Question

The angle of elevation of the top of a tower is observed to be 60°. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45°. Find:

1. the height of the tower,
2. its horizontal distance from the points of observation.

Answer 1

In ΔABC

`tan 60^circ = (h + 30)/x`

`xsqrt(3) = h + 30`  ...(1)

In ΔADE

`tan 45^circ = h/x`

`1 = h/x`

h = x    ...(2)

From equation (1)

`xsqrt(3) = x + 30`

`x(sqrt(3) - 1) = 30`

`x = 30/(sqrt(3) - 1) xx (sqrt(3) + 1)/(sqrt(3) + 1)`

`x = (30(sqrt(3) + 1))/(3 - 1)`

= `(30(1.732 + 1))/2`

= 15 × 2.732

= 40.980 m

From equation (2)

h = 40.98 m

Height of tower =  h + 30

= 40.98 + 30

= 70.98 m

= 71 m (approx.)

Horizontal distance = x = 40.98 m = 41 m.