WAEC 2010 · Paper 2 · Q11

(a)
A ship $P$ is $3\text{ km}$ due east of a harbour. Another ship $Q$ is also $3\text{ km}$ from the harbour but on a bearing of $042^\circ$ from the harbour. (i) Find the distance between the two ships. (ii) Find the bearing of ship $Q$ from ship $P$ .
Separate values with commas, e.g. 3, −2
Model answer
(b)
A motorist travelled $300\text{ km}$ at an average speed of $75\text{ km/h}$ and returned at an average speed of $v\text{ km/h}$ . If his average speed for the whole journey is $60\text{ km/h}$ , find $v$ .

Worked solution (try it first)

Sketch the harbour $H$ with $P$ due east (bearing $090^\circ$ ) and $Q$ on $042^\circ$ .
The angle between them at $H$ is $\angle QHP = 90^\circ - 42^\circ$
$= 48^\circ$ .
$HP = HQ = 3\text{ km}$ , so triangle $HPQ$ is isosceles and the base angles are $\angle HQP = \angle HPQ$
$= \frac12(180^\circ - 48^\circ)$
$= 66^\circ$ .
Sine rule: $\dfrac{|PQ|}{\sin 48^\circ} = \dfrac{3}{\sin 66^\circ}$ , so $|PQ| = \dfrac{3 \sin 48^\circ}{\sin 66^\circ}$
$= 2.44\text{ km}$ .

Average speed is total distance over total time.
The whole journey is $600\text{ km}$ at $60\text{ km/h}$ , so it takes $\frac{600}{60} = 10$ hours.
The outward trip takes $\frac{300}{75} = 4$ hours, so the return takes $10 - 4 = 6$ hours.
So $v = \frac{300}{6} = 50$ .