1. **Problem:** A ship travels from A to B (3 km on a bearing of 040°) and then from B to C (5 km on a bearing of 120°). Find the distance AC. 2. **Formula and rules:** To find the distance between points A and C, we use the Law of Cosines. The Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab \cos(C)$$ where $a$ and $b$ are sides of a triangle and $C$ is the included angle. 3. **Find the angle between the two legs:** The bearing from A to B is 040°, and from B to C is 120°. The angle at B between these two paths is: $$\theta = 120^\circ - 40^\circ = 80^\circ$$ 4. **Apply Law of Cosines:** Let $AB = 3$ km, $BC = 5$ km, and angle $B = 80^\circ$. $$AC^2 = 3^2 + 5^2 - 2 \times 3 \times 5 \times \cos(80^\circ)$$ 5. **Calculate:** $$AC^2 = 9 + 25 - 30 \times \cos(80^\circ)$$ $$AC^2 = 34 - 30 \times 0.1736 = 34 - 5.208 = 28.792$$ 6. **Distance AC:** $$AC = \sqrt{28.792} \approx 5.37 \text{ km}$$ --- 1. **Problem:** Find the true bearing of point C from point A. 2. **Explanation:** The true bearing from A to C is the angle measured clockwise from north at A to the line AC. 3. **Find the angle at A:** The angle at B is 80°, so the angle at A is: $$\alpha = 180^\circ - 80^\circ - \beta$$ where $\beta$ is the angle at C. But we can find the direction of AC by vector addition or by using the Law of Sines. 4. **Use Law of Sines:** $$\frac{AC}{\sin(80^\circ)} = \frac{BC}{\sin(\alpha)} = \frac{AB}{\sin(\beta)}$$ 5. **Calculate angle at A:** $$\sin(\alpha) = \frac{BC \sin(80^\circ)}{AC} = \frac{5 \times 0.9848}{5.37} = 0.916$$ $$\alpha = \arcsin(0.916) \approx 66.7^\circ$$ 6. **Calculate bearing:** The bearing from A to B is 040°, so the bearing from A to C is: $$40^\circ + 66.7^\circ = 106.7^\circ$$ --- 1. **Problem:** Two ships leave a port at the same time. Ship A travels 10 km on a bearing of 060°, and Ship B travels 15 km on a bearing of 150°. How far apart are the two ships? 2. **Find angle between paths:** $$\theta = 150^\circ - 60^\circ = 90^\circ$$ 3. **Apply Law of Cosines:** $$d^2 = 10^2 + 15^2 - 2 \times 10 \times 15 \times \cos(90^\circ)$$ Since $\cos(90^\circ) = 0$: $$d^2 = 100 + 225 = 325$$ 4. **Distance:** $$d = \sqrt{325} \approx 18.03 \text{ km}$$ --- 1. **Problem:** A plane flies 200 km on a bearing of 030°, then 300 km on a bearing of 100°. How far is the plane from its starting point? 2. **Find angle between legs:** $$\theta = 100^\circ - 30^\circ = 70^\circ$$ 3. **Apply Law of Cosines:** $$d^2 = 200^2 + 300^2 - 2 \times 200 \times 300 \times \cos(70^\circ)$$ 4. **Calculate:** $$d^2 = 40000 + 90000 - 120000 \times 0.3420 = 130000 - 41040 = 88960$$ 5. **Distance:** $$d = \sqrt{88960} \approx 298.27 \text{ km}$$ --- 1. **Problem:** A scout hikes 4 km on a bearing of 020° and then 6 km on a bearing of 130°. What is the final bearing of the scout from the starting point? 2. **Find angle between legs:** $$\theta = 130^\circ - 20^\circ = 110^\circ$$ 3. **Use Law of Cosines to find distance from start:** $$d^2 = 4^2 + 6^2 - 2 \times 4 \times 6 \times \cos(110^\circ)$$ $$d^2 = 16 + 36 - 48 \times (-0.3420) = 52 + 16.416 = 68.416$$ $$d = \sqrt{68.416} \approx 8.27 \text{ km}$$ 4. **Use Law of Sines to find angle at start:** $$\sin(\alpha) = \frac{6 \sin(110^\circ)}{8.27} = \frac{6 \times 0.9397}{8.27} = 0.681$$ $$\alpha = \arcsin(0.681) \approx 42.9^\circ$$ 5. **Final bearing:** $$20^\circ + 42.9^\circ = 62.9^\circ$$ --- **Final answers:** 1. Distance AC $\approx 5.37$ km 2. True bearing of C from A $\approx 106.7^\circ$ 3. Distance between ships $\approx 18.03$ km 4. Distance plane from start $\approx 298.27$ km 5. Final bearing of scout $\approx 62.9^\circ$