A ship travels 100 km on a bearing of 045°, then turns to a bearing of 120° and travels 180 km to its destination. (i) Sketch the path travelled by the ship. (ii) Find the distance of ship from its original position. (iii) Find the bearing from the original position.

• Start at the origin, O. Draw a North line.
• From O, draw a line segment OA, representing 100 km, at an angle of $045^\circ$ clockwise from the North line.
• At point A, draw another North line.
• From A, draw a line segment AB, representing 180 km, at an angle of $120^\circ$ clockwise from the North line at A.
• Point B is the destination. The path is O to A, then A to B.

The North line at O and the North line at A are parallel. The line OA is a transversal.
The angle between the North line at A and the line AO (pointing towards O) is $45^\circ$ (alternate interior angle to the $045^\circ$ bearing).
The angle from the North line at A, clockwise to AB is $120^\circ$.
The angle from the North line at A, clockwise to the South line at A is $180^\circ$.
So, the angle from the South line at A, counter-clockwise to AB is $180^\circ - 120^\circ = 60^\circ$.
Therefore, $\angle OAB = 45^\circ + 60^\circ = 105^\circ$.

Using the Cosine Rule in $\triangle OAB$:

Substitute the given values: $OA = 100$ km, $AB = 180$ km, $\angle OAB = 105^\circ$.

The distance of the ship from its original position is $\boxed{\text{227.41 km}}$.

Using the Sine Rule in $\triangle OAB$:

Substitute the values: $AB = 180$ km, $\angle OAB = 105^\circ$, $OB = 227.4148$ km.

The bearing of B from O is the initial bearing of A from O plus $\angle AOB$.
Bearing $= 045^\circ + \angle AOB$
Bearing $= 45^\circ + 49.86^\circ$
Bearing $= 94.86^\circ$
Rounded to one decimal place, the bearing is $\boxed{094.9^\circ}$.