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

$$Angulardifference=43{}^{\circ }+17{}^{\circ }=60{}^{\circ }$$

$$Distancebetweenthepoints=\frac{\theta }{360}\times 2\pi R$$

$Distancebetweenthepoints=\frac{60}{360}\times \frac{2}{1}\times \frac{22}{7}\times \frac{6400}{1}=6,704.76km$

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

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

$$\alpha =43{}^{\circ }+17{}^{\circ }=60{}^{\circ }$$

$$R=6,400km$$

$$\pi =3.142$$

$$\theta =56{}^{\circ }$$

$d=\frac{60}{360}\times 2\times 3.142\times 6400\times cos56{}^{\circ }=3,748km$

 

Distance from Point $A$to $B$is along a small circle, latitude $40{}^{\circ }N$, thus:

$\alpha =28{}^{\circ }+25{}^{\circ }=53{}^{\circ }$

$$R=6,400km$$

$$\pi =3.142$$

$$\theta =40{}^{\circ }$$

$$d=\frac{53}{360}\times 2\times 3.142\times 6400\times cos40{}^{\circ }=4,535.69km$$

Distance from Point $B$to $C$is along a great circle, longitude $25{}^{\circ }W$, thus:

$Angulardifference=40{}^{\circ }+15{}^{\circ }=55{}^{\circ }$

$$Distancebetweenthepoints=\frac{\theta }{360}\times 2\pi R$$

$$Distancebetweenthepoints=\frac{55}{360}\times \frac{2}{1}\times \frac{22}{7}\times \frac{6400}{1}=6,146.03km$$

a. Total distance from point $A$ to $B$ to $C$ = $4,535.69km+6,146.03km=10,681.72km$

Total distance in nautical miles = $\frac{10,681.72}{1.86}=5,742.86nauticalmiles$

b. Average speed = $\frac{distance}{time}=\frac{5,742.86}{49}=117.20knots$ ($1nauticalmileperhour=1knot$)