WAEC 2008 · Paper 2 · Q12

(a)
In the diagram, $|PR| = |RQ|$ , $|RS| = 10\text{ cm}$ , $\angle RPS = 70^\circ$ and $\angle PQR = 30^\circ$ . Calculate $|PS|$ .
Not to scale.
(b)
Two points $A$ and $B$ lie on the parallel of latitude $60^\circ\text{N}$ . $A$ lies on longitude $20^\circ\text{E}$ and $B$ is $1500\text{ km}$ due east of $A$ . Calculate the: (i) radius of the parallel of latitude on which they lie; (ii) longitude on which point $B$ lies, correct to the nearest degree. $\left[\text{Take }\pi = 3.142\text{, radius of the earth} = 6400\text{ km}\right]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$|PR| = |RQ|$ , so triangle $PQR$ is isosceles and $\angle QPR = \angle PQR = 30^\circ$ .
$\angle PRS$ is an exterior angle of triangle $PQR$ , so it equals the sum of the two opposite interior angles: $30^\circ + 30^\circ = 60^\circ$ .
In triangle $PRS$ : $\angle PSR = 180^\circ - 70^\circ - 60^\circ$
$= 50^\circ$ (needed only as a check).
Sine rule in triangle $PRS$ : $\frac{PS}{\sin 60^\circ} = \frac{RS}{\sin 70^\circ}$ .
$PS = \frac{10 \sin 60^\circ}{\sin 70^\circ}$
$= \frac{10 \times 0.8660}{0.9397}$
$= 9.216$ .
So $|PS| = 9.22\text{ cm}$ .

The radius of the parallel of latitude $60^\circ$ is $R\cos 60^\circ = 6400 \times 0.5$
$= 3200\text{ km}$ .

Let $\theta$ be the difference in longitude.
The arc along the parallel is $\frac{\theta}{360} \times 2\pi r$ , so $\frac{\theta}{360} \times 2 \times 3.142 \times 3200 = 1500$ .
$\theta = \frac{1500 \times 360}{2 \times 3.142 \times 3200}$
$= 26.85^\circ$ .
$B$ is east of $A$ , so add: $20^\circ + 26.85^\circ = 46.85^\circ$ .
$B$ is on longitude $47^\circ\text{E}$ , to the nearest degree.