From the top of a 50 m high tower, the angles of depression of the top and bottom

Let H be the height of pole, makes an angle of depression from top of tower to top and bottom of poles are 45° and 60° respectively.

Let $A B=H, C E=h, A D=x$ and $D E=50 \mathrm{~m} . \angle C B E=45^{\circ}$ and $\angle D A E=60^{\circ}$.

Here we have to find height of pole.

The corresponding figure is as follows

$\ln \triangle A D E$

$\Rightarrow \quad \tan A=\frac{D E}{A D}$

$\Rightarrow \quad \tan 60^{\circ}=\frac{50}{x}$

$\Rightarrow \quad \sqrt{3}=\frac{3000}{x}$

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

Again in $\triangle B C E$

$\Rightarrow \quad \tan B=\frac{C E}{B C}$

$\Rightarrow \quad \tan 45^{\circ}=\frac{h}{x}$

$\Rightarrow \quad 1=\frac{h}{x}$

$\Rightarrow \quad h=\frac{50}{\sqrt{3}}$

$\Rightarrow \quad h=28.87$

Therefore $H=50-h$

$\Rightarrow \quad H=50-28.87$

$\Rightarrow \quad H=21.13$

Hence height of pole is $21.13 \mathrm{~m}$.