Three points A, B and C lie on a horizontal ground are such that AB = 24 m and BC = 53 m.
Point B is due north of C. The bearing of A from B is 122°.

1. Find the included angle at $B$ between $BA$ and $BC$.

Bearing of $C$ from $B$ is $180^\circ$.

Bearing of $A$ from $B$ is $122^\circ$.

Hence,
$\angle ABC=180^\circ-122^\circ=58^\circ.$

2. Apply the cosine rule to find $AC$.

Using the cosine rule,
$AC^2=AB^2+BC^2-2(AB)(BC)\cos \angle ABC.$

$AC^2=24^2+53^2-2(24)(53)\cos 58^\circ$

$AC^2=576+2809-2544\cos 58^\circ$

$AC=45.8\text{ m} \;\text{(to 3 s.f.)}.$

1. Find the horizontal and vertical components needed from $A$ to $C$.

Since the bearing of $A$ from $B$ is $122^\circ$, the angle between $BA$ and the south direction is
$180^\circ-122^\circ=58^\circ.$
Horizontal distance of $A$ east of $B$:
$24\sin58^\circ.$
Vertical distance of $A$ south of $B$:
$24\cos58^\circ.$
Since $BC=53$ m, from $A$ to $C$ the southward distance is
$53-24\cos58^\circ.$

2. Find the angle that $AC$ makes with the south direction.

Let $\theta$ be the angle west of south from $A$ to $C$.
$\tan\theta=\frac{24\sin58^\circ}{53-24\cos58^\circ}$
$\theta=\tan^{-1}\left(\frac{24\sin58^\circ}{53-24\cos58^\circ}\right)$
$\theta=26.8^\circ \text{ (to 1 d.p.)}$

3. Convert the angle west of south into a three-figure bearing.

The direction from $A$ to $C$ is $26.8^\circ$ west of south.

Therefore the bearing of $C$ from $A$ is
$180^\circ+26.8^\circ=206.8^\circ.$
To the nearest degree,
$206.8^\circ\approx207^\circ.$

1. Find the included angle at $B$ between $BA$ and $BC$.

Since $BC$ is due south from $B$, the bearing of $C$ from $B$ is $180^\circ$.

The bearing of $A$ from $B$ is $122^\circ$.

Hence
$\angle ABC=180^\circ-122^\circ=58^\circ.$

2. Use the triangle area formula with two sides and the included angle.

$\text{Area of }\triangle ABC=\frac12(AB)(BC)\sin\angle ABC$

$=\frac12(24)(53)\sin58^\circ$

$=539.363\ldots$

So the area is $539\text{ m}^2$ to 3 significant figures.