Answer:
### Part (a): Scale Drawing
To create a scale drawing, follow these steps:
1. **Determine the Scale**: Given scale is 10 cm = 30 km. Therefore, 1 km = \(\frac{10}{30}\) cm = \(\frac{1}{3}\) cm.
2. **Convert Distances**:
- First leg (24 km): \[ 24 \times \frac{1}{3} = 8 \text{ cm} \]
- Second leg (18 km): \[ 18 \times \frac{1}{3} = 6 \text{ cm} \]
3. **Draw the Journey**:
- Start at a point, mark it as \(A\).
- Using a protractor, draw a line from \(A\) at a bearing of 060°.
- Measure and draw a line of 8 cm along this bearing, marking the end point as \(B\).
- From \(B\), use the protractor to draw a line at a bearing of 160°.
- Measure and draw a line of 6 cm along this bearing, marking the end point as \(C\).
### Part (b): Distance from Starting Point to End Point
We can use the Law of Cosines to find the distance from the starting point \(A\) to the end point \(C\):
1. **Find the Angle**:
- The change in bearing from 060° to 160° creates an angle of 100° between the two legs of the journey.
2. **Use the Law of Cosines**:
- Let \(AC\) be the distance between the starting point and the end point.
\[ AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\theta) \]
\[ AC^2 = 24^2 + 18^2 - 2 \times 24 \times 18 \times \cos(100^\circ) \]
\[ AC^2 = 576 + 324 - 2 \times 24 \times 18 \times (-0.1736) \] (since \(\cos(100^\circ) = -0.1736\))
\[ AC^2 = 576 + 324 + 149.2992 \]
\[ AC^2 = 1049.2992 \]
\[ AC \approx \sqrt{1049.2992} \approx 32.4 \text{ km} \]
So, the ship is approximately 32 km from its starting point.
### Part (c): Bearing to Return to Starting Point
To find the bearing on which the ship should sail to return to its starting point:
1. **Calculate the angle \( \alpha \) at \( B \)**:
- Using the sine rule:
\[ \frac{\sin(BAC)}{BC} = \frac{\sin(\theta)}{AC} \]
\[ \frac{\sin(BAC)}{18} = \frac{\sin(100^\circ)}{32.4} \]
\[ \sin(BAC) = 18 \times \frac{\sin(100^\circ)}{32.4} \]
\[ \sin(BAC) = 18 \times \frac{0.9848}{32.4} \]
\[ \sin(BAC) = 0.5467 \]
\[ BAC \approx \sin^{-1}(0.5467) \approx 33.1^\circ \]
2. **Determine the Bearing**:
- The initial bearing was 060°. Adding the calculated angle:
\[ \text{Bearing} = 060^\circ + 33.1^\circ = 93.1^\circ \]
- This is the interior bearing, to return to the start we take the reciprocal bearing:
\[ 93.1^\circ - 180^\circ = 273.1^\circ \]
So, the ship should sail on a bearing of approximately 273° to return to its starting point.
