P, Q, R and S are four villages located on an island, where Q is due north of R. PQ = 924 m, QR = 640 m, PR = 1124 m, RS = 583 m. Angle RPS = 26° and angle PSR is obtuse.

1. Check whether triangle $PQR$ is right-angled using the three given side lengths.

In $\triangle PQR$,
$PQ^2+QR^2=924^2+640^2$
$=853\,776+409\,600$
$=1\,263\,376$
Also,
$PR^2=1124^2=1\,263\,376$
Hence,
$PQ^2+QR^2=PR^2.$

2. Conclude the right angle is at $Q$.

Since $PR$ is the longest side and $PQ^2+QR^2=PR^2$, $PR$ is the hypotenuse.

Therefore,
$\angle PQR=90^\circ,$
so $PQ \perp QR$.

3. Use the given north-south direction to justify that $P$ is due east of $Q$.

Since $Q$ is due north of $R$, the line $QR$ is a north-south line.

As $PQ \perp QR$, $PQ$ is an east-west line.

From the diagram, $P$ lies to the east of $Q$.

Therefore, $P$ is due east of $Q$.

1. Find the bearing of $P$ from $R$ using the right-angled triangle $PQR$.

From part (a), $PQ \\perp QR$, with $Q$ due north of $R$ and $P$ due east of $Q$.\n\nIn right-angled triangle $PQR$,\n$\\tan \\angle QRP=\\frac{PQ}{QR}=\\frac{924}{640}$\n$\\angle QRP=55.3^\circ \\text{ correct to 1 d.p.}$\nSo the bearing of $P$ from $R$ is $055.3^\circ$.

2. Use the sine rule in triangle $PRS$ to find the obtuse angle $\\angle PSR$.

In triangle $PRS$,\n$\\frac{\\sin \\angle PSR}{PR}=\\frac{\\sin \\angle RPS}{RS}$\n$\\frac{\\sin \\angle PSR}{1124}=\\frac{\\sin 26^\circ}{583}$\n$\\sin \\angle PSR=\\frac{1124\\sin 26^\circ}{583}=0.844\\ldots$\nHence,\n$\\angle PSR=57.6^\circ \\text{ or } 122.4^\circ$\nSince $\\angle PSR$ is obtuse,\n$\\angle PSR=122.4^\circ.$

3. Find $\\angle PRS$ and hence the bearing of $S$ from $R$.

Using the angle sum of triangle $PRS$,\n$\\angle PRS=180^\circ-26^\circ-122.4^\circ=31.6^\circ.$\nFrom the diagram, $S$ is clockwise from $P$ about $R$. Therefore,\n$\\text{bearing of }S\\text{ from }R=55.3^\circ+31.6^\circ=86.9^\circ.$\nAs a three-figure bearing, this is\n$087^\circ.$

1. Find $\angle PSR$ and hence $\angle PRS$ in triangle $PRS$.

In $\triangle PRS$, using the sine rule,
$\frac{\sin \angle PSR}{1124}=\frac{\sin 26^\circ}{583}$
$\sin \angle PSR=\frac{1124\sin 26^\circ}{583}$
Since $\angle PSR$ is obtuse,
$\angle PSR\approx 122.3^\circ$
Hence
$\angle PRS=180^\circ-26^\circ-122.3^\circ\approx31.7^\circ.$

2. Find the length $PS$ using the sine rule.

Using the sine rule again,
$\frac{PS}{\sin 31.7^\circ}=\frac{583}{\sin 26^\circ}$
$PS=\frac{583\sin31.7^\circ}{\sin26^\circ}$
$PS\approx698\text{ m}.$

3. Find the straight-line distance $QS$.

From part (a), $PQ\perp QR$, so in $\triangle PQR$,
$\tan \angle QPR=\frac{QR}{PQ}=\frac{640}{924}$
$\angle QPR\approx34.7^\circ.$
Therefore
$\angle QPS=34.7^\circ+26^\circ=60.7^\circ.$
In $\triangle PQS$, using the cosine rule,
$QS^2=924^2+698^2-2(924)(698)\cos60.7^\circ$
$QS\approx843\text{ m}=0.843\text{ km}.$

4. Use speed to calculate the time and round to the nearest ten seconds.

Speed $=9.5\text{ km/h}$.
$\text{Time}=\frac{0.843}{9.5}\text{ h}$
$\text{Time}\approx0.0887\text{ h}$
$0.0887\times60\approx5.32\text{ min}$
This is approximately $5$ min $19$ s, which rounds to the nearest ten seconds as
$5\text{ min }20\text{ s}.$