WAEC 2010 · Paper 2 · Q9

In the diagram, $|AB| = 8\text{ km}$ , $|BC| = 13\text{ km}$ , the bearing of $A$ from $B$ is $310^\circ$ and the bearing of $B$ from $C$ is $230^\circ$ . Calculate, correct to 3 significant figures:

(a)
the distance $AC$ ;
(b)
the bearing of $C$ from $A$ ;
(c)
how far east of $B$ , $C$ is.

Worked solution (try it first)

(a)

Find $\angle ABC$ .
From $B$ , $A$ is on $310^\circ$ and $C$ is on $230^\circ - 180^\circ = 050^\circ$ (the back bearing of $230^\circ$ ).
Going round from $050^\circ$ to $310^\circ$ is $260^\circ$ , so the angle inside the triangle is $360^\circ - 260^\circ = 100^\circ$ .
Cosine rule: $AC^2 = 8^2 + 13^2 - 2(8)(13)\cos 100^\circ$
$= 233 + 36.12$
$= 269.12$ .
So $AC = \sqrt{269.12} = 16.4\text{ km}$ to 3 significant figures.

(b)

Sine rule: $\sin \angle CAB = \dfrac{13 \sin 100^\circ}{16.40}$
$= 0.7805$ , so $\angle CAB = 51.3^\circ$ .
The bearing of $B$ from $A$ is $310^\circ - 180^\circ = 130^\circ$ .
$C$ is $51.3^\circ$ further round towards north: $130^\circ - 51.3^\circ = 78.7^\circ$ .
The bearing of $C$ from $A$ is $079^\circ$ to the nearest degree.

(c)

$BC$ makes $50^\circ$ with the north line at $B$ (bearing $050^\circ$ ), so the eastward distance is $13 \sin 50^\circ = 13 \times 0.7660$
$= 9.96\text{ km}$ .

Report a problem with this question