1. **State the problem:** A ship sails 95 km on a bearing of 140°, then 102 km on a bearing of 260°, and then returns directly to the start. We need to find the distance and bearing of the return journey.
2. **Convert bearings to standard angles for calculation:** Bearings are measured clockwise from north. To convert bearing \(\theta\) to a standard angle \(\alpha\) measured counterclockwise from the positive x-axis (east), use \(\alpha = 90° - \theta\).
- For 140° bearing: \(\alpha_1 = 90° - 140° = -50°\) or equivalently \(310°\).
- For 260° bearing: \(\alpha_2 = 90° - 260° = -170°\) or equivalently \(190°\).
3. **Calculate the components of each leg:**
For leg 1 (95 km, 310°):
$$x_1 = 95 \cos 310° = 95 \cos(-50°) = 95 \times 0.6428 = 61.07\,\text{km}$$
$$y_1 = 95 \sin 310° = 95 \sin(-50°) = 95 \times (-0.7660) = -72.77\,\text{km}$$
For leg 2 (102 km, 190°):
$$x_2 = 102 \cos 190° = 102 \times (-0.9848) = -100.46\,\text{km}$$
$$y_2 = 102 \sin 190° = 102 \times (-0.1736) = -17.71\,\text{km}$$
4. **Find the resultant position after the two legs:**
$$x_r = x_1 + x_2 = 61.07 - 100.46 = -39.39\,\text{km}$$
$$y_r = y_1 + y_2 = -72.77 - 17.71 = -90.48\,\text{km}$$
5. **Determine the return vector (starting point to current position):** The ship returns directly to start, so the return vector is the negative of the resultant:
$$x_{return} = 39.39\,\text{km}$$
$$y_{return} = 90.48\,\text{km}$$
6. **Calculate the distance of the return journey:**
$$d = \sqrt{x_{return}^2 + y_{return}^2} = \sqrt{39.39^2 + 90.48^2} = \sqrt{1551.92 + 8186.54} = \sqrt{9738.46} \approx 98.68\,\text{km}$$
7. **Calculate the bearing of the return journey:**
The angle \(\phi\) relative to east (x-axis):
$$\phi = \arctan \left(\frac{y_{return}}{x_{return}}\right) = \arctan \left(\frac{90.48}{39.39}\right) \approx 66.9°$$
Convert back to bearing:
$$\text{bearing} = 90° - \phi = 90° - 66.9° = 23.1°$$
Since both components are positive, the vector is in the first quadrant, confirming the bearing is \(23.1°\) from north.
**Final answers:** The distance of the return journey is approximately **98.68 km** and the bearing is approximately **23.1°**.
Ship Return
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