Question

A vertical Tower is 20 m high .A man standing at some distance from the tower knows 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?

Answer 1

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Let AB be the tower and C be the position of the man.
Let ∠ACB = θNow, cos θ = 0.5⇒cos θ = cos 60°⇒θ = 60°Now,tan 60° = ABBC⇒3√ = 24BC⇒BC = 243√⇒BC =243√ = 243√3 = 83√ = 8 × 1.732 = 13.856 m So, the man is standing at a distance of 13.856 m from the foot of the tower.

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Answer 2

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.