From the bottom of a pole of height h, the angle of elevation of the top of a tower is `alpha`. The pole subtends an angle `beta` at the top of the tower. find the height of the tower.

Options

A. `(h sin alpha sin (alpha - beta))/(sin beta)`
B. `(h sin alpha cos (alpha + beta))/(cos beta)`
C. `(h sin alpha cos (alpha - beta))/(sin beta)`
D. `(h sin alpha sin (alpha +beta))/(cos beta)`

Answer

C

Explanation

Let PQ be the tower and OA be the pole .
In `triangleOPQ`, we have,
tan `alpha =(PQ)/(OP)=(PQ)/x`
`rArr PQ = x tan alpha `
`rArr h +QR = x tan alpha `
`rArr QR=x tan alpha -h`…(i)

In `angleAPQ`, we have
`tan (alpha - beta) =(QR)/x`
`rArr tan (alpha - beta)=(x tan alpha - h)/x ` [ Using (i)]
`rArr tan (alpha-beta)=tan alpha -h/x `
`rArr h/x = tan alpha - tan (alpha -beta) rArr x=h/(tan alpha -tan (alpha -beta))`
`therefore PQ=x tan alpha =(h tan alpha )/(tan alpha -tan (alpha -beta))=(h sin alpha cos (alpha -beta))/(sin beta)`