Question

A tower stands vertically on the ground. From a point on the ground 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?

Answer 1

Let  AB be the tower of height, hm and C be the point on the ground, makes an angle of elevation 60° with the top of tower AB.

In a triangle ABC, given that BC = 20 m and ∠C = 60°

Now we have to find the height of tower AB, so we use trigonometrical ratios.

In the triangle ABC

`=> tan C = "Opposite side (AB)"/("Adjacent side (BC)")`

i.e `tan 60^@ = (AB)/20`

`=> AB = 20sqrt3`

`= 20 sqrt3`

∴ Height of tower H = `20sqrt3m`