A boy 1.6 m tall is 20 m away from a tower and observes that the angle of elevation of the top of the tower is 60⁰. Find the height of the tower.

Solution:

Consider AB as the boy and TR as the tower

AB = 1.6 m

Take TR = h

From the point A construct AE parallel to BR

ER = AB = 1.6 m

TE = h – 1.6

AE = BR = 20 m

In right triangle TAE

tan θ = TE/AE

Substituting the values

tan 60⁰ = (h – 1.6)/ 20

So we get

√3 = (h – 1.6)/ 20

h – 1.6 = 20√3

h = 20√3 + 1.6

h = 20 (1.732) + 1.6

By further calculation

h = 34.640 + 1.6

h = 36.24

Hence, the height of the tower is 36.24 m.