From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Answer: Option C

Let AB be the tower.

Then, ∠APB = 30° and AB = 100m
$$\frac{AB}{AP}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}$$
⇒ AP = AB × √3 = 100 × √3
⇒ AP = 100 × 1.73 = 173m