WAEC 2013 · Paper 2 · Q12

(a)
An aeroplane flies due north from a town $T$ on the equator at a speed of $950\text{ km}$ per hour for 4 hours to another town $P$ . It then flies eastwards to town $Q$ on longitude $65^\circ$ E. If the longitude of $T$ is $15^\circ$ E, (i) represent this information in a diagram; (ii) calculate the: (I) latitude of $P$ , correct to the nearest degree; (II) distance between $P$ and $Q$ , correct to 4 significant figures. $\left[\text{Take }\pi = \frac{22}{7}, \text{ radius of the earth} = 6400\text{ km}\right]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Draw the earth with the equator and the meridian $15^\circ$ E.
Mark $T$ on the equator, $P$ due north of it on the same meridian, and $Q$ east of $P$ on the same latitude, at $65^\circ$ E.

(I)** Flying north for 4 hours covers $950 \times 4 = 3800$ km along a meridian, a great circle of radius 6400 km.
If the latitude of $P$ is $\theta$ : $\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400 = 3800$ , so $\theta \approx 34.0^\circ$ .
$P$ is at latitude $34^\circ$ N.
(II) $P$ and $Q$ are on latitude $34^\circ$ N, and the difference in longitude is $65^\circ - 15^\circ = 50^\circ$ .
$|PQ| = \frac{50}{360} \times 2 \times \frac{22}{7} \times 6400\cos 34^\circ$
$\approx 4632$ km (4 significant figures).