**Question 2:**

*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.

**(a)**State the location of point

*K.*

**(b)**

*Q*is 2220 nautical miles from

*P*, measured along the same meridian.

Calculate the value of θ.

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

*P*due west to

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

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

*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.

**(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

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

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