**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.

Solution:

**(a)**

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

Longitude of *K*= (180^{o} – 40^{o}) W = 140^{o} W

Therefore, **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)

Distance from *P *to *R*

= (40 + 10) × 60 × cos 25^{o}

= 50 × 60 × cos 25^{o}

= **2718.92 n.m. **

(d)

Total distance travelled

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

= 2220 + 2718.92

= 4938.92 nautical miles

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