Question

12. 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 a 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 4 significant figures. [Take π=22/7, Radius of the earth=6400km]

Accepted Answer

(a) The diagram shows a globe with the equator marked, town T on the equator at longitude 15°E, town P directly north of T at some latitude θ°N (still on longitude 15°E, since it flew due north), and town Q at the same latitude as P but on longitude 65°E (having flown eastwards along the latitude/parallel from P).

(b)(i) Distance TP (along a great circle/meridian) = speed × time = 950 × 4 = 3800km. Distance along a meridian: dist = (θ/360°)×2πR, where θ is the angle (latitude difference) subtended at the earth's centre. 3800 = (θ/360)×2×(22/7)×6400. Solving: θ = (3800×360×7)/(2×22×6400) = 34°N (nearest degree). So the latitude of P is 34°N.
(ii) The distance PQ is along the parallel of latitude 34°N, between longitudes 15°E and 65°E, a difference of 50° in longitude. Distance PQ = (θ/360°)×2πR×cosα, where α=34° (the latitude) and θ=50° (longitude difference). PQ = (50/360)×(2×22/7)×6400×cos34° = 4632km (correct to 4 significant figures).

WAEC Mathematics 2013 Theory — Question 13

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