The diagram shows the positions of three towns, A, B and C. AB is 73 km, BC is 62 km and AC is 56 km. A is due North of C.

1. Identify the angle needed in the triangle

Since $A$ is due North of $C$, the line $AC$ is a north-south line.

Find $\angle BAC$, the angle between $AB$ and $AC$.

2. Use the cosine rule to calculate $\angle BAC$

By the cosine rule,
$BC^2=AB^2+AC^2-2(AB)(AC)\cos \angle BAC$
$62^2=73^2+56^2-2(73)(56)\cos \angle BAC$
$\cos \angle BAC=\frac{73^2+56^2-62^2}{2(73)(56)}$
$\cos \angle BAC=\frac{4621}{8176}$

3. Evaluate the angle and convert it to a bearing

$\angle BAC=\cos^{-1}\left(\frac{4621}{8176}\right)$
$\angle BAC\approx55.6^\circ$

From $A$, the direction of $C$ is due South, which has bearing $180^\circ$.

Since $B$ is to the east of $AC$ as shown in the diagram,
$\text{bearing of }B\text{ from }A=180^\circ-55.6^\circ=124.4^\circ$

To the nearest degree, the bearing is
$124^\circ.$