Longitude and Latitude - SS3 Mathematics Past Questions and Answers - page 1
Find the distance between this pair of points on the earth surface: \(A(43{^\circ}N,\ \ 50{^\circ}E)\) and \(B(17{^\circ}S,\ 50{^\circ}E)\). Take \(R = 6,400km\)
6,550km
3,748km
3,412.90km
6,704.76
\[Angular\ difference\ = \ 43{^\circ} + 17{^\circ} = 60{^\circ}\]
\[Distance\ between\ the\ points = \frac{\theta}{360} \times 2\pi R\]
\(Distance\ between\ the\ points = \frac{60}{360} \times \frac{2}{1} \times \frac{22}{7} \times \frac{6400}{1} = 6,704.76\ km\)
Find the distance between this pair of points on the earth surface: \(P(56{^\circ}N,\ \ 43{^\circ}W)\) and \(Q(56{^\circ}N,\ \ 17{^\circ}E)\). Take \(R = 6,400km\) and \(\pi = 3.142\)
6,550km
3,748km
3,412.90km
6,704.76km
Given \(P(56{^\circ}N,\ \ 43{^\circ}W)\) and \(Q(56{^\circ}N,\ \ 17{^\circ}E)\):
\[d = \frac{\alpha}{360} \times 2\pi R\cos\theta\]
\[\alpha = 43{^\circ} + 17{^\circ} = 60{^\circ}\]
\[R = 6,400km\]
\[\pi = 3.142\]
\[\theta = 56{^\circ}\]
\(d = \frac{60}{360} \times 2 \times 3.142 \times 6400{\times cos}{56{^\circ}} = 3,748\ km\)
A ship sails from point \(A(40{^\circ}N,28{^\circ}E)\) to point \(B(40{^\circ}N,\ 25{^\circ}W)\) and then to point \(C(15{^\circ}S,\ 25{^\circ}W)\). Take the radius of the earth, \(R = 6400\ km\) and \(\pi = 3.142\). Calculate:
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The distance from \(A\) to \(B\) and then to \(C\) in nautical miles (\(1\ nautical\ mile\ = \ 1.86km\))
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The average speed of the ship in knots if the whole journey takes \(49\) hours (\(1\ nautical\ mile\ per\ hour = 1\ knot\))
note: nautical miles and\ knots are units associated with sea and air travel
Distance from Point \(A\ \)to \(B\ \)is along a small circle, latitude \(40{^\circ}N\), thus:
\(\alpha = 28{^\circ} + 25{^\circ} = 53{^\circ}\)
\[R = 6,400km\]
\[\pi = 3.142\]
\[\theta = 40{^\circ}\]
\[d = \frac{53}{360} \times 2 \times 3.142 \times 6400{\times cos}{40{^\circ}} = 4,535.69\ km\]
Distance from Point \(B\ \)to \(C\ \)is along a great circle, longitude \(25{^\circ}W\), thus:
\(Angular\ difference\ = \ 40{^\circ} + 15{^\circ} = 55{^\circ}\)
\[Distance\ between\ the\ points = \frac{\theta}{360} \times 2\pi R\]
\[Distance\ between\ the\ points = \frac{55}{360} \times \frac{2}{1} \times \frac{22}{7} \times \frac{6400}{1} = 6,146.03\ km\]
a. Total distance from point \(A\) to \(B\) to \(C\) = \(4,535.69\ km + \ 6,146.03\ km = 10,681.72\ km\)
Total distance in nautical miles = \(\frac{10,681.72}{1.86} = 5,742.86\ nautical\ miles\)
b. Average speed = \(\frac{distance}{time} = \frac{5,742.86}{49} = 117.20\ knots\) (\(1\ nautical\ mile\ per\ hour = 1\ knot\))