Question

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of tower PQ. [Take `sqrt(3) = 1.73`.]

Answer 1

We have

XY = 40 m, ∠PXQ = 60° and ∠MYQ = 45°

Let PQ = h

Also, MP = XY = 40 m, MQ = PQ – MP = h – 40

In ΔMYQ,

` tan 45^circ = (MQ)/(MY)`

`⇒ 1 = (h - 40)/(MY)`

⇒ MY = h – 40

⇒ PX = MY = h – 40   ...(1)

Now, in ΔMXQ,

`tan 60^circ = (PQ)/(PX)`

`⇒ sqrt(3) = h/(h-40)`   ...[From (i)]

`⇒ h sqrt(3) - 40sqrt(3) = h `

`⇒ h sqrt(3) - h = 40sqrt(3)`

`⇒ h(sqrt(3) - 1) = 40sqrt(3)`

`⇒ h = (40sqrt(3))/((sqrt(3) - 1))`

`⇒ h = (40sqrt(3))/((sqrt (3) - 1)) xx ((sqrt(3) + 1))/((sqrt(3) + 1))`

`⇒ h = (40sqrt(3)(sqrt(3) + 1))/((3 - 1))`

`⇒ h = (40sqrt(3)(sqrt(3) + 1))/2`

`⇒ h = 20sqrt(3)(sqrt(3) + 1)`

`⇒ h = 60 + 20sqrt(3)`

`⇒ h = 60 + 20 xx 1.73`

`⇒ h = 60 + 34.6`

∴ h = 94.6 m

So, the height of the tower PQ is 94.6 m.