Question

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. The height of the tower is?

• A. $\frac{h\sin \alpha \sin (\alpha -\beta )}{\sin \beta }$
• B. $\frac{h\sin \alpha \cos (\alpha +\beta )}{\cos \beta }$
• C. $\frac{h\sin \alpha \cos (\alpha -\beta )}{\sin \beta }$
• D. $\frac{h\sin \alpha \sin (\alpha +\beta )}{\cos \beta }$

Answer 1

The correct option is C $\frac{h\sin \alpha \cos (\alpha -\beta )}{\sin \beta }$

Let PQ be the tower and OA be the pole.

In $\Delta$OPQ, we have

$\tan \alpha =\frac{PQ}{OP}=\frac{PQ}{x}$

$\Rightarrow PQ=x\tan \alpha$

$\Rightarrow h+QR=x\tan \alpha$

$\Rightarrow QR=x\tan \alpha -h$ .........(i)

In $\Delta$ARQ, we have

$\tan (\alpha -\beta )=\frac{QR}{x}$

$\Rightarrow \tan (\alpha -\beta )=\frac{x\tan \alpha -h}{x}$ [using Eq. (i)]

$\Rightarrow \tan (\alpha -\beta )=\tan \alpha -\frac{h}{x}$

$\Rightarrow \frac{h}{x}=\tan \alpha -\tan (\alpha -\beta )$

$\Rightarrow x=\frac{h}{\tan \alpha -\tan (\alpha -\beta )}$

$\therefore PQ=x\tan \alpha$

$=\frac{h\tan \alpha }{\tan \alpha -\tan (\alpha -\beta )}=\frac{h\times \frac{\sin \alpha }{\cos \alpha }}{\frac{\sin \alpha \cos (\alpha -\beta )-\sin (\alpha -\beta )}{\sin \alpha \cos (\alpha -\beta )}}=\frac{h\sin \alpha \cos (\alpha -\beta )}{\sin \beta }$.