**Question 1:**

Diagram below shows four points

*P*,*Q*,*R*and*M*, on the surface of the earth.*P*lies on longitude of 70^{o}W.*QR*is the diameter of the parallel of latitude of 40^{o }N.*M*lies 5700 nautical miles due south of*P*.**(a)**Find the position of

*R*.

**(b)**Calculate the shortest distance, in nautical miles, from

*Q*to

*R*, measured along the surface of the earth.

**(c)**Find the latitude of

*M*.

**(d)**An aeroplane took off from

*R*and flew due west to

*P*along the parallel of latitude with an average speed of 660 knots.

Calculate the time, in hours, taken for the flight.

*Solution:***(a)**

Latitude of

*R*= latitude of*Q*= 40^{o}NLongitude of

*Q*= (70^{o}– 25^{o}) W = 45^{o}WLongitude of

*R*= (180^{o}– 45^{o}) E = 135^{o}ETherefore,

**position of***R*= (40^{o}N, 135^{o}E).

**(b)**

Shortest distance from

*Q*to*R*= (180 – 40 – 40) x 60

= 100 × 60

=

**6000 nautical miles**

**(c)**

**(d)**

**Question 2:**

*P*(25

^{o}N, 35

^{o}E),

*Q*(25

^{o}N, 40

^{o}W),

*R*and

*S*are four points on the surface of the earth.

*PS*is the diameter of the common parallel of latitude 25

^{o}N.

(a) Find the longitude of

*S.*

(b)

*R*lies 3300 nautical miles due south of

*P*measured along the surface of the earth.

Calculate the latitude of

*R*.

(c) Calculate the shortest distance, in nautical mile, from

*P*to

*S*measured along the surface of the earth.

(d) An aeroplane took off from

*R*and flew due north to

*P.*Then, it flew due west to

*Q*.

The total time taken for the whole flight was 12 hours 24 minutes.

(i) Calculate the distance, in nautical mile, from

*P*due west

*Q*measured along the common parallel of latitude.

(ii) Calculate the average speed, in knot, of the whole flight.

Solution:

Solution:

**(a)**

Longitude of

*S*= (180

^{o}– 35

^{o}) W = 145

^{o}W

**(b)**

$\begin{array}{l}\angle POR=\frac{3300}{60}\\ \text{}={55}^{o}\\ \text{Latitudeof}R={\left(55-25\right)}^{o}\\ \text{}={30}^{o}S\end{array}$

**(c)**

Shortest distance from

*P*to

*S*

= (65 + 65) x 60

= 130 x 60

= 7800 nautical miles

**(d)(i)**Distance of

*PQ*

= (35 + 40) x 60 x cos 25

^{o}= 75 x 60 x cos 25

^{o}= 4078.4 nautical miles

$\begin{array}{l}\left(\text{ii}\right)\\ \text{Totaldistancetravelled}\\ RP+PQ\\ =3300+4078.4\\ =7378.4\text{nauticalmiles}\\ \\ \text{Averagespeed=}\frac{\text{Totaldistancetravelled}}{\text{Timetaken}}\\ \text{}=\frac{7378.4}{12.4}\leftarrow \overline{)\text{12hours24min}=12+\frac{12}{60}=12+0.4}\\ \text{}=595.0\text{knot}\end{array}$