1. **Stating the problem:** We have points A, B, and C with bearings and distances given: - Bearing of B from A is 060°. - Bearing of C from A is 120°. - Distance AB = 12 km. - Distance AC = 10 km. We need to find the distance BC and the bearing of C from B. 2. **Drawing and labeling the triangle ABC:** - Place point A at the origin. - Point B is located 12 km from A at a bearing of 060°, which means 60° clockwise from north. - Point C is located 10 km from A at a bearing of 120°. 3. **Applying the Cosine Rule to find BC:** - The angle between AB and AC at A is the difference in bearings: $$120^\circ - 60^\circ = 60^\circ$$ - Using the Cosine Rule: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(60^\circ)$$ - 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. **Determining the bearing of C from B:** - First, find coordinates of B and C relative to A: - B: $$x_B = 12 \times \sin(60^\circ) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \approx 10.39$$ - $$y_B = 12 \times \cos(60^\circ) = 12 \times 0.5 = 6$$ - C: $$x_C = 10 \times \sin(120^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66$$ - $$y_C = 10 \times \cos(120^\circ) = 10 \times (-0.5) = -5$$ - Vector from B to C: - $$\Delta x = x_C - x_B = 8.66 - 10.39 = -1.73$$ - $$\Delta y = y_C - y_B = -5 - 6 = -11$$ - Calculate the angle from north (bearing) using: $$\theta = \arctan\left(\frac{|\Delta x|}{|\Delta y|}\right) = \arctan\left(\frac{1.73}{11}\right) \approx 9^\circ$$ - Since $$\Delta x < 0$$ and $$\Delta y < 0$$, the direction is in the southwest quadrant, so bearing from B to C is: $$180^\circ + 9^\circ = 189^\circ$$ **Final answers:** - Distance BC is approximately 11.14 km. - Bearing of C from B is approximately 189°.