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