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:

A. 149 m

B. 156 m

C. 173 m

D. 200 m

Answer: Option C

Solution(By Examveda Team)

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