The angle of elevation of the top of a hill from a point is α. After walking b meters towards the top up a slope inclined at an angle β to the horizontal, the angle of elevation of the top becomes γ. Then the height of the hill is
A
bsinαsin(γ−β)sin(γ−β)
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B
bsinαsin(γ−α)sin(γ−β)
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C
bsin(γ−α)sin(γ−β)
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D
bsin(γ−β)sin(γ−α)
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Solution
The correct option is Absinαsin(γ−β)sin(γ−β) Let BC be the hill and C be the top of hill. ∠CAB=α PA=b From right angle △PAD sinβ=a2b⇒a2=bsinβ.....(1) also from △CQP,a1=PCsinγ....(2) ∠PCA=γ−α ∴bsin(γ−a)=PCsin(α−β)⇒PC=bsin(α−β)sin(γ−a) from (2) a1=bsin(α−β)sinγsin(γ−a)...(3) Height of hill =bsin(γ−a)[sinβsin(γ−a)+sin(α−β)sin(γ−a)sinγ] =bsinαsin(γ−a)[sinγcosβ−sinβcosγ]=bsinαsin(γ−β)sin(γ−β)