Question

From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower.

Answer 1

Let OP be the tower and points A and B be the positions of the cars.

We have,

AB = 100 m, ∠OAP = 60° and ∠OBP = 45°

Let OP = h

In ΔAOP,

`tan 60^circ = (OP)/(OA)`

`⇒ sqrt(3) = h/(OA)`

`⇒ OA = h/sqrt(3)`

Also, in ΔBOP,

`tan 45^circ = (OP)/(OB)`

`⇒ 1 = h/(OB)`

⇒ OB = h

Now, OB – OA = 100

`⇒ h - h/sqrt(3) = 100`

` ⇒ (h sqrt(3) - h)/sqrt(3) = 100`

`⇒(h(sqrt(3) - 1))/sqrt(3) = 100`

`h = (100sqrt(3))/((sqrt(3) - 1)) xx ((sqrt(3) + 1))/((sqrt(3) + 1))`

`h = (100 sqrt(3)(sqrt(3) + 1))/((3 - 1))`

`⇒ h = (100(3 + sqrt(3)))/2`

⇒ h = 50(3 + 1.732)

⇒ h = 50(4.732)

∴ h = 236.6 m 

So, the height of the tower is 236.6 m.