A surveyor leaves her base camp and drives 42km on a bearing of 032⁰ . She then drives 28km on a bearing of 154⁰, how far is she then from her base camp and what is her bearing from it ?

Let the base camp be point A.
The surveyor drives 42 km on a bearing of $032^\circ$ to point B.
From point B, she drives 28 km on a bearing of $154^\circ$ to point C.
We need to find the distance AC and the bearing of C from A.

First, let's find the interior angle $\angle ABC$.
Draw a North line at B.
The line BA (from B to A) is on a back bearing from the first leg. The bearing of AB is $032^\circ$. The angle from the North line at B, measured clockwise to BA, is $032^\circ + 180^\circ = 212^\circ$.
The line BC (from B to C) is on a bearing of $154^\circ$. This is the angle from the North line at B, measured clockwise to BC.
The interior angle $\angle ABC$ is the difference between these two angles:

$$\angle ABC = 212^\circ - 154^\circ = 58^\circ$$

In $\triangle ABC$:
$AB = 42$ km
$BC = 28$ km
$\angle ABC = 58^\circ$
Using the Cosine Rule: $AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos(\angle ABC)$

$$AC^2 = 42^2 + 28^2 - 2(42)(28) \cos(58^\circ)$$ $$AC^2 = 1764 + 784 - 2352 \cos(58^\circ)$$ $$AC^2 = 2548 - 2352(0.529919...)$$ $$AC^2 = 2548 - 1246.307$$ $$AC^2 = 1301.693$$ $$AC = \sqrt{1301.693}$$ $$AC \approx 36.079 \text{ km}$$

Rounding to one decimal place:

$$\boxed{AC = 36.1 \text{ km}}$$

To find the bearing of C from A, we need to find the angle $\angle BAC$ and add it to the initial bearing of AB ($032^\circ$).
Using the Sine Rule in $\triangle ABC$: $\frac{\sin(\angle BAC)}{BC} = \frac{\sin(\angle ABC)}{AC}$

$$\frac{\sin(\angle BAC)}{28} = \frac{\sin(58^\circ)}{36.079}$$ $$\sin(\angle BAC) = \frac{28 \times \sin(58^\circ)}{36.079}$$ $$\sin(\angle BAC) = \frac{28 \times 0.848048}{36.079}$$ $$\sin(\angle BAC) = \frac{23.745344}{36.079}$$ $$\sin(\angle BAC) \approx 0.65800$$ $$\angle BAC = \arcsin(0.65800)$$ $$\angle BAC \approx 41.15^\circ$$

The bearing of C from A is the bearing of AB plus $\angle BAC$:

$$\text{Bearing} = 032^\circ + 41.15^\circ$$ $$\text{Bearing} = 73.15^\circ$$

Rounding to the nearest degree and expressing as a three-figure bearing:

$$\boxed{\text{Bearing} = 073^\circ}$$