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.