P and Q are two points on latitude 55°N and their longitudes are 33°W and 20°E respectively. Calculate the distance between P and Q measured along (a) the p...

The supplied school reference is inconsistent with the question. The question gives longitudes \(33^\circ\text{W}\) and \(20^\circ\text{E}\), so the difference in longitude is:

\[ 33^\circ+20^\circ=53^\circ \]

The use of \(23^\circ\text{E}\), giving \(56^\circ\), does not match the stated data. Therefore the distances must be calculated using \(53^\circ\), not \(56^\circ\).

(a) Distance along the parallel of latitude

A parallel is a circle smaller than the Equator. At latitude \(55^\circ\), its radius is:

\[ r=R\cos55^\circ \] \[ r=6400\cos55^\circ \]

The required arc is \(\frac{53}{360}\) of this parallel’s circumference:

\[ \begin{aligned} d &=\frac{53}{360}\times 2\pi R\cos55^\circ\\ &=\frac{53}{360}\times2\times\frac{22}{7}\times6400\cos55^\circ\\ &\approx\frac{53}{360}\times23075.1\\ &\approx 3397\text{ km} \end{aligned} \]

Distance along the parallel of latitude: \(\boxed{3397\text{ km}}\)

(b) Distance along a great circle

The shortest route over the Earth’s surface is an arc of a great circle. Let \(\theta\) be the angle at the Earth’s centre subtended by \(P\) and \(Q\). For two points at the same latitude:

\[ \cos\theta = \sin^2 55^\circ+\cos^2 55^\circ\cos53^\circ \] \[ \begin{aligned} \cos\theta &\approx (0.8192)^2+(0.5736)^2(0.6018)\\ &\approx0.8690 \end{aligned} \] \[ \theta\approx\cos^{-1}(0.8690)\approx29.66^\circ \]

The great-circle distance is therefore:

\[ \begin{aligned} d &=\frac{\theta}{360}\times2\pi R\\ &=\frac{29.66}{360}\times2\times\frac{22}{7}\times6400\\ &\approx3314\text{ km} \end{aligned} \]

Distance along the great circle: \(\boxed{3314\text{ km (approximately)}}\)

The great-circle distance is slightly shorter than the distance along the parallel because a parallel other than the Equator is not a great circle. In examination questions, first find the longitude difference carefully: longitudes on opposite sides of the Greenwich meridian are added.