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, 40m vertically above X, the angle of elevation is 45° . Find the height of tower PQ.

Answer 1

We have
XY = 40m,∠PXQ = 60° and ∠MYQ = 45°
Let PQ = h
Also, MP = XY = 40m, MQ =  PQ  - MP = h - 40
In ΔMYQ,

` tan 45° = (MQ)/(MY)`

`⇒1 = (h-40)/(MY)`

⇒ MY = H - 40

⇒ PX = MY = h - 40                       ................(1)

Now , in ΔMXQ,

`tan 60° = (PQ)/(PX)`

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

`⇒ h sqrt(3) - 40 sqrt(3) = h `

`⇒ h sqrt(3) -h = 40 sqrt(3)`

`⇒ h  (sqrt(3)-1) = 40 sqrt(3)`

`⇒ h = (40sqrt(3))/((sqrt(3)-1))`

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

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

`⇒h = (40 sqrt(3) ( sqrt(3) +1))/2`

`⇒h= 20sqrt(3) (sqrt(3)+1)`

`⇒ h= 60+20 sqrt(3)`

`⇒ h= 60+20xx 1.73`

`⇒h = 60+ 34.6`

∴ h = 94.6m

So, the height of the tower PQ is 94. 6 m.