A plane flies due west from point A (x°N, 15°E) to point B (x°N, 75°W). If the distance between A and B is 5005 km, calculate the position of point B. [R=6370 km, π=22/7]

Calculate

Point A is ($x^\circ$N, $15^\circ$E).
Point B is ($x^\circ$N, $75^\circ$W).
The aircraft flew due West, so the latitude $x^\circ$ remains constant.
The difference in longitude, $\theta$, is the sum of the longitudes since one is East and the other is West.

$$\theta = 15^\circ + 75^\circ = 90^\circ$$

The distance AB is given as 5005 km.
The formula for distance along a parallel of latitude is $D = \frac{\theta}{360^\circ} \times 2\pi R \cos x^\circ$.
Given $R = 6370$ km and $\pi = \frac{22}{7}$.

$$5005 = \frac{90}{360} \times 2 \times \frac{22}{7} \times 6370 \times \cos x^\circ$$ $$5005 = \frac{1}{4} \times 2 \times \frac{22}{7} \times 6370 \times \cos x^\circ$$ $$5005 = \frac{1}{2} \times \frac{22}{7} \times 6370 \times \cos x^\circ$$ $$5005 = \frac{11}{7} \times 6370 \times \cos x^\circ$$ $$5005 = 11 \times 910 \times \cos x^\circ$$ $$5005 = 10010 \times \cos x^\circ$$ $$\cos x^\circ = \frac{5005}{10010}$$ $$\cos x^\circ = \frac{1}{2}$$ $$x^\circ = \cos^{-1}\left(\frac{1}{2}\right)$$ $$x^\circ = 60^\circ$$

So, the latitude of point B is $60^\circ$N.
The position of point B is:

$$\boxed{60^\circ\text{N, } 75^\circ\text{W}}$$

From point B ($60^\circ$N, $75^\circ$W), the aircraft flew due south. This means the longitude remains $75^\circ$W.
The aircraft flew for 3 hours 40 minutes at an average speed of 910 km/h.
Time taken for journey B to C = 3 hours 40 minutes = $3 + \frac{40}{60}$ hours = $3 + \frac{2}{3}$ hours = $\frac{11}{3}$ hours.
Distance BC = Speed $\times$ Time

$$\text{Distance BC} = 910 \text{ km/h} \times \frac{11}{3} \text{ h} = \frac{10010}{3} \text{ km}$$

The distance due south is along a meridian (a great circle). The formula for distance along a great circle is $D = \frac{\alpha}{360^\circ} \times 2\pi R$, where $\alpha$ is the difference in latitude.

$$\frac{10010}{3} = \frac{\alpha}{360} \times 2 \times \frac{22}{7} \times 6370$$ $$\frac{10010}{3} = \frac{\alpha}{360} \times 44 \times 910$$ $$\frac{10010}{3} = \frac{\alpha}{360} \times 40040$$ $$\alpha = \frac{10010}{3} \times \frac{360}{40040}$$ $$\alpha = \frac{10010}{3} \times \frac{360}{4 \times 10010}$$ $$\alpha = \frac{1}{3} \times \frac{360}{4}$$ $$\alpha = \frac{1}{3} \times 90^\circ$$ $$\alpha = 30^\circ$$

The difference in latitude is $30^\circ$. Since the aircraft flew south from $60^\circ$N, the new latitude is $60^\circ - 30^\circ = 30^\circ$N.
The position of point C is:

$$\boxed{30^\circ\text{N, } 75^\circ\text{W}}$$