1. **State the problem:** A ship moves 20 nautical miles (NM) north and then 35 NM east. We need to find the bearing it should take to head directly back to the starting point. 2. **Set up the coordinate system:** Consider the starting point as the origin $(0,0)$. 3. **Find the ship's current position:** After moving north 20 NM, the ship is at $(0,20)$. Then it moves 35 NM east, so the new position is $(35,20)$. 4. **Calculate the distance back to the start:** The ship needs to return along the vector from $(35,20)$ to $(0,0)$, which is $\langle -35, -20 \rangle$. 5. **Calculate the angle of the bearing:** Bearing is measured clockwise from north. Find the angle $\theta$ the vector makes with the north direction (positive y-axis). The bearing angle from north is given by $\theta = \arctan\left(\frac{\text{east component}}{\text{north component}}\right)$. Here, moving back, east component is $-35$ and north component is $-20$, but for angle measurement, take absolute values: $$ \theta = \arctan\left(\frac{35}{20}\right) $$ 6. **Calculate $\theta$: ** $$ \theta = \arctan\left(1.75\right) \approx 60.26^\circ $$ 7. **Determine the bearing:** Since the ship is southwest relative to the origin when going back, its bearing is $180^\circ - 60.26^\circ = 119.74^\circ$ from north clockwise. But bearings are usually given clockwise from the north position, so the bearing is $180^\circ + 60.26^\circ = 240.26^\circ$. 8. **Final answer:** The ship should take a bearing of approximately $240.3^\circ$ to head directly back to the starting point.