An aeroplane flies due North from a town T on the equator at a speed of 950km per hour for 4 hours to another town P. It then flies eastwards to town Q on longitude 65°E. If the longitude of T is 15°E,
(a) represent this information in a diagram ;
(b) calculate the : (i) latitude of P, correct to the nearest degree ; (ii) distance between P and Q, correct to four significant figures. [Take \(\pi = \frac{22}{7}\); Radius of the earth = 6400km].
Explanation
(a) see diagram above.
(b)(i) \(Speed = \frac{distance}{time}\)
\(950 = \frac{d}{4} \implies d = 950 \times 4\)
\(d = 3800km\)
\(d = \frac{\theta}{360} \times 2\pi r\)
\(3800 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400\)
\(3800 = \frac{\theta \times 281600}{2520}\)
\(\theta = \frac{2520 \times 3800}{281600}\)
\(\theta = 34.01°\)
\(\theta = 34°N\) (to the nearest degree)
(ii) Distance between P and Q, correct to four significant figures.
Longitude difference = 65° - 15° = 50°
Using, \(d = \frac{\theta}{360} \times 2 \pi r\)
where \(r = R \cos \theta\)
\(d = \frac{\theta}{360} \times 2 \pi R \cos \theta\)
= \(\frac{50}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 34.01°\)
= \(4631.53 km\)
\(\approxeq 4632km\) (to 4 significant figures).