WAEC 2017 · Paper 2 · Q3

(a)
The angle of depression of a point $P$ on the ground from the top $T$ of a building is $23.6^\circ$ . If the distance from $P$ to the foot of the building is $50\text{ m}$ , calculate, correct to the nearest metre, the height of the building.
(b)
In the diagram, $PQT$ and $SRU$ are parallel lines, $QS \parallel TR$ , $|SR| = 6\text{ cm}$ and $|RU| = 10\text{ cm}$ . If the area of $\triangle TRU = 45\text{ cm}^2$ , calculate the area of the trapezium $QTUS$ .

Worked solution (try it first)

Draw the building $FT$ and the point $P$ on the ground 50 m from $F$ .
The angle of depression from $T$ equals the angle of elevation from $P$ (alternate angles), so $\angle FPT = 23.6^\circ$ .
Then $\tan 23.6^\circ = \frac{h}{50}$ , so $h = 50 \times 0.4369 \approx 21.8$ m, which is 22 m to the nearest metre.

The height of $\triangle TRU$ on the base $RU$ is the distance between the parallel lines: $\frac12 \times 10 \times h = 45$ , so $h = 9$ cm.
$QTRS$ has both pairs of opposite sides parallel, so it is a parallelogram and $|QT| = |SR| = 6$ cm.
The parallel sides of the trapezium are $QT = 6$ cm and $SU = 6 + 10 = 16$ cm.
Area $= \frac12(6 + 16) \times 9$
$= 99\text{ cm}^2$ .