Question 9 (12 marks):
A(25^o N, 35^o E), B(25^o N, 40^o W), C and D are four points which lie on the surface of the earth. AD is the diameter of the common parallel latitude 25^o N .
(a) Find the longitude of D.

(b) C lies 3300 nautical miles due south of A measured along the surface of the earth.
Calculate the latitude of C.

(c) Calculate the shortest distance, in nautical mile, from A to D measured along the surface of the earth.

(d) An aeroplane took off from C and flew due north to point A. 
The total time taken for the whole flight was 12 hours 24 minutes.

(i) Calculate the distance, in nautical mile, from A due west to B measured along the common parallel of latitude.

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


Solution:
(a) 
Longitude of D = (180^o – 35^o)W
= 145^oW

(b)
$\begin{array}{l}\angle AOC=\frac{3300}{60}={55}^o\\ \text{Latitude of }C\\ ={\left(55-25\right)}^{o}S\\ ={30}^{o}S\end{array}$

(c) 
Shortest distance of A to D
= (65^o + 65^o) × 60’
= 130^o × 60’
= 7800 nautical miles

(d)(i) 
Distance from A to B
= (35^o + 40^o) × 60’ × cos 25^o
= 75^o × 60’ × cos 25^o
= 4078.4 nautical miles

(d)(ii)
$\begin{array}{l}\text{Total distance travelled}\\ =CA+AB\\ =3300+4078.4\\ =7378.4\text{ nautical miles}\\ \\ \text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}\\ =\frac{7378.4}{12.4}\text{ knot}\\ =595.0\text{ knot}\end{array}$

Question 8 (12 marks):
Diagram 8 shows four points, G, H, I and J on the surface of the Earth. JI is the diameter of the parallel of latitude 50^o N. O is the centre of the Earth.

Diagram 8

(a) State the location of G.

(b) Calculate the shortest distance, in nautical mile, from I to J measured along the surface of the Earth.

(c) Calculate the shortest distance, in nautical mile, from G to H measured along the common parallel of latitude.

(d) An aeroplane took off from I and flew due south to point P. The average speed of the journey was 800 knots. The time taken for the flight was 5.25 hours.
Calculate the latitude of P.


Solution:
(a)
Location of G = (70^o S, 20^o W)

(b)
∠ JOI
= 180^o – 50^o – 50^o 
= 80^o
Distance of I to J
= 80^o × 60’
= 4800 nautical miles

(c)
Distance of G to H
= (20^o + 120^o) × 60’ × cos 70^o
= 140^o × 60’ × cos 70^o
= 2872.97 nautical miles

(d)
$\begin{array}{l}\text{Average speed}=\frac{\text{Total distance travelled}}{\text{Total time taken}}\\ 800=\frac{x}{5.25}\\ x=4200\text{ nautical miles}\\ I\text{ to }P=4200\\ \\ \text{Difference between parallel}=y\\ y\times 60=4200\\ y={70}^o\\ \\ \text{Thus, latitude of }P\\ ={70}^o-{50}^o\\ ={20}^{o}S\end{array}$

Question 7 (12 marks):
Diagram 7 in the answer space shows the locations of points J, L and M, which lie on the surface of the earth. O is the centre of the earth. The longitude of M is 30^o W. K is another point on the surface of the earth such that KJ is the diameter of the common parallel of latitude 45^o S.

(a)(i) Mark and label point K on Diagram 7 in the answer space.

(ii) Hence, state the longitude of point K.

(b) L lies due north of M and the shortest distance from M to L measured along the surface of the earth is 7500 nautical miles.
Calculate the latitude of L.

(c) Calculate the distance, in nautical mile, from K due east to M measured along the common parallel of latitude.

(d) An aeroplane took off from K and flew due east to M along the common parallel of latitude. The average speed of the aeroplane for the flight was 750 knots.
Calculate the total time, in hour, taken for the whole flight.



Answer:



Solution:
(a)(i)


(a)(ii)
Longitude of point K = 130^oW

(b)
$\begin{array}{l}\angle LOM\times 60=7500\\ \angle LOM=\frac{7500}{60}\\ \angle LOM={125}^o\\ \\ \text{Latitude of }L={125}^o-{45}^o\\ ={80}^{o}N\end{array}$

(c)
KM = (130^o – 30^o) × 60 × cos 45^o
= 4242.64 nautical miles

(d)
$\begin{array}{l}\text{Time}=\frac{\text{Distance}}{\text{speed}}\\ =\frac{4242.64}{750}\\ =5.66\text{ hours}\end{array}$