Question

A person observed the angle of elevation of the top of a tower as ${30}^o$. He walked $50m$ towards the foot of the angle of elevation of the top of the tower as ${60}^o$. Find the height of the tower.

Answer 1

Let $AB$ be the tower and distance between the points $C$ and $D$ be $50m.$

In $△ABC$

$\tan {60}^o=\frac{AB}{BC}$

$\Rightarrow \sqrt{3}=\frac{AB}{BC}$

$\Rightarrow BC=\frac{AB}{\sqrt{3}}...........\left(1\right)$

In $△ABD$

$\tan {30}^o=\frac{AB}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{BD}$

$\Rightarrow BD=AB\sqrt{3}................\left(2\right)$

And,

$CD=BD-BC$

$\Rightarrow 50=AB\sqrt{3}-\frac{AB}{\sqrt{3}}$

$\Rightarrow 50=AB(\sqrt{3}-\frac{1}{\sqrt{3}})$

$\Rightarrow 50=AB\times \frac{2}{\sqrt{3}}$

$\Rightarrow AB=25\sqrt{3}m$