WAEC 2010 · Paper 2 · Q12

Longitude & latitude

$P$ (lat $40^\circ$ N, long $18^\circ$ W) and $Q$ (lat $40^\circ$ N, long $78^\circ$ W) are two cities on the surface of the earth. Calculate the: $\left[\text{Take }\pi = \frac{22}{7}\text{ and the radius of the earth} = 6400\text{ km}\right]$

(a)
radius of the parallel of latitude on which $P$ and $Q$ lie, correct to the nearest $10\text{ km}$ ;
(b)
length of the minor arc $PQ$ , correct to the nearest $100\text{ km}$ ;
(c)
vertical distance between the centre of the earth and the centre of the small circle on which $P$ and $Q$ lie, correct to the nearest km.

Worked solution (try it first)

(a)

The radius of the parallel of latitude $\theta$ is $R\cos\theta$ : $r = 6400 \cos 40^\circ$
$= 6400 \times 0.7660$
$= 4902.7\text{ km}$ .
To the nearest $10\text{ km}$ , $r = 4900\text{ km}$ .

(b)

Both places are west, so the difference in longitude is $78^\circ - 18^\circ = 60^\circ$ .
Arc $PQ = \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 4902.7$
$= 5136\text{ km}$ .
To the nearest $100\text{ km}$ , the arc is $5100\text{ km}$ .

(c)

In the cross-section, the centre $X$ of the small circle is directly above the centre $O$ of the earth, and $OP$ makes $40^\circ$ with the equator, so $|OX| = R \sin 40^\circ$ .
$|OX| = 6400 \times 0.6428$
$= 4114\text{ km}$ to the nearest km.

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