Question

A man 1.8 metres tall, looking down from the top of a telephone tower sees the top of a building 10 metres high at an angle of depression 40° and the foot of the building at an angle of depression 60°. What is the height of the tower? How far is it away from the building?

Answer 1

Let AF be the height of the man.

Let FC be the height of the telephone tower on which the man is standing.

Let ED be the height of the building.

Let CD be the distance between the telephone tower and the building.

Let ∠XAE be the angle of depression made by the man’s eye while seeing the top of the building.

Let ∠XAD be the angle of depression made by the man’s eye while seeing the top of the building.

The figure for the given situation with dimensions can be made as follows:

In ΔAEB:

Now, in ΔACD:

From equation (1) and equation (2), we get:

∴ Height of the telephone tower = (10 + x) m

= (10 + 7.6) m

= 17.6 m

Putting the value of x in equation (1), we get:

Thus, the height of the telephone tower and the distance of the telephone tower from the building are 17.6 m and 11.2 m respectively.