An aeroplane flies due west for 3 hours from P (lat. 50°N, long. 60°W) to a point Q at an average speed of 600km/h. The aeroplane then flies due south from Q to a point Y 500km away. Calculate, correct to 3 significant figures,
(a) the longitude of Q ;
(b) the latitude of Y . [Take the radius of the earth = 6400km and (pi = frac{22}{7})].
Explanation
(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° approxeq 25.1°)
(therefore) The longitude of Q = (60 + theta)
= (60 + 25.1 = 85.1°W)
(b) The distance (Q) along the meridian
(frac{theta}{360} times 2pi 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°)
(therefore text{The latitude of Y} = 50° – 4.47° = 45.53°)
(approxeq 45.5°N)