1. **Problem Statement:** You are given three islands A, B, and C with bearings and distances: - Bearing of B from A is 060° - Bearing of C from A is 120° - Distance AB = 12 km - Distance AC = 10 km You need to find: - The distance BC using the Cosine Rule - The bearing of C from B 2. **Drawing and Labeling Triangle ABC:** - Place point A at the origin. - Draw line AB at 60° from north (clockwise), length 12 km. - Draw line AC at 120° from north, length 10 km. 3. **Applying the Cosine Rule to find BC:** - The angle between AB and AC at A is $120^\circ - 60^\circ = 60^\circ$. - Cosine Rule formula: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\theta)$$ where $\theta = 60^\circ$. 4. **Calculate BC:** $$BC^2 = 12^2 + 10^2 - 2 \times 12 \times 10 \times \cos(60^\circ)$$ $$BC^2 = 144 + 100 - 240 \times 0.5$$ $$BC^2 = 244 - 120 = 124$$ $$BC = \sqrt{124} \approx 11.14 \text{ km}$$ 5. **Finding the bearing of C from B:** - Find coordinates of B and C assuming A at origin: - $B = (12 \sin 60^\circ, 12 \cos 60^\circ) = (12 \times 0.866, 12 \times 0.5) = (10.39, 6)$ - $C = (10 \sin 120^\circ, 10 \cos 120^\circ) = (10 \times 0.866, 10 \times (-0.5)) = (8.66, -5)$ - Vector from B to C: $$\vec{BC} = (8.66 - 10.39, -5 - 6) = (-1.73, -11)$$ - Calculate bearing from B to C: - Bearing is measured clockwise from north. - Angle $\alpha = \arctan\left(\frac{|\Delta x|}{|\Delta y|}\right) = \arctan\left(\frac{1.73}{11}\right) \approx 9^\circ$. - Since $\Delta x$ is negative and $\Delta y$ is negative, vector points to southwest quadrant. - Bearing from north clockwise: $180^\circ + 9^\circ = 189^\circ$. **Final answers:** - Distance BC $\approx 11.14$ km - Bearing of C from B $\approx 189^\circ$