Longitude and Latitude - SS3 Mathematics Past Questions and Answers - page 1

\[Angular\ difference\ = \ 43{^\circ} + 17{^\circ} = 60{^\circ}\]

\[Distance\ between\ the\ points = \frac{\theta}{360} \times 2\pi R\]

\(Distance\ between\ the\ points = \frac{60}{360} \times \frac{2}{1} \times \frac{22}{7} \times \frac{6400}{1} = 6,704.76\ km\)

Given \(P(56{^\circ}N,\ \ 43{^\circ}W)\) and \(Q(56{^\circ}N,\ \ 17{^\circ}E)\):

\[d = \frac{\alpha}{360} \times 2\pi R\cos\theta\]

\[\alpha = 43{^\circ} + 17{^\circ} = 60{^\circ}\]

\[R = 6,400km\]

\[\pi = 3.142\]

\[\theta = 56{^\circ}\]

\(d = \frac{60}{360} \times 2 \times 3.142 \times 6400{\times cos}{56{^\circ}} = 3,748\ km\)

 

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 = \(\frac{distance}{time} = \frac{5,742.86}{49} = 117.20\ knots\) (\(1\ nautical\ mile\ per\ hour = 1\ knot\))