WAEC 2016 · Paper 2 · Q7

$X(60^\circ\text{N}, 12^\circ\text{E})$ and $Y(60^\circ\text{N}, 42^\circ\text{E})$ are points on the earth's surface.

(a)
Illustrate this information in a diagram.
Model answer
A clear sketch is enough (it need not be to scale), but it must show every given fact: draw the earth with centre $O$ , the equator, and the circle of latitude $60^\circ$ N, with $X$ and $Y$ on it. The $60^\circ$ angle is at $O$ , from the equator plane up to the latitude. A second sketch of the $60^\circ$ N circle seen from above helps: its radius is $6400\cos 60^\circ = 3200$ km, and $X$ and $Y$ are $42^\circ - 12^\circ = 30^\circ$ apart at its centre. The chord $XY$ is a straight line; the distance "along the common latitude" is the arc.
(b)
Taking $\pi = 3.142$ and the radius of the earth $= 6400\text{ km}$ , calculate the: (i) length of the chord $XY$ , correct to the nearest 10 km; (ii) angle that the chord $XY$ subtends at the centre of the earth, correct to one decimal place; (iii) distance between $X$ and $Y$ along their common latitude.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Draw the earth with the equator and the circle of latitude $60^\circ$ N.
Mark $X$ at $12^\circ$ E and $Y$ at $42^\circ$ E on that circle, and the angle $60^\circ$ at the centre of the earth up to the latitude.
The circle of latitude $60^\circ$ N has radius $6400\cos 60^\circ = 3200$ km, and the difference in longitude is $42^\circ - 12^\circ = 30^\circ$ .

In the circle of latitude, the chord $XY$ subtends $30^\circ$ at its centre: $XY = 2 \times 3200 \times \sin 15^\circ$
$\approx 1656.4$ km, which is 1660 km to the nearest 10 km.

The same chord seen from the centre of the earth (radius 6400 km): $2 \times 6400 \times \sin\frac{\alpha}{2} = 1656.4$ , so $\sin\frac{\alpha}{2} \approx 0.1294$ , $\frac{\alpha}{2} \approx 7.44^\circ$ and $\alpha \approx 14.9^\circ$ .

Along the latitude: $\frac{30}{360} \times 2 \times 3.142 \times 3200 \approx 1675.73$ km.