(a) \(\frac{r}{R} = \cos 50\) \(r = 6400 \cos 50 = 4,113.84 km\) \(\approxeq 4,114km\) The distance PQ along the parallel of latitude = speed (km/h) \(\times\) time (h) = 600km/h x 3h = 1800km \(\frac{\theta}{360} \times 2 \times \frac{22}{7} \times r = 1800\) \(\theta = \frac{1800 \times 7 \times 360}{44 \times 4113.84}\) \(\theta = 25.059^\circ \approxeq 25.1^\circ\) \(\therefore\) The longitude of Q = \(60 + \theta\) = \(60 + 25.1 = 85.1^\circW\) (b) The distance (Q) along the meridian \(\frac{\theta}{360} \times 2\pi R\) \(500 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400\) \(\theta = \frac{500 \times 360 \times 7}{44 \times 6400}\) \(\theta = 4.47^\circ\) \(\therefore \text{The latitude of Y} = 50^\circ - 4.47^\circ = 45.53^\circ\) \(\approxeq 45.5^\circN\)