4. Two boats left the harbour at the same time.One travelled at 10km/h on a bearing of N47^circ E . The other travelled at 8km/h on a bearing of N79^circ E .How far apart were the boats after 45 min? Round the distance to the nearest tenth of a kilometre.

Here's how to solve this problem:1. **Convert time to hours:** 45 minutes is equal to 45/60 = 0.75 hours.2. **Calculate distance traveled by each boat:** * Boat 1: Distance = Speed × Time = 10 km/h × 0.75 h = 7.5 km * Boat 2: Distance = Speed × Time = 8 km/h × 0.75 h = 6 km3. **Find the angle between the boats' paths:** The difference in their bearings is 79° - 47° = 32°.4. **Use the cosine law to find the distance between the boats:** The cosine law states that c² = a² + b² - 2ab*cos(C), where a and b are the sides of a triangle, C is the angle between them, and c is the side opposite angle C. In our case: * a = 7.5 km * b = 6 km * C = 32° So, c² = 7.5² + 6² - 2 * 7.5 * 6 * cos(32°) c² = 56.25 + 36 - 90 * 0.848 c² ≈ 56.25 + 36 - 76.32 c² ≈ 15.93 c ≈ √15.93 c ≈ 3.99 km5. **Round to the nearest tenth:** The boats are approximately 4.0 km apart.Therefore, the correct answer is approximately km.