WAEC 2018 · Paper 2 · Q12

(a)
A man starts from a point $X$ and walks $285\text{ m}$ to $Y$ on a bearing of $078^\circ$ . He then walks due south to a point $Z$ which is $307\text{ m}$ from $X$ . (i) Illustrate the information on a diagram. (ii) Find, correct to the nearest whole number, the: (I) bearing of $X$ from $Z$ ; (II) distance between $Y$ and $Z$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

The north line at $X$ and the south line at $Y$ are parallel, so $\angle XYZ = 78^\circ$ (alternate angles).
Sine rule for the angle at $Z$ : $\sin\angle YZX = \frac{285\sin 78^\circ}{307}$
$= \frac{278.77}{307}$
$\approx 0.9081$ , so $\angle YZX \approx 65.2^\circ$ (acute, because it is opposite the shorter of the two known sides).
(I) At $Z$ , $Y$ is due north, and $X$ is $65.2^\circ$ round anticlockwise (towards the west).
Bearing of $X$ from $Z$ $= 360^\circ - 65.2^\circ$
$= 294.8^\circ$
$\approx 295^\circ$ .
(II) The third angle is $\angle YXZ = 180^\circ - 78^\circ - 65.2^\circ$
$= 36.8^\circ$ .
Sine rule: $|YZ| = \frac{307\sin 36.8^\circ}{\sin 78^\circ}$
$\approx \frac{183.9}{0.9781}$
$\approx 188$ m.