In the diagram, /PQ/ = 8 m, /QR/ = 13 m, the bearing of Q from P is 050° and the bearing of R from Q is 130°. Calculate correct to three significant figures, PR (ii) the bearing of R from P; Calculate the shortest distance between point Q and PR, hence the area of triangle PQR. (WASSCE)

The bearing of $Q$ from $P$ is $050^\circ$. This means the angle from the North line at $P$, clockwise to $PQ$, is $50^\circ$.
Since the North lines at $P$ and $Q$ are parallel, the alternate interior angle between the North line at $Q$ (pointing North) and $QP$ is $50^\circ$.
The bearing of $R$ from $Q$ is $130^\circ$. This is the angle from the North line at $Q$, clockwise to $QR$.
The angle from the North line at $Q$, clockwise to the South line at $Q$ is $180^\circ$.
Therefore, the angle between the South line at $Q$ and $QR$ (measured anti-clockwise from South) is $180^\circ - 130^\circ = 50^\circ$.
The angle $\angle PQR$ is the sum of the angle between $QP$ and the North line at $Q$ (west side) and the angle between $QR$ and the South line at $Q$ (west side).

$$\angle PQR = 50^\circ + 50^\circ = 100^\circ$$

The Cosine Rule states $c^2 = a^2 + b^2 - 2ab \cos C$. Here, $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)$.

$$PR^2 = 8^2 + 13^2 - 2(8)(13)\cos(100^\circ)$$ $$PR^2 = 64 + 169 - 208(-0.1736)$$ $$PR^2 = 233 + 36.11878$$ $$PR^2 = 269.11878$$ $$PR = \sqrt{269.11878} \approx 16.40489 \text{ m}$$

Rounding to three significant figures:

$$PR \approx \boxed{\text{16.4 m}}$$

Part (a)(ii): Calculate the bearing of R from P

The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B}$. Here, $\frac{QR}{\sin(\angle QPR)} = \frac{PR}{\sin(\angle PQR)}$.

$$\frac{13}{\sin(\angle QPR)} = \frac{16.40489}{\sin(100^\circ)}$$ $$\sin(\angle QPR) = \frac{13 \sin(100^\circ)}{16.40489}$$ $$\sin(\angle QPR) = \frac{13 \times 0.98480775}{16.40489} \approx 0.77980$$ $$\angle QPR = \arcsin(0.77980) \approx 51.25^\circ$$

The bearing of $Q$ from $P$ is $050^\circ$. The bearing of $R$ from $P$ is the bearing of $Q$ from $P$ plus $\angle QPR$.

$$\text{Bearing of R from P} = 50^\circ + 51.25^\circ = 101.25^\circ$$

Rounding to three significant figures:

$$\text{Bearing of R from P} \approx \boxed{101^\circ}$$

Part (b): Calculate the shortest distance between point Q and line PR, hence the area of triangle PQR.

The shortest distance is the altitude (height) from $Q$ to the base $PR$. Let this distance be $h$.
In the right-angled triangle formed by $Q$, $P$, and the foot of the perpendicular from $Q$ to $PR$, we have:

$$h = PQ \sin(\angle QPR)$$ $$h = 8 \sin(51.25^\circ)$$ $$h = 8 \times 0.7800 \approx 6.240 \text{ m}$$

Rounding to three significant figures:

$$\text{Shortest distance} \approx \boxed{6.24 \text{ m}}$$

The area of a triangle can be calculated using the formula: Area $= \frac{1}{2}ab\sin C$.

$$\text{Area of } \triangle PQR = \frac{1}{2} (PQ)(QR)\sin(\angle PQR)$$ $$\text{Area of } \triangle PQR = \frac{1}{2} (8)(13)\sin(100^\circ)$$ $$\text{Area of } \triangle PQR = 52 \times 0.98480775$$ $$\text{Area of } \triangle PQR \approx 51.20999 \text{ m}^2$$

Rounding to three significant figures:

$$\text{Area of } \triangle PQR \approx \boxed{51.2 \text{ m}^2}$$