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].

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An aeroplane flies due North from a town T on the equator at a speed of 950km per hour ...

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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

π = \frac{22}{7}

; Radius of the earth = 6400km].

Explanation

(a)
(b)(i)

S p e e d = \frac{d i s t a n c e}{t i m e}

950 = \frac{d}{4} ⟹ d = 950 \times 4

d = 3800 k m

d = \frac{θ}{360} \times 2 π r

3800 = \frac{θ}{360} \times 2 \times \frac{22}{7} \times 6400

3800 = \frac{θ \times 281600}{2520}

θ = \frac{2520 \times 3800}{281600}

θ = 34.01 °

θ = 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{θ}{360} \times 2 π r

where

r = R \cos θ

d = \frac{θ}{360} \times 2 π R \cos θ

\frac{50}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 34.01 °

4631.53 k m

≊ 4632 k m

(to 4 significant figures).

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