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

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.


Therefore, position of A = (50^o N, 165^o E).

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

(c)
Distance of KA
= 54 x 60 x cos 50^o
= 2082.6 nautical miles

(d)
$\begin{array}{l}\text{Total distance}=AK+KQ\\ =2082.6+\left(50\times 60\right)\\ =5082.6\text{ nautical miles}\\ \\ \text{Total time }=\frac{5082.6}{550}\\ =9.241\text{ hours}\end{array}$

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.


(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)
$\begin{array}{l}\left(\text{i}\right)\\ \text{Distance travel from }B\text{ to }D\\ =420\times 6\frac{1}{2}\leftarrow \boxed{\begin{array}{l}\text{ Distance travelled}\\ \text{ = average speed }\times \text{ time taken }\end{array}}\\ =2730\text{ nautical miles}\\ \\ \left(\text{ii}\right)\\ \text{Difference in latitude between }B\text{ to }D\\ =\frac{2730}{60}\\ ={45.5}^{\text{o}}\\ \\ \therefore \text{Latitude of }D=\left({53}^{\text{o}}-{45.5}^{\text{o}}\right)N\\ ={7.5}^{\text{o}}N\end{array}$

Question 4:
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:
(a)
Longitude of S = (180^o – 35^o) W = 145^o W

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

(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

$\begin{array}{l}\left(\text{ii}\right)\\ \text{Total distance travelled}\\ RP+PQ\\ =3300+4078.4\\ =7378.4\text{ nautical miles}\\ \\ \text{Average speed =}\frac{\text{Total distance travelled}}{\text{Time taken}}\\ =\frac{7378.4}{12.4}\leftarrow \boxed{\text{12 hours 24 min}=12+\frac{12}{60}=12+0.4}\\ =595.0\text{ knot}\end{array}$

Question 3:
Diagram below shows the locations of points A (34^o S, 40^o W) and B (34^o S, 80^o E) which lie on the surface of the earth. AC is a diameter of the common parallel of latitude 34^o S.


(a) State the longitude of C.

(b) Calculate the distance, in nautical mile, from A due east to B, measured along the common parallel of latitude 34^o S.

(c) K lies due north of A and the shortest distance from A to K measured along the surface of the earth is 4440 nautical miles.
Calculate the latitude of K.

(d) An aeroplane took off from B and flew due west to A along the common parallel of latitude. Then, it flew due north to K. The average speed for the whole flight was 450 knots.
Calculate the total time, in hours, taken for the whole flight.

Solution:
(a)
Longitude of C = (180^o – 40^o) E = 140^o E

(b)
Distance of AB
= (40 + 80) x 60 x cos 34^o
= 120 x 60 x cos 34^o
= 5969 nautical miles

(c)
$\begin{array}{l}\angle AOK=\frac{4440}{60}\\ ={74}^o\\ \text{Latitude of }K={\left(74-34\right)}^{o}N\\ ={40}^{o}N\end{array}$

(d)
$\begin{array}{l}\text{Total distance travelled}\\ BA+AK\\ =5969+4440\\ =10409\text{ nautical miles}\\ \\ \text{Total time taken =}\frac{\text{Total distance travelled}}{\text{Average speed}}\\ =\frac{10409}{450}\\ =23.13\text{ hours}\end{array}$