WAEC 2025 · Paper 2 · Q10✱✱

(a)
A man travels from $X$ on a bearing of $060^\circ$ to $Y$ , $20\text{ km}$ away. From $Y$ he travels to $Z$ on a bearing of $195^\circ$ . If $Z$ is directly east of $X$ , find, correct to three significant figures: (i) $|YZ|$ ; (ii) $|ZX|$ .
Separate values with commas, e.g. 3, −2
(b)
An aircraft flies due south from latitude $36^\circ\text{N}$ to latitude $36^\circ\text{S}$ along the same longitude. (i) Find the distance travelled, correct to three significant figures. (ii) If its speed is $800\text{ km/h}$ , find the time taken, to the nearest hour. $\left[\pi = \frac{22}{7}, R = 6400\text{ km}\right]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Break each leg into north and east parts.
$X$ to $Y$ on $060^\circ$ , 20 km: north $20\cos 60^\circ = 10$ km, east $20\sin 60^\circ \approx 17.32$ km.
$Y$ to $Z$ on $195^\circ$ is $15^\circ$ west of due south.
$Z$ is due east of $X$ , so $Z$ is level with $X$ : the leg $YZ$ must come 10 km back south.

Both places are on the same line of longitude, so the aircraft flies along a great circle through an angle of $36^\circ + 36^\circ = 72^\circ$ .
Distance $= \frac{72}{360} \times 2 \times \frac{22}{7} \times 6400$
$\approx 8045.7$ km, which is 8050 km to three significant figures.

Time $= \frac{8045.7}{800} \approx 10.06$ , which is 10 hours to the nearest hour.