A tower stands vertically on the ground.

Solution:

Given a tower at the ground such that at a distance of 20 m away from foot of tower,

the top of tower makes an angle of  with the ground.

We have to find the height of the tower.

Let the tower be AB

Height of the tower = h

Distance of the tower from point C where top of the tower makes an angle of  = 20 m

Since B is the foot of the tower, CB = 20 m

$\angle A C B=60^{\circ}$

In $\triangle A B C$

$\frac{A B}{B C}=\tan 60^{\circ}$

$\frac{h}{20}=\sqrt{3}$

$h=20 \sqrt{3}$

Height of the tower $=20 \sqrt{3}$