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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 ?

Heights & Distances

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Answer

Let angle of elevation be θ.

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 ? Heights and Distances, Concise Mathematics Solutions ICSE Class 10.

According to question,

⇒ cos θ = 0.53

⇒ cos θ = cos 58°

⇒ θ = 58°.

⇒ cos2 θ = 0.2809

⇒ 1 - sin2 θ = 0.2809

⇒ sin2 θ = 1 - 0.2809

⇒ sin2 θ = 0.7191

⇒ sin θ = 0.7191\sqrt{0.7191}

⇒ sin θ = 0.848

⇒ tan θ = sin θcos θ=0.8480.53\dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{0.848}{0.53} = 1.6

∴PerpendicularBase=1.6⇒ABBC=1.6⇒20BC=1.6⇒BC=201.6=12.5 m.\therefore \dfrac{\text{Perpendicular}}{\text{Base}} = 1.6 \\[1em] \Rightarrow \dfrac{AB}{BC} = 1.6 \\[1em] \Rightarrow \dfrac{20}{BC} = 1.6 \\[1em] \Rightarrow BC = \dfrac{20}{1.6} = 12.5 \text{ m}.

The man is standing at a distance of 12.5 meters.

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