WAEC 2020 · Paper 2 · Q3

(a)
An aeroplane flies $500\text{ km}$ from town $P$ on a bearing of $053^\circ$ to town $Q$ . It then flies $700\text{ km}$ to town $R$ on a bearing of $165^\circ$ . (i) Illustrate the information with a diagram. (ii) Calculate, correct to three significant figures, the distance between $P$ and $R$ .
(b)
In the diagram, $\overline{XY}$ is the diameter of the circle $WXYZ$ , $\overline{XY} \parallel \overline{WZ}$ and $\angle ZXY = 28^\circ$ . Find $\angle XWZ$ .

Worked solution (try it first)

At $Q$ , the direction back to $P$ is $053^\circ + 180^\circ = 233^\circ$ and the direction to $R$ is $165^\circ$ , so $\angle PQR = 233^\circ - 165^\circ$
$= 68^\circ$ .
Cosine rule: $|PR|^2 = 500^2 + 700^2 - 2(500)(700)\cos 68^\circ$
$= 740\,000 - 700\,000 \times 0.3746$
$\approx 477\,780$ .
So $|PR| \approx 691$ km (3 significant figures).

$XY$ is a diameter, so $\angle XZY = 90^\circ$ (angle in a semicircle).
In triangle $XYZ$ : $\angle XYZ = 180^\circ - 90^\circ - 28^\circ$
$= 62^\circ$ .
$WXYZ$ is a cyclic quadrilateral, so $\angle XWZ = 180^\circ - \angle XYZ$
$= 118^\circ$ (opposite angles).