A man travels 5 km due east of a town A to town B, and then travels 7 km on a bearing of 162° from B to another town C. Calculate the: (a) Shortest distance from A to C, correct to the nearest km (b) Bearing of C from A, correct to the nearest degree

Let town A be the origin.
A man travels 5 km due east from A to B. So, the distance $AB = 5$ km.
From B, he travels 7 km to C on a bearing of $162^\circ$. So, the distance $BC = 7$ km.
We need to find the angle $\angle ABC$.
Draw a North line at B. Since A is due west of B (because B is due east of A), the line BA points in the direction of $270^\circ$ from the North line at B.
The bearing of C from B is $162^\circ$.
The angle $\angle ABC$ is the angle between the line BA and the line BC.

$$\angle ABC = 270^\circ - 162^\circ = 108^\circ$$

The Cosine Rule states $AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos(\angle ABC)$.

$$AC^2 = 5^2 + 7^2 - 2(5)(7) \cos(108^\circ)$$ $$AC^2 = 25 + 49 - 70 \cos(108^\circ)$$ $$AC^2 = 74 - 70(-0.30901699)$$ $$AC^2 = 74 + 21.6311893$$ $$AC^2 = 95.6311893$$ $$AC = \sqrt{95.6311893}$$ $$AC \approx 9.7791209 \text{ km}$$

The Sine Rule states $\frac{\sin A}{a} = \frac{\sin B}{b}$.

$$\frac{\sin(\angle BAC)}{BC} = \frac{\sin(\angle ABC)}{AC}$$ $$\frac{\sin(\angle BAC)}{7} = \frac{\sin(108^\circ)}{9.7791209}$$ $$\sin(\angle BAC) = \frac{7 \sin(108^\circ)}{9.7791209}$$ $$\sin(\angle BAC) = \frac{7 \times 0.9510565}{9.7791209}$$ $$\sin(\angle BAC) \approx 0.68081$$ $$\angle BAC = \arcsin(0.68081)$$ $$\angle BAC \approx 42.90^\circ$$

Town B is due east of Town A. This means the line AB is along the East direction from A. The angle from the North line at A to the East direction (line AB) is $90^\circ$.
The bearing of C from A is the angle measured clockwise from the North line at A to the line AC.
Bearing of C from A = $90^\circ + \angle BAC$.

$$\text{Bearing} = 90^\circ + 42.90^\circ = 132.90^\circ$$