From Toronto To Kabul

The geodesic on a sphere is the great circle distance, which means a circle on the surface of the sphere whose center coincides with the center of the sphere.

Given this fact, the distance, $d$ , will be equal to the central angle between the two points, $\theta$ , times the radius of the sphere, $r$ . In this case, that gives us $d = 6371 \times \theta$ (assuming the angle is expressed in radians).

To find the central angle between two points on the surface of the sphere using latitude $(L)$ and longitude $(\lambda)$ , we use the spherical law of cosines:

$\theta = \arccos \left( \sin L_1 \sin L_2 + \cos L_1 \cos L_2 \cos (\lambda_1 - \lambda_2) \right)$

Since the latitude and longitude are currently in D $^\circ$ M'S'' (degrees, minutes, seconds) we'll want to convert them to radians (turning S and W into negative numbers to represent the different between them appropriately):

*Toronto: * $L_1 = 43^\circ39'\text{N} = 43^\circ + \frac{39}{60}^\circ = 43.65^\circ \approx .7618362 \text{ radians} \\ \lambda_1 = 79^\circ23'\text{W} = -(79^\circ + \frac{23}{60} = -79.38\overline{33}^\circ \approx -1.3855005 \text{ radians}$

*Kabul: * $L_2 = 34^\circ31'\text{N} = 34^\circ + \frac{31}{60}^\circ = 34.51\overline{66}^\circ \approx .60242948 \text{ radians} \\ \lambda_2 = 69^\circ12'\text{E} = 69^\circ + \frac{12}{60} = 69.2^\circ \approx 1.2077678 \text{ radians}$

Applying the above to the spherical law of cosines:

$\begin{aligned} \theta &= \arccos \left( \sin (.76183) \sin (.602429) + \cos (.76183) \cos (.602429)) \cos (2.59327) \right) \\ &\approx 1.688734 \text{ radians} \end{aligned}$

Thus the geodesic between these two cities is $d \approx 6371 \times 1.688734 \approx 10759$ km.

Note: For those who chose 10783 because Wolfram (or otherwise) told you so, the discrepancy is because Earth isn't perfectly spherical.

Use Haversine formula , set distance, $d$ as the subject

$\large d = 2r \arcsin \left ( \sqrt{ \sin^2 \left ( \frac {\Delta \phi}{2} \right ) + \cos ( \phi_1 ) \cos ( \phi_2 ) \sin^2 \left ( \frac {\Delta \lambda }{2} \right ) } \right )$

$r = 6371$ , $\Delta \phi = 43^\circ 39' - 34^\circ 31'$ , $\Delta \lambda = 72^\circ 23' + 69^\circ 12'$

Convert measurements to radians and solve it, you should get $\boxed{10759}$ round up to the nearest integer.