Question

Two men are on opposite sides of a tower. The measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 metres, find the distance between the two men. [Take √3 = 1.732]

Answer 1

Draw a figure based on given instructions:

Let the positions of the two men are A and B. Let PQ represent the tower.

∠QAP = 30° and ∠QBP = 45°.

Given: QP = 50 m.

We get two right-angled triangles ∆QPB and ∆QPA.

Now,

From right △QPA:

tan 30° = PQ/AP

1/√3 = 50/AP

AP = 50√3 = 86.6

From right ∆QPB:

tan45° = PQ/PB

1 = 50/PB

PB = 50

Now, AB = AP + PB

= 86.6 + 50

= 136.6

Therefore, the distance between the two men is 136.6 meters.