Question

Let a vertical tower $AB$ of height $2h$ stands on a horizontal ground. Let from a point $P$ on the ground a man can see upto height $h$ of the tower with an angle of elevation $2\alpha$. When from $P$, he moves a distance $d$ in the direction of $\overrightarrow{AP}$, he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $d=\sqrt{7}h$, then $\tan \alpha$ is equal to

• A. $\sqrt{5}-2$
• B. $\sqrt{3}-1$
• C. $\sqrt{7}-2$
• D. $\sqrt{7}-\sqrt{3}$

Answer 1

Answer: (C)

$\tan 2\alpha =\frac{h}{x}$
and $\tan \alpha =\frac{2h}{x+\sqrt{7}h}$
$\tan \alpha =\frac{2h}{h\cot 2\alpha +\sqrt{7}h}$
$\tan \alpha =\frac{2}{\frac{\left(1-{\tan }^2\alpha \right)}{2\tan \alpha }+\sqrt{7}}$
Put $\tan \alpha =t\&$ simplify
$\Rightarrow \tan \alpha =\sqrt{7}-2$