1. **Problem statement:** We have two points 25 km apart east-west. A ship sails 20 km from the first point on a bearing of 045°. We need to find the bearing of the second point from the ship. 2. **Understanding bearings:** Bearings are measured clockwise from the north direction. A bearing of 045° means the ship moves northeast, 45° clockwise from north. 3. **Set up coordinate system:** Place the first point at the origin $(0,0)$ and the second point 25 km east at $(25,0)$. 4. **Find ship's coordinates:** The ship sails 20 km at 045°, so its coordinates are: $$x = 20 \times \sin 45^\circ = 20 \times \frac{\sqrt{2}}{2} = 10\sqrt{2}$$ $$y = 20 \times \cos 45^\circ = 20 \times \frac{\sqrt{2}}{2} = 10\sqrt{2}$$ So ship is at $(10\sqrt{2}, 10\sqrt{2})$. 5. **Vector from ship to second point:** $$\Delta x = 25 - 10\sqrt{2}$$ $$\Delta y = 0 - 10\sqrt{2} = -10\sqrt{2}$$ 6. **Calculate angle from north to vector:** The bearing is measured clockwise from north, so we find the angle $\theta$ between the vector and north (positive y-axis). Calculate angle $\alpha$ between vector and east (x-axis): $$\tan \alpha = \frac{|\Delta y|}{\Delta x} = \frac{10\sqrt{2}}{25 - 10\sqrt{2}}$$ Numerically: $$10\sqrt{2} \approx 14.142$$ $$25 - 14.142 = 10.858$$ $$\tan \alpha = \frac{14.142}{10.858} \approx 1.303$$ $$\alpha = \arctan(1.303) \approx 52.5^\circ$$ 7. **Determine bearing:** Since $\Delta x > 0$ and $\Delta y < 0$, the vector points southeast quadrant. Bearing from north clockwise is: $$180^\circ - \alpha = 180^\circ - 52.5^\circ = 127.5^\circ$$ 8. **Final answer:** The bearing of the second point from the ship is approximately **127.5°**.