1. **State the problem:** Melur walks along the path A to B, then B to C, then C to A. We know distances AB = 3.6 km, AC = 8.4 km, and bearings: B from A is 054°, C from B is 132°. We need to find the total time taken walking at 6 km/h. 2. **Find length BC:** We have triangle ABC with sides AB = 3.6 km, AC = 8.4 km, and need BC. 3. **Find angle at B:** Bearing of B from A is 054°, so line AB makes 54° from north. Bearing of C from B is 132°, so line BC makes 132° from north. The angle ABC is the difference between these bearings: $$\angle ABC = 132^\circ - 54^\circ = 78^\circ$$ 4. **Use Law of Cosines to find BC:** $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle ABC)$$ $$BC^2 = 3.6^2 + 8.4^2 - 2 \times 3.6 \times 8.4 \times \cos(78^\circ)$$ Calculate each term: $$3.6^2 = 12.96$$ $$8.4^2 = 70.56$$ $$\cos(78^\circ) \approx 0.2079$$ So, $$BC^2 = 12.96 + 70.56 - 2 \times 3.6 \times 8.4 \times 0.2079$$ $$BC^2 = 83.52 - 12.57 = 70.95$$ $$BC = \sqrt{70.95} \approx 8.42 \text{ km}$$ 5. **Calculate total distance walked:** $$\text{Total distance} = AB + BC + CA = 3.6 + 8.42 + 8.4 = 20.42 \text{ km}$$ 6. **Calculate total time:** Speed = 6 km/h $$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{20.42}{6} \approx 3.403 \text{ hours}$$ 7. **Convert time to hours and minutes:** $$0.403 \times 60 = 24.18 \text{ minutes}$$ So, total time is approximately 3 hours 24 minutes. **Final answer:** Melur takes about **3 hours 24 minutes** to complete the walk.