1. **Problem statement:** An aeroplane flies from Lilongwe on a bearing of 031° for 20 km, then changes course and flies 140 km on a bearing of 074°. Find: a) The distance of Lilongwe from the aeroplane. b) The bearing of Lilongwe from the aeroplane. 2. **Formula and rules:** - We use the Law of Cosines to find the distance between two points when two sides and the included angle are known: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ - Bearings are measured clockwise from North. 3. **Step 1: Find the angle between the two legs of the journey.** The first bearing is 031°, the second is 074°. The angle between the two legs is: $$\theta = 74^\circ - 31^\circ = 43^\circ$$ 4. **Step 2: Apply Law of Cosines to find distance Lilongwe to plane (side opposite angle \theta).** Let side $a = 20$ km, side $b = 140$ km, angle $C = 43^\circ$. $$c^2 = 20^2 + 140^2 - 2 \times 20 \times 140 \times \cos(43^\circ)$$ $$c^2 = 400 + 19600 - 5600 \times \cos(43^\circ)$$ Calculate $\cos(43^\circ) \approx 0.7314$: $$c^2 = 400 + 19600 - 5600 \times 0.7314 = 20000 - 4095.84 = 15904.16$$ $$c = \sqrt{15904.16} \approx 126.13 \text{ km}$$ 5. **Step 3: Find the bearing of Lilongwe from the aeroplane.** Use the Law of Sines to find the angle opposite side $a$: $$\frac{\sin A}{a} = \frac{\sin C}{c}$$ $$\sin A = \frac{a \sin C}{c} = \frac{20 \times \sin 43^\circ}{126.13}$$ Calculate $\sin 43^\circ \approx 0.6820$: $$\sin A = \frac{20 \times 0.6820}{126.13} = \frac{13.64}{126.13} \approx 0.1081$$ $$A = \arcsin(0.1081) \approx 6.21^\circ$$ 6. **Step 4: Calculate the bearing from the aeroplane to Lilongwe.** The aeroplane's final bearing is 074°. The angle between the line from aeroplane to Lilongwe and the second leg is $A = 6.21^\circ$. Bearing from aeroplane to Lilongwe: $$074^\circ - 6.21^\circ = 67.79^\circ$$ **Final answers:** - a) Distance from Lilongwe to aeroplane is approximately **126.13 km**. - b) Bearing of Lilongwe from aeroplane is approximately **068°**.