1. **State the problem:** An aircraft flies 500 km on a bearing of 100 degrees, then 600 km on a bearing of 160 degrees. We need to find the distance and bearing from the starting point to the finishing point. 2. **Understanding bearings and vectors:** Bearings are measured clockwise from the north. We can represent each leg of the journey as a vector in the plane using trigonometry. 3. **Convert bearings to standard angles:** Standard angle from the positive x-axis (east) is $\theta = 90^\circ - \text{bearing}$. - For 100 degrees bearing: $\theta_1 = 90^\circ - 100^\circ = -10^\circ$ - For 160 degrees bearing: $\theta_2 = 90^\circ - 160^\circ = -70^\circ$ 4. **Calculate vector components:** - First leg vector: $\vec{v_1} = 500(\cos(-10^\circ), \sin(-10^\circ))$ - Second leg vector: $\vec{v_2} = 600(\cos(-70^\circ), \sin(-70^\circ))$ 5. **Evaluate components:** - $\cos(-10^\circ) = \cos 10^\circ \approx 0.9848$ - $\sin(-10^\circ) = -\sin 10^\circ \approx -0.1736$ - $\cos(-70^\circ) = \cos 70^\circ \approx 0.3420$ - $\sin(-70^\circ) = -\sin 70^\circ \approx -0.9397$ So, - $\vec{v_1} = (500 \times 0.9848, 500 \times -0.1736) = (492.4, -86.8)$ km - $\vec{v_2} = (600 \times 0.3420, 600 \times -0.9397) = (205.2, -563.8)$ km 6. **Sum vectors to find resultant:** $$\vec{R} = \vec{v_1} + \vec{v_2} = (492.4 + 205.2, -86.8 - 563.8) = (697.6, -650.6)$$ km 7. **Calculate distance from start:** $$d = \sqrt{697.6^2 + (-650.6)^2} = \sqrt{486,672 + 423,280} = \sqrt{909,952} \approx 953.4 \text{ km}$$ 8. **Calculate bearing of resultant vector:** - Angle from east axis: $\alpha = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{-650.6}{697.6}\right) \approx -43.3^\circ$ - Convert back to bearing (measured clockwise from north): $$\text{bearing} = 90^\circ - \alpha = 90^\circ - (-43.3^\circ) = 133.3^\circ$$ 9. **Final answer:** The aircraft's finishing point is approximately 953.4 km away on a bearing of 133.3 degrees from the starting point.