Question

A 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 the man 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:

Step-by-step explanation:

In the figure, AB is the vertical tower of height 20 m. C denotes the

position of the man.

Let ACB = theta denotes the angle of elevation. Then cos = 0.5 = 1/2

i.e. θ = 60°

Therefore, distance of the man from the foot of the tower = BC

tan60 = AB/BC

√3 = 20/BC

BC = 20/√3 cm