The diagram shows the positions of 3 points A, B, and C, at sea level. AB = 4.83 km and BC = 7.24 km. The bearing of B from A is 064° and the bearing of C from B is 127°.

1. Find the angle $ABC$ using the given bearings.

The bearing of $B$ from $A$ is $064^\circ$, so the bearing of $A$ from $B$ is
$064^\circ + 180^\circ = 244^\circ.$
The bearing of $C$ from $B$ is $127^\circ$.

Hence
$\angle ABC = 244^\circ - 127^\circ = 117^\circ.$

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

Using the cosine rule,
$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos \angle ABC.$
Substitute $AB=4.83$, $BC=7.24$ and $\angle ABC=117^\circ$:
$AC^2 = 4.83^2 + 7.24^2 - 2(4.83)(7.24)\cos 117^\circ.$

3. Evaluate and round to the required shown value.

$AC = \sqrt{4.83^2 + 7.24^2 - 2(4.83)(7.24)\cos 117^\circ}$
$AC = 10.368\ldots$
Therefore,
$AC = 10.368\text{ km}.$

1. Find angle $ACB$ using the sine rule.

From part (i), $AC=10.368\text{ km}$ and $\angle ABC=117^\circ$.

Using the sine rule,
$\frac{\sin \angle ACB}{AB}=\frac{\sin \angle ABC}{AC}$
$\frac{\sin \angle ACB}{4.83}=\frac{\sin 117^\circ}{10.368}$
$\sin \angle ACB=\frac{4.83\sin117^\circ}{10.368}$
$\angle ACB=24.5^\circ\text{ approximately}$

2. Use bearings to find the bearing of A from C.

The bearing of $C$ from $B$ is $127^\circ$.

So the bearing of $B$ from $C$ is
$127^\circ+180^\circ=307^\circ.$

At $C$, the angle between $CB$ and $CA$ is $24.5^\circ$.

Therefore the bearing of $A$ from $C$ is
$307^\circ-24.5^\circ=282.5^\circ.$

To the nearest degree,
$282.5^\circ\approx 282^\circ.$

1. Express the area of triangle $ABC$ using the two known sides and the included angle at $B$.

From the bearings, $\angle ABC = 117^\circ$.

$\text{Area of }\triangle ABC = \frac12(4.83)(7.24)\sin 117^\circ$

2. Equate this area to $\frac12 \times AC \times$ perpendicular height and solve for the shortest distance.

Let the shortest distance from $B$ to $AC$ be $h$ km.

Using $AC = 10.368$ km from part (i),

$\frac12(10.368)h = \frac12(4.83)(7.24)\sin117^\circ$

$h = \frac{(4.83)(7.24)\sin117^\circ}{10.368}$

$h = 3.005\ldots$

$h \approx 3.01\text{ km}$

1. Set up the right-angled triangle using the angle of depression.

Let the object be $O$, directly below $C$.

Since $C$ is directly below the horizontal line from $B$ to sea level, $CO$ is vertical and $BC$ is horizontal, so $\triangle BCO$ is right-angled at $C$.

The angle of depression from $B$ is $12^\circ$, so
$\angle CBO = 12^\circ.$

2. Use the tangent ratio to find the vertical distance from $C$ to the object.

Using $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$,
$\tan 12^\circ = \frac{CO}{7.24}.$

$CO = 7.24\tan 12^\circ = 1.5389\ldots$

$CO \approx 1.54\text{ km}.$