WAEC 2023 · Paper 2 · Q10

In a town, Chief $X$ resides $60\text{ m}$ away on a bearing of $057^\circ$ from the palace $P$ , while Chief $Y$ resides on a bearing of $150^\circ$ from the same palace $P$ . The residences of $X$ and $Y$ are $180\text{ m}$ apart.

(a)
Illustrate the information in a diagram.
Model answer
A clear sketch is enough (it need not be to scale), but it must show every given fact: draw a north line at $P$ , with $X$ on bearing $057^\circ$ (60 m) and $Y$ on bearing $150^\circ$ , and $XY = 180$ m. The angle $XPY = 150^\circ - 57^\circ = 93^\circ$ . Add a north line at $Y$ for the bearing of $X$ from $Y$ in (b). $PY$ is unknown (it works out to about 167 m).
(b)
Find, correct to three significant figures, the: (i) bearing of $X$ from $Y$ ; (ii) distance between $P$ and $Y$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Draw north at the palace $P$ .
Chief $X$ is 60 m on $057^\circ$ and Chief $Y$ on $150^\circ$ .
Join $X$ and $Y$ (180 m).
The angle at $P$ is $\angle XPY = 150^\circ - 57^\circ$
$= 93^\circ$ .

(b)(i)

Sine rule for the angle at $Y$ : $\sin\angle PYX = \frac{60\sin 93^\circ}{180}$
$\approx 0.3329$ , so $\angle PYX \approx 19.4^\circ$ .
At $Y$ , the direction back to $P$ is $150^\circ + 180^\circ = 330^\circ$ , and $X$ is $19.4^\circ$ further round clockwise.
Bearing of $X$ from $Y$ $= 330^\circ + 19.4^\circ$
$\approx 349^\circ$ .

(ii)

The third angle is $\angle PXY = 180^\circ - 93^\circ - 19.4^\circ$
$= 67.6^\circ$ .
Sine rule: $|PY| = \frac{180\sin 67.6^\circ}{\sin 93^\circ}$
$\approx 167$ m.

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