Question

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer 1

Let AB be the tower and the angle of elevation from point C (on ground) is 30°. 

In ∆ABC, 

\(\frac{AB}{ BC} = tan 30°\)

\(\frac{AB}{ 30 }= \frac{1}{ \sqrt3}\)

\(AB = \frac{30}{ \sqrt3} = 10\sqrt3\,m\)

Therefore, the height of the tower is \(10\sqrt3\,m\).