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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 ?

We have, cos θ = 0.53, let distance of the man from the foot of the tower be x. AB = 20m, BC = x, then AC = √[x2 + (20)2] = √[x2 + 400] cos θ = BC/AC = x/√[x2 + 400] Or, 0.53 = x/√[x2 + 400] Or, (0.53)2 = x2/[x2 + 400] [Squaring both sides] Or, 0.2809 = x2/[x2 + 400] Or, x2 = 0.2809x2 + 112.36 Or, x2 – 0.2809x2 = 112.36 Or, 0.7191x2 = 112.36 Or, x2 = 112.36/0.7191 = 12.5 m [Ans.]

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1.distance = 72m let height = h cos α = 0.53 sin α = √1-0.53² = 0.848 tan α = sinα/cosα = 1.6 tanα = h/72 ⇒h = 72×1.6=115.2m

2.h = 20 cos α = 0.53 distance from tower = d tan α = 20/d ⇒d = 20/tan α = 20/1.6 = 12.5m