Airport B is 500 km away from airport A on a bearing of 064°. An aircraft leaves airport A at 22 45 on Monday to fly to airport B. Its speed during the flight is 375 km/h.

1. Calculate the flight time from A to B.

$\text{Time}=\frac{500}{375}=\frac{4}{3}\text{ h}=1\text{ h }20\text{ min}$

2. Add the flight time to the departure time.

Departure time: $22\ 45$ on Monday

$22\ 45+1\text{ h }20\text{ min}=00\ 05$

Since the time passes midnight, the arrival is on Tuesday.

1. Find the angle between $BP$ and $BQ$ at $B$.

Since $AB$ has bearing $064^\circ$, the reverse direction $BP$ has bearing
$064^\circ+180^\circ=244^\circ.$
$BQ$ is due north, so the angle between $BP$ and $BQ$ is
$360^\circ-244^\circ=116^\circ.$

2. Use the cosine rule to calculate $PQ$.

In triangle $PBQ$, $PB=300$ km, $BQ=120$ km and $\angle PBQ=116^\circ$.

By the cosine rule,
$PQ^2=300^2+120^2-2(300)(120)\cos116^\circ.$
$PQ=\sqrt{300^2+120^2-2(300)(120)\cos116^\circ}$
$PQ=368.7\ldots$
So
$PQ\approx 369\text{ km}.$

1. Find the angle at $P$ in triangle $PBQ$.

From part (b), $PQ = 369\text{ km}$.

The angle at $B$ is $116^\circ$, so by the sine rule,
$\frac{\sin \angle BPQ}{120}=\frac{\sin 116^\circ}{369}$
$\sin \angle BPQ=\frac{120\sin116^\circ}{369}$
$\angle BPQ=17.0^\circ\text{, correct to 1 d.p.}$

2. Calculate the bearing of $Q$ from $P$.

The bearing of $B$ from $P$ is $064^\circ$.

Since $Q$ is north of $B$, $PQ$ is $17.0^\circ$ to the left of $PB$.

$\text{Bearing of }Q\text{ from }P=64^\circ-17.0^\circ=47.0^\circ$

Written as a three-figure bearing: $047^\circ$.