A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?
A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?

Solution:

Consider AB as the tower

Take a man C stands at a distance x m from the foot of the tower

cos θ = 0.53

We know that

Height of the tower AB = 20 m

cos θ = 0.53

So we get

θ = 580

Let us take

tan θ = AB/CB

Substituting the values

tan 580 = 20/x

So we get

1.6003 = 20/x

By cross multiplication

x = 20/1.6003

x = 12.49 = 12.5 m

Hence, the height of the tower is 12.5 m.