Question

A vertical tower is 20 m high. A man standing at some distance from the tower know that the cosine of the angel of elevation of the top of the tower is 0.53.how far he is standing from foot the tower.

Answer 1

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]

=x² – 0.2809x² = 112.36

Or,

0.7191x² = 112.36

x² = 112.36/0.7191

= 12.5 m.

Answer 2

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