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Question
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?
Solution
cos θ = 0.53
A man is standing at C
Let CB = x and height of tower AB = 20 m
In right ΔABC; we have
`tan theta = (AB)/(CB) = 20/x` ...(i)
cos θ = 0.53
From the table of cosines, we find that
θ = 58° (approx.)
Now substituting the value of θ in (i)
`tan 58^circ = 20/x`
`\implies 1.6 = 20/x`
∴ `x = 20/1.6`
= `(20 xx 10)/16`
= `25/2`
= 12.5 m
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