Question

From the top of a vertical tower, the angles 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 =100m, ∠OAP = 60° and ∠OBP = 45°
Let OP = h
In ΔAOP,

` tan 60° = (OP)/(OA)`

`⇒ sqrt(3) = h/(OA)`

`⇒OA = h/sqrt(3)`

Also, in Δ BOP ,

` tan 45° = (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) -h))/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 =236.6m 

So, the height of the tower is 236.6 m.
Disclaimer. The answer given in the textbook is incorrect. The same has been rectified above.