A person observes other angle of elevation of the peak of a hill from a station to be `alpha` . He walks `cm e t e r s` along a slope inclined at the angle `beta` and finds the angle of elevation of the peak of the hill be `gammadot` Show that the height of the peak above the ground is `(csinalphasin(gamma-beta))/((singamma-alpha))` .

Options

A. `(b sin alpha sin (gamma-beta))/(sin (gamma-alpha))`
B. `(b sin alpha sin (gamma -alpha))/(sin (gamma-beta))`
C. `(b sin (gamma-beta))/(sin (gamma - alpha ))`
D. `(sin (gamma-beta))/(b sin alpha sin (gamma-alpha ))`

Answer

A

Explanation

Let P be the top of hill of height h and A be a point on the ground such that the angle of elevation of top from A is `alpha`. After walking b metres towards the top up of a slope inclined at an angle `beta` to the horizon i.e. at point B, the angle of elevation of P is `gamma`.

In `triangleBCP`, we have
`tan gamma =(PC)/(BC) rArr tan gamma =(h -b sin beta)/(OA-b cos beta)` ....(i)
In `triangleAOP`, we have
`tan alpha =h/(OA) rArr OA =h cot alpha`
Substituting the value of OA is (i), we get
`tan gamma =(h -b sin beta)/(h cot alpha - b cos beta)`
`rArr h (cot alpha tan gamma -1) =b (tan gamma cos beta-sin beta)`
`rArr h(sin (gamma-alpha ))/(sin alpha cos gamma)=(b sin (gamma-beta))/(cos gamma)`
`rArr h=(b sin (gamma -beta)sin alpha ) /(sin (gamma -alpha ))`