WAEC 2014 · Paper 2 · Q12

(a)
Given that $5\cos(x + 8.5)^\circ - 1 = 0$ , $0^\circ \le x \le 90^\circ$ , calculate, correct to the nearest degree, the value of $x$ .
(b)
The bearing of $Q$ from $P$ is $150^\circ$ and the bearing of $P$ from $R$ is $015^\circ$ . If $Q$ and $R$ are $24\text{ km}$ and $32\text{ km}$ respectively from $P$ : (i) represent this information in a diagram; (ii) calculate the distance between $Q$ and $R$ , correct to two decimal places; (iii) find the bearing of $R$ from $Q$ , correct to the nearest degree.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

At $P$ , draw north: $Q$ is on $150^\circ$ , 24 km away.
The bearing of $P$ from $R$ is $015^\circ$ , so the bearing of $R$ from $P$ is the back bearing, $015^\circ + 180^\circ = 195^\circ$ : $R$ is 32 km from $P$ on $195^\circ$ .
The angle between the two lines at $P$ is $\angle QPR = 195^\circ - 150^\circ$
$= 45^\circ$ .

Cosine rule: $|QR|^2 = 32^2 + 24^2 - 2(32)(24)\cos 45^\circ$
$= 1600 - 1086.1$
$\approx 513.9$ .
So $|QR| \approx 22.67\text{ km}$ .

Use the cosine rule for the angle at $Q$ (it avoids any doubt about an obtuse angle): $\cos\angle PQR = \frac{24^2 + 22.67^2 - 32^2}{2 \times 24 \times 22.67}$
$\approx \frac{65.9}{1088.2}$
$\approx 0.0606$ , so $\angle PQR \approx 86.5^\circ$ .
At $Q$ , the direction back to $P$ is $150^\circ + 180^\circ = 330^\circ$ , and $R$ is $86.5^\circ$ further round anticlockwise.
Bearing of $R$ from $Q$ $= 330^\circ - 86.5^\circ$
$= 243.5^\circ$
$\approx 243^\circ$ .