Question

The shadow of a tower standing on a level plane is found to be 50 m longer when sun's elevation is ${60}^0$ than when it is ${30}^{\circ }$. The height of the tower is $25\sqrt{p}$, then $p$ is

Answer 1

Let $h=$height of the tower, $l=$length of the shadow
From data, $\tan {60}^{\circ }=\sqrt{3}=\frac{l}{h}$
and $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}=\frac{l-50}{h}$
Solving two equations,
$\sqrt{3}(\sqrt{3}h-50)=h$
$50\sqrt{3}=2h$
$h=25\sqrt{3}$