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:<br /><br />1. **Convert time to hours:** 45 minutes is equal to 45/60 = 0.75 hours.<br /><br />2. **Calculate distance traveled by each boat:**<br /> * Boat 1: Distance = Speed × Time = 10 km/h × 0.75 h = 7.5 km<br /> * Boat 2: Distance = Speed × Time = 8 km/h × 0.75 h = 6 km<br /><br />3. **Find the angle between the boats' paths:** The difference in their bearings is 79° - 47° = 32°.<br /><br />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:<br /> * a = 7.5 km<br /> * b = 6 km<br /> * C = 32°<br /><br /> So, c² = 7.5² + 6² - 2 * 7.5 * 6 * cos(32°)<br /> c² = 56.25 + 36 - 90 * 0.848<br /> c² ≈ 56.25 + 36 - 76.32<br /> c² ≈ 15.93<br /> c ≈ √15.93<br /> c ≈ 3.99 km<br /><br />5. **Round to the nearest tenth:** The boats are approximately 4.0 km apart.<br /><br />Therefore, the correct answer is approximately $\boxed{4.0}$ km.<br />