WAEC 2023 · Paper 2 · Q5

Elevation, depression & bearings

A boy stands at a point $M$ on the same horizontal level as the foot, $T$ , of a vertical building. He observes an object on the top, $P$ , of the building at an angle of elevation of $66^\circ$ . He moves directly backwards to a new point $C$ and observes the same object at an angle of elevation of $53^\circ$ . If $|MT| = 50\text{ m}$ :

(a)
Illustrate the information in a diagram.
Model answer
A clear sketch is enough (it need not be to scale), but it must show every given fact: draw the building $PT$ standing vertically on level ground, with a right angle at $T$ . $M$ is 50 m from $T$ , with the angle of elevation of $P$ equal to $66^\circ$ . $C$ is further back on the same line, with elevation $53^\circ$ (smaller, because it is further away). Mark the height $h$ and the unknown $|MC|$ .
(b)(i)
Calculate, correct to one decimal place, the height of the building;
(b)(ii)
$|MC|$ .

Worked solution (try it first)

(a)

Draw the building $TP$ upright, with $M$ 50 m from its foot and $C$ further back on the same line.
The angles of elevation of $P$ are $66^\circ$ from $M$ and $53^\circ$ from $C$ .

(b)(i)

In triangle $MTP$ : $|PT| = 50\tan 66^\circ$
$\approx 50 \times 2.2460$
$\approx 112.3$ m.

(ii)

In triangle $CTP$ : $\tan 53^\circ = \frac{112.3}{|CT|}$ , so $|CT| = \frac{112.3}{1.3270} \approx 84.63$ m.
So $|MC| = 84.63 - 50 \approx 34.6$ m.

Report a problem with this question