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 the tower PQ and the distance XQ. 

Answer 1

In the figure, PQ is the tower.

In ΔPQX,

∴ `"h"/"x" = tan60^circ = sqrt(3)`

⇒ h = `sqrt(3)`x   ...(1)

In ΔQRY,

`("h" - 40)/"x" = tan 45^circ = 1`

⇒ h = 40 + x   ...(2)

From (1) and (2),

`sqrt(3)`x = 40 + x

⇒ `(sqrt(3) - 1)"x" = 40`

⇒ `"x" = 40/(sqrt(3) - 1) = (40(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1)) = 40/2(sqrt(3) + 1) = 20(sqrt(3) + 1)`

∴ `"h" = 40 + 20(sqrt(3) + 1) = 20sqrt(3) + 60 = 20(sqrt(3) + 3) = 20 xx 4.732 = 94.64`

Thus , the height of the tower PQ is 94.64 m.

Again, in ΔPQX,

∴ `"h"/"XQ" = sin60^circ = 1/sqrt(2)`

⇒ `"XQ" = sqrt(2)"h" = 1.414 xx 94.64 = 109.3`m

Thus , the distance XQ is 109.3m.