Question

A man standing on a horizontal plane, observes the angle of elevation of the top of a tower to be $\alpha$ . After walking a distance equal to double the height of the tower, the angle of elevation becomes $2\alpha$ , then $\alpha$ is:-

• A. $\frac{\pi }{18}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (B)

Let $PQ$ be the tower of height $h$ .
Given, $AB=2h$ , $\angle PAQ=\alpha$ and $\angle PBQ=2\alpha$
$\therefore \angle APB=\angle PBQ-\angle PAQ=2\alpha -\alpha$

In $\Delta ABP,\angle BAP=\angle APB=\alpha$
$\therefore PB=AB=2h$
In $\Delta PQB$ ,
$sin2\alpha =\frac{PQ}{PB}=\frac{h}{2h}=\frac{1}{2}$
$\Rightarrow sin2\alpha =sin\frac{\pi }{6}$
$\therefore 2\alpha =\frac{\pi }{6}$
$\Rightarrow \alpha =\frac{\pi }{12}$