Question

Two men are on opposite side of tower. They 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 `sqrt(3) = 1.732`.]

Answer 1

Let CD be the tower and A and B be the positions of the two men standing on the opposite sides.
Thus, we have:
∠DAC = 30°,  ∠DBC = 45° and CD = 50 m
Let AB =  xmand BC =  ymsuch that AC = (x - y)m.

In the right ΔDBC,we have:

`(CD)/(BC) = tan 45° = 1`

`⇒ 50/y = 1`

⇒  y = 50 m

In the right  ΔACD,we have:

`(CD)/(AC) = tan 30° = 1/ sqrt(3) `

`⇒  50/ ((x-y)) = 1/ sqrt(3)`

`x -y = 50 sqrt(3)`

On putting y  = 50in the above equation, we get:

`x -50= 50sqrt(3)`

`⇒  x = 50 + 50 sqrt(3) = 50 ( sqrt(3) +1) = 136.6m `

∴Distance between the two men = AB = x = 136.6m