**Question 3:**

*P*(25

^{o}S, 40

^{o}E),

*Q*(θ

^{o}N, 40

^{o}E),

*R*(25

^{o}S, 10

^{o}W) and

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

*PK*is the diameter of the earth.

**State the location of point**

(a)

(a)

*K.*

(b)

(b)

*Q*is 2220 nautical miles from

*P*, measured along the same meridian.

Calculate the value of θ.

**Calculate the distance, in nautical mile, from**

(c)

(c)

*P*due west to

*R*, measured along the common parallel of latitude.

**An aeroplane took off from**

(d)

(d)

*Q*and flew due south to

*P.*Then, it flew due west to

*R*. The average speed of the aeroplane was 600 knots.

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

Solution:

Solution:

**(a)**

As

*PK*is the diameter of the earth, therefore latitude of*K*= 25^{o}NLongitude of

*K*= (180^{o}– 40^{o}) W = 140^{o}WTherefore,

**location of***K*= (25^{o}N, 140^{o}W).

**(b)**

Let the centre of the earth be

*O.*
$\begin{array}{l}\angle POQ=\frac{2220}{60}\\ \text{}={37}^{o}\\ {\theta}^{o}={37}^{o}-{25}^{o}={12}^{o}\\ \therefore \text{The value of}\theta \text{is 12}\text{.}\end{array}$

(c)

(c)

Distance from

*P*to*R*= (40 + 10) × 60 × cos 25

^{o}= 50 × 60 × cos 25

^{o}=

**2718.92 n.m.**

(d)

(d)

Total distance travelled

= distance from

*Q*to*P*+ distance from*P*to*R*= 2220 + 2718.92

= 4938.92 nautical miles

$\begin{array}{l}\text{Time taken =}\frac{\text{total distance from}Q\text{to}R}{\text{average speed}}\\ \text{}=\frac{4938.92}{600}\\ \text{}=8.23\text{hours}\end{array}$

$\begin{array}{l}\text{Time taken =}\frac{\text{total distance from}Q\text{to}R}{\text{average speed}}\\ \text{}=\frac{4938.92}{600}\\ \text{}=8.23\text{hours}\end{array}$

**Question 4:**

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)**

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

**(d)**

$\begin{array}{l}\text{Totaldistancetravelled}\\ BA+AK\\ =5969+4440\\ =10409\text{nauticalmiles}\\ \\ \text{Totaltimetaken=}\frac{\text{Totaldistancetravelled}}{\text{Averagespeed}}\\ \text{}=\frac{10409}{450}\\ \text{}=23.13\text{hours}\end{array}$