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)
(b)(i) (Speed = frac{distance}{time})
(950 = frac{d}{4} implies d = 950 times 4)
(d = 3800km)
(d = frac{theta}{360} times 2pi 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).