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Question
Vertical tower is 20m 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
Given, cos θ = 0.53
Let the man is standing at a distance of 'x' m from the foot of the tower.
cos θ = `"BC"/"AC" = x/(sqrt(x^2 + 400))`
0.53 = `x/(sqrt(x^2 + 400))`
⇒ `(0.53)^2 = x^2/(sqrt(x^2 + 400))`
⇒ 0.2809 x2 + 112.36 = x2
⇒ x2 - 0.2809 x2 = 112.36
⇒ x2 = `(112.36)/(0.7191)`
⇒ x2 = 156.25
⇒ x = 12.5 metres
∴ The man is standing from the foot of the tower be 12.5 meter.
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