1. **State the problem:** An aeroplane flies due north from Ikeja airport for 500 km, then flies on a bearing of 060° for 300 km, and finally flies over a road junction. We need to find the distance from Ikeja airport to the road junction. 2. **Understand the problem:** The plane's path forms two legs: first 500 km north, then 300 km at 60° bearing (which is 60° east of north). We want the straight-line distance from the start (Ikeja airport) to the road junction. 3. **Set up the coordinate system:** Let Ikeja airport be at the origin (0,0). The first leg is due north, so after 500 km, the plane is at point A with coordinates: $$A = (0, 500)$$ 4. **Calculate the second leg's components:** The plane flies 300 km at 60° bearing from point A. Bearing 060° means 60° east of north, so: - North component: $300 \times \cos 60^\circ$ - East component: $300 \times \sin 60^\circ$ Calculate these: $$300 \times \cos 60^\circ = 300 \times 0.5 = 150$$ $$300 \times \sin 60^\circ = 300 \times \frac{\sqrt{3}}{2} = 150\sqrt{3}$$ 5. **Find coordinates of the road junction (point B):** $$B = (0 + 150\sqrt{3}, 500 + 150) = (150\sqrt{3}, 650)$$ 6. **Calculate the distance from Ikeja airport (0,0) to point B:** Use the distance formula: $$d = \sqrt{(150\sqrt{3} - 0)^2 + (650 - 0)^2} = \sqrt{(150\sqrt{3})^2 + 650^2}$$ Calculate inside the square root: $$(150\sqrt{3})^2 = 150^2 \times 3 = 22500 \times 3 = 67500$$ $$650^2 = 422500$$ Sum: $$67500 + 422500 = 490000$$ 7. **Final distance:** $$d = \sqrt{490000} = 700$$ **Answer:** The distance from Ikeja airport to the road junction is 700 km.