Question

The angle of elevation of the top 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 of the top is 45°. Calculate the height of the tower.

Answer 1

Let PQ be the tower of height H m and an angle of elevation of the top of tower PQ from point X is 60°. An angle of elevation at 40 m vertical from point X is 45°.

Let PQ = H m and SX = 40m. OX = x, ∠PST = 45°, ∠PXQ = 60°

Here we have to find height of tower.

The corresponding figure is as follows

We use trigonometric ratios.

In ΔPST

`=> tan 45^@ = h/x`

`=> 1 = h/x`

`=> x = h`

Again in ΔPST

`=> tan 456@  = h/x`

`=> 1 = h/x`

`=> x = h`

Again in ΔPXQ

`=> tan 60^@ = (h + 40)/x`

`=> sqrt3 = (h + 40)/x`

`=> h + 40 = sqrt3h`

`=> h(sqrt3 - ) = 40`

`=> h = 40/(sqrt3 - 1)`

`=> h = 54.64`

Therefore `H = 54.64 + 40`

`=> H = 94.64`

Hence the height of tower uis 94.64 m