WAEC 2023 · Paper 2 · Q10✱✱

In a large library, Adamu is seated $15\text{ m}$ away from Uju on a bearing of $305^\circ$ . Bola is $25\text{ m}$ away from Adamu on a bearing of $065^\circ$ .

(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: start at Uju ( $U$ ) with a north line. Adamu ( $A$ ) is 15 m away on bearing $305^\circ$ ( $55^\circ$ west of north). At $A$ draw another north line; Bola ( $B$ ) is 25 m away on bearing $065^\circ$ . The angle $UAB = 60^\circ$ (the back-bearing to $U$ is $125^\circ$ , and $125^\circ - 65^\circ = 60^\circ$ ). $UB$ is the distance to find.
(b)
Calculate the distance between Bola and Uju.
(c)
Calculate the bearing of Bola from Uju.

Try it on a graph

Plot the curves, move them, and read values off the graph.

Worked solution (try it first)

(a)

Draw north at Uju $U$ and put Adamu $A$ 15 m away on $305^\circ$ .
Draw north at $A$ and put Bola $B$ 25 m from $A$ on $065^\circ$ .
Join $U$ and $B$ .

(b)

At $A$ , the direction back to $U$ is $305^\circ - 180^\circ = 125^\circ$ and the direction to $B$ is $065^\circ$ , so $\angle UAB = 125^\circ - 65^\circ$
$= 60^\circ$ .
Cosine rule: $|UB|^2 = 15^2 + 25^2 - 2(15)(25)\cos 60^\circ$
$= 225 + 625 - 375$
$= 475$ .
So $|UB| \approx 21.8$ m.

(c)

Cosine rule for the angle at $U$ : $\cos\angle AUB = \frac{15^2 + 21.79^2 - 25^2}{2 \times 15 \times 21.79}$
$\approx \frac{74.8}{653.7}$
$\approx 0.1144$ , so $\angle AUB \approx 83.4^\circ$ .
At $U$ , Adamu is on $305^\circ$ and Bola is $83.4^\circ$ further round clockwise: $305^\circ + 83.4^\circ = 388.4^\circ$ , which is $028^\circ$ after taking away $360^\circ$ .

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