1. **Problem statement:** An aircraft moves from station P(25°N, 15°E) to station Q(25°N, 23°W) along the parallel of latitude, then from Q to R(35°S, 23°W) along the meridian. We need to find: (i) The total distance from P to R. (ii) The average speed if the journey takes 12 hours. 2. **Relevant formulas and rules:** - Distance along a parallel of latitude: $$d = R \times \cos(\phi) \times \Delta \lambda$$ where $\phi$ is the latitude, $\Delta \lambda$ is the difference in longitude in radians, and $R$ is Earth's radius. - Distance along a meridian: $$d = R \times \Delta \phi$$ where $\Delta \phi$ is the difference in latitude in radians. - Convert degrees to radians: $$\text{radians} = \text{degrees} \times \frac{\pi}{180}$$ - Average speed: $$\text{speed} = \frac{\text{total distance}}{\text{time}}$$ 3. **Calculate distance from P to Q:** - Latitude $\phi = 25^\circ$ (same for P and Q) - Longitude difference: from 15°E to 23°W is $15 + 23 = 38^\circ$ - Convert $38^\circ$ to radians: $$38 \times \frac{3.142}{180} = 0.663 \text{ radians}$$ - Calculate distance: $$d_{PQ} = 6400 \times \cos(25^\circ) \times 0.663$$ - Calculate $\cos(25^\circ)$: $$\cos(25^\circ) = 0.9063$$ - So, $$d_{PQ} = 6400 \times 0.9063 \times 0.663 = 3847.5 \text{ km}$$ 4. **Calculate distance from Q to R:** - Longitude is constant at 23°W - Latitude difference: from 25°N to 35°S is $25 + 35 = 60^\circ$ - Convert $60^\circ$ to radians: $$60 \times \frac{3.142}{180} = 1.047 \text{ radians}$$ - Calculate distance: $$d_{QR} = 6400 \times 1.047 = 6700.8 \text{ km}$$ 5. **Total distance from P to R:** $$d_{total} = d_{PQ} + d_{QR} = 3847.5 + 6700.8 = 10548.3 \text{ km}$$ 6. **Calculate average speed:** - Total time = 12 hours - Average speed: $$\text{speed} = \frac{10548.3}{12} = 879.0 \text{ km/hr}$$ **Final answers:** (i) Total distance from P to R is approximately **10548 km**. (ii) Average speed of the aircraft is approximately **879 km/hr**.