Question

A vertical Tower is 20 m high. A man standing at some distance from the tower knows that the cosine of angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of 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]

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

"

Answer 2

Answer:

12.5 m

Step-by-step explanation:

height of tower = AB = 20 m

angle of elevation = θ

distance of man from foot of tower = BC

cos θ = 0.53

=> cos θ = cos 58° (using natural cosine table)

=> θ = 58°

tan θ = AB/BC

=> tan 58° = 20/BC

=> 1.6 = 20/BC

=> BC = 20/1.6 = 12.5 m