**Question 5:**

*A*(53

^{o}N, 84

^{o}E),

*B*(53

^{o}N, 25

^{o}W),

*C*and

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

*AC*is the diameter of the parallel of latitude 53

^{o}N.

**(a)**State the location of

*C.*

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

*A*to

*C*measured along the surface of the earth.

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

*A*due east

*B*measured along the common parallel of latitude.

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

*B*and flew due south to

*D.*The average speed of the flight was 420 knots and the time taken was 6½ hours.

Calculate

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

*B*to

*D*measured along the meridian.

**(ii)**the latitude of

*D*.

*Solution:***(a)**

Latitude of

*C*= 53

^{o}N

Longitude of

*C*= (180

^{o}– 84

^{o}) E = 96

^{o}E

Therefore location of

*C*= (53

^{o}N, 96

^{o}E)

**(b)**

Shortest distance from

*A*to

*C*

= (180 – 53 – 53) x 60

= 74 x 60

= 4440 nautical miles

**(c)**

Distance from

*A*to

*B*

= (84 – 25) x 60 x cos 53

^{o}= 59 x 60 x cos 53

^{o}= 2130.43 nautical miles

**(d)**

$\begin{array}{l}\left(\text{i}\right)\\ \text{Distancetravelfrom}B\text{to}D\\ =420\times 6\frac{1}{2}\leftarrow \overline{)\begin{array}{l}\text{Distancetravelled}\\ \text{=averagespeed}\times \text{timetaken}\end{array}}\\ =2730\text{nauticalmiles}\\ \\ \left(\text{ii}\right)\\ \text{Differenceinlatitudebetween}B\text{to}D\\ =\frac{2730}{60}\\ ={45.5}^{\text{o}}\\ \\ \therefore \text{Latitudeof}D=\left({53}^{\text{o}}-{45.5}^{\text{o}}\right)N\\ \text{}={7.5}^{\text{o}}N\end{array}$

**Question 6:**

Diagram below shows the locations of points

*P*,

*Q*,

*R*,

*A*,

*K*and

*C*, on the surface of the earth.

*O*is the centre of the earth.

(a) Find the location of

*A*.

(b) Given the distance

*QR*is 3240 nautical miles, find the longitude of

*Q*.

(c) Calculate the distance, in nautical miles of

*KA*, measured along the common parallel latitude.

(d) An aeroplane took off from

*A*and flew due west to

*K*along the common parallel of latitude. Then, it flew due south to

*Q*. The average speed of the aeroplane was 550 knots.

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

*Solution:***(a)**

Longitude of

*A*= (180

^{o}– 15

^{o}) = 165

^{o}E

Latitude of

*A*= 50

^{o}N

**position of**

*A*= (50^{o}N, 165^{o}E).**(b)**

$\begin{array}{l}\angle QOR=\frac{3240}{60}\\ \text{}={54}^{o}\\ \therefore \text{Longitudeof}Q=({165}^{o}-{54}^{o})E\\ \text{}={111}^{o}E\end{array}$

**(c)**

Distance of

*KA*

= 54 x 60 x cos 50

^{o}= 2082.6 nautical miles

**(d)**

$\begin{array}{l}\text{Totaldistance}=AK+KQ\\ \text{}=2082.6+\left(50\times 60\right)\\ \text{}=5082.6\text{nauticalmiles}\\ \\ \text{Totaltime}=\frac{5082.6}{550}\\ \text{}=9.241\text{hours}\end{array}$