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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?

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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? Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

Given,

cos θ = 0.53

From table of cosines, we have,

θ = 58°

Let AB be the height of tower = 20 m and distance of man form foot of tower be CB = x

In right angled triangle ABC,

⇒tan⁡θ=PerpendicularBase⇒tan⁡θ=(ABCB)⇒tan⁡58°=20x⇒1.6=20x⇒x=201.6⇒x=12.5 m.\Rightarrow \tan \theta = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan \theta = \Big(\dfrac{AB}{CB}\Big) \\[1em] \Rightarrow \tan 58° = \dfrac{20}{x} \\[1em] \Rightarrow 1.6 = \dfrac{20}{x} \\[1em] \Rightarrow x = \dfrac{20}{1.6} \\[1em] \Rightarrow x = 12.5 \text{ m}.

Hence, distance of man form foot of tower is 12.5 m.

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