A ship sails from Port A on a bearing 065° to Port B a distance 54 km. It then sails on the bearing of 155° from B to Port C a distance of 80 km. Find correct to one decimal place: (i) The distance be

Draw a North line at Port A and Port B. Since these North lines are parallel, we can use properties of parallel lines.
The bearing of B from A is $065^\circ$. This means the angle between the North line at A and the line segment AB is $65^\circ$.
The alternate interior angle between the line segment AB and the South line at B is also $65^\circ$.
The bearing of C from B is $155^\circ$. This is the angle measured clockwise from the North line at B to the line segment BC.
The angle between the North line at B and the South line at B is $180^\circ$.
Therefore, the angle between the South line at B and the line segment BC is $180^\circ - 155^\circ = 25^\circ$.
The angle $\angle ABC$ is the sum of the angle between AB and the South line at B, and the angle between the South line at B and BC.

$$\angle ABC = 65^\circ + 25^\circ = 90^\circ$$

In $\triangle ABC$, we have $AB = 54 \text{ km}$, $BC = 80 \text{ km}$, and $\angle ABC = 90^\circ$.
Since $\angle ABC = 90^\circ$, we can use the Pythagorean theorem (a special case of the Cosine Rule).

$$AC^2 = AB^2 + BC^2$$ $$AC^2 = (54 \text{ km})^2 + (80 \text{ km})^2$$ $$AC^2 = 2916 \text{ km}^2 + 6400 \text{ km}^2$$ $$AC^2 = 9316 \text{ km}^2$$ $$AC = \sqrt{9316} \text{ km}$$ $$AC \approx 96.5194 \text{ km}$$

Rounding to one decimal place:

$$AC \approx 96.5 \text{ km}$$

The distance between A and C is $\boxed{96.5 \text{ km}}$.

First, find the angle $\angle BAC$ in $\triangle ABC$. Since it's a right-angled triangle:

$$\tan(\angle BAC) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB}$$ $$\tan(\angle BAC) = \frac{80}{54}$$ $$\angle BAC = \arctan\left(\frac{80}{54}\right)$$ $$\angle BAC \approx 55.99^\circ$$

The bearing of C from A is the sum of the bearing of B from A and the angle $\angle BAC$.
Bearing of C from A = $065^\circ + \angle BAC$
Bearing of C from A = $065^\circ + 55.99^\circ$
Bearing of C from A = $120.99^\circ$
Rounding to one decimal place:
Bearing of C from A $\approx 121.0^\circ$
The bearing of C from A is $\boxed{121.0^\circ}$.