From the top of a tower $50m$ high 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.

The correct option is D. 21.12 m

In figure,

$AB=$ height of the tower.

$CD=$ height of the pole.

Angles of depressions are ${45}^{\circ },{60}^{\circ }$ respectively.

In $△ADE$

$\tan {45}^{\circ }=\frac{AE}{ED}$

$\Rightarrow 1=\frac{AE}{ED}$

Therefore, in $△ADE,AE=DE=x$

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

$\Rightarrow \sqrt{3}=\frac{50m}{x}$

$\Rightarrow x=\frac{50m}{\sqrt{3}}$

$\Rightarrow x=\frac{50m}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow x=\frac{50\sqrt{3}}{3}m$

$\Rightarrow x=\frac{50\times 1.732}{3}m$

$\Rightarrow x=28.87m$

Now, height of the pole $=50m-28.87m=21.12m$