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. **Model and Triangle Setup:** Draw triangle ABC with A as the reference point. - Place A at the origin. - B is located 12 km from A at 60° bearing. - C is located 10 km from A at 120° bearing. 3. **Using the Cosine Rule to find BC:** The angle between AB and AC at A is the difference in bearings: $$\theta = 120^\circ - 60^\circ = 60^\circ$$ Cosine Rule formula: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\theta)$$ Substitute values: $$BC^2 = 12^2 + 10^2 - 2 \times 12 \times 10 \times \cos(60^\circ)$$ Calculate: $$BC^2 = 144 + 100 - 240 \times 0.5 = 244 - 120 = 124$$ Therefore: $$BC = \sqrt{124} = 2\sqrt{31} \approx 11.14 \text{ km}$$ 4. **Finding the bearing of C from B:** - Find coordinates of B and C relative to A: - B: $$(12 \cos 60^\circ, 12 \sin 60^\circ) = (6, 10.39)$$ - C: $$(10 \cos 120^\circ, 10 \sin 120^\circ) = (-5, 8.66)$$ - Vector from B to C: $$\overrightarrow{BC} = (x_C - x_B, y_C - y_B) = (-5 - 6, 8.66 - 10.39) = (-11, -1.73)$$ - Calculate the angle of vector BC relative to north (y-axis): $$\theta = \arctan\left(\frac{\Delta x}{\Delta y}\right) = \arctan\left(\frac{-11}{-1.73}\right) = \arctan(6.36) \approx 81.1^\circ$$ Since both $\Delta x$ and $\Delta y$ are negative, vector BC points to the southwest quadrant, so bearing from B is: $$180^\circ + 81.1^\circ = 261.1^\circ$$ **Final answers:** - Distance BC is approximately 11.14 km - Bearing of C from B is approximately 261°