Question

From the top of a $50$ m high tower, the angles of depression of the top and bottom of a pole are observed to be ${45}^{\circ }$ and ${60}^{\circ }$ respectively. Find the height of the pole.

Answer 1

Here $CD$ is the tower and $AB$ is the pole.

In $△CDB$, $\tan {60}^o=\frac{CD}{BC}$

$\sqrt{3}=\frac{50}{BC}$

$BC=\frac{50}{\sqrt{3}}m$

$AE=BC$

$\therefore$ $AE=\frac{50}{\sqrt{3}}m$

In $△ADE,\tan {45}^o=\frac{DE}{AE}$

$\Rightarrow$ $1=\frac{DE}{\frac{50}{\sqrt{3}}}$

$\therefore$ $DE=\frac{50}{\sqrt{3}}m$

$\therefore$ Height of the pole is $\frac{50}{\sqrt{3}}m$