JAMB 1986 · UME · Q34

Elevation, depression & bearings

The angle of elevation of the top of a vertical tower 50 m high from a point $X$ on the ground is $30^\circ$ . From a point $Y$ on the opposite side of the tower, the angle of elevation of the top is $60^\circ$ . Find the distance between $X$ and $Y$ .

Worked solution (try it first)

$X$ and $Y$ are on opposite sides, so $XY$ is the sum of their distances from the foot of the tower.
From $X$ : $\tan30^\circ = \frac{50}{d}$ , so $d = \frac{50}{\tan30^\circ}$
$= 50\sqrt3$
$\approx 86.60$ m.
From $Y$ : $d = \frac{50}{\tan60^\circ}$
$= \frac{50}{\sqrt3}$
$\approx 28.87$ m.
So $XY = 86.60 + 28.87 = 115.47$ m, option D.

Report a problem with this question