# Napier's Rules applied to spherical distance calculations

I was in the middle of writing the same old geographic distance calculation using the Haversine formula when it occurred to me: shouldn't there be simpler way to do this? Haversine is of course derived from the Law of Cosines. But in thinking about this problem, I ran across Napier's Rules for right-angled spherical triangles. It seems like Napier's Rules should apply, after all Latitude small circles and Longitude great circles always intersect at right angles, so you should be able to draw a right-angled spherical triangle where the hypotenuse connects any two points (unless the case devolves into a line or single point).

So if I'm applying Napier's Rules correctly, if our delta latitude and delta longitude in radians are $a$ and $b$ respectively, the angle $c$ should trace the arc between the two points. So Napier says:

$\sin(\pi-c) = \cos(a) \cos(b)$

Which should then simplify to:

$c = \arcsin(\cos(a) \cos(b))$

But when I try to verify this with a couple test points, the result doesn't match the Haversine formula. Is there a mistake in my (admittedly rusty) algebra or is my mistake in assuming I can apply Napier's Rules to this problem?

Napier's rule can be applied to solve problems in spherical geometry and I suggest you may apply your angles in such an equation. Sine of the middle part = product of cosines of opposite parts.So $\sin(90-c)=\cos a x \cos b$, considering a as the angle to be found. $\cos a = \cos c$ divided by $\cos b$. where the angle between 100 and 110 is $b$ , and the distance in terms of an angle between the centre of the circle 001 and 110 is $c$. Keep in mind the principles of stereographic projection while converting angles to distance and vice-versa.