Question

A pole 6m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is 60° and the angle of depression of the point P from the top of the tower is 45°. Find the height of the tower and the distance of point P from the foot of the tower. (Use `sqrt3` = 1.73)

Answer 1

Consider QR as the tower, PQ as the pole on it. Given, ∠PAR = 60° and ∠QAR = 45°

Height of pole = BC = 6 m

let height of tower = h m

Distance of point P = x m

In ΔBAP

`(BA)/(AP) = tan45°`

`h/x = 1

⇒ h = x

In ΔCAP

`(CA)/(AP) = tan60°`

`(6+h)/x = sqrt3`

6 + x = √3x

6 = √3x − x

`x = 6/(sqrt3 - 1)xx((sqrt3+1)/(sqrt3+1))`

`x = (6 (sqrt3 + 1))/((sqrt3)^2-(1)^2) = (6(1.73+1))/(3-1)`

`= (cancel6 (2.73))/cancel2 = 8.19 m`

h = x = 8.19 m

Height of tower = 8.19 m

Distance of point p = 8.19 m