WAEC 2017 · Paper 2 · Q9

(a)
An aeroplane flies 100 km from town $A$ on a bearing of $330^\circ$ to town $B$ . It then flies 300 km due west to town $C$ . (i) Illustrate this information in a diagram. (ii) Calculate the: (I) distance between $A$ and $C$ , correct to two decimal places; (II) bearing of $C$ from $A$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Draw north at $A$ and draw $AB$ , 100 km on $330^\circ$ ( $30^\circ$ west of north).
Draw north at $B$ and draw $BC$ , 300 km due west ( $270^\circ$ ).
Join $C$ to $A$ .

(I)** At $B$ , the direction back to $A$ is $330^\circ - 180^\circ = 150^\circ$ and the direction to $C$ is $270^\circ$ , so $\angle ABC = 270^\circ - 150^\circ$
$= 120^\circ$ .
Cosine rule: $|AC|^2 = 100^2 + 300^2 - 2(100)(300)\cos 120^\circ$
$= 10\,000 + 90\,000 + 30\,000$
$= 130\,000$ .
So $|AC| = \sqrt{130\,000} \approx 360.56$ km.
(II) Sine rule: $\sin\angle BAC = \frac{300\sin 120^\circ}{360.56}$
$\approx \frac{259.81}{360.56}$
$\approx 0.7206$ , so $\angle BAC \approx 46.1^\circ$ .
At $A$ , $B$ is on $330^\circ$ and $C$ is $46.1^\circ$ further round anticlockwise.
Bearing of $C$ from $A$ $= 330^\circ - 46.1^\circ$
$\approx 284^\circ$ .