测得山顶P的仰角为a，沿倾斜角为β的斜坡向上走a米到达点B，在B处测得山顶P的仰角为γ，则山高PQ为asinαsin(γ−β)sin(γ−α)asinαsin(γ−β)sin(γ−α)米．

Question

如图：在山脚下A测得山顶P的仰角为a，沿倾斜角为β的斜坡向上走a米到达点B，在B处测得山顶P的仰角为γ，则山高PQ为

(asinαsin(γ−β))/(sin(γ−α))

(asinαsin(γ−β))/(sin(γ−α))

米．

Answer 1

过B作BC⊥PQ，交PQ于点C，过B作BD⊥AQ，交AQ于点D，
可得BD=CQ，
在△PAB中，∠PAB=α-β，∠BPA=（(π)/(2)-α）-（(π)/(2)-γ）=γ-α，
∴(PB)/(sin(α−β))=(a)/(sin(γ−α))，即PB=(asin(α−β))/(sin(γ−α))，
则PQ=PC+CQ=PB•sinγ+asinβ
=(asin(α−β))/(sin(γ−α))•sinγ+asinβ
=(asin(α−β)sinγ+asinβsin(γ−α))/(sin(γ−α))
=(asinαcosβsinγ −acosαsinβsinγ+asinβsinγcosα−asinβcosγsinα)/(sin(γ−α))
=(asinα(cosβsinγ −sinβcosγ))/(sin(γ−α))
=(asinαsin(γ−β))/(sin(γ−α))．
故答案为：(asinαsin(γ−β))/(sin(γ−α))