Please forgive me if this is the wrong kind of question, but I need someone to verify or refute my work.
One leg of a triangle has length, $b$ (base), resulting from angle theta swept out by a ray along the sphere's equator. If the sphere has radius $R$, then the length of $b$ is $\theta R$.
The vertical leg of a triangle has length, a (altitude), resulting from angle, phi, swept out by a ray along the sphere's longitude up from the equator. a and b make the right angle gamma (see picture below) If the sphere has radius $R$, then the length of a is $\phi \cdot R$
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the Pythagorean approximation in Euclidean space is hypotenuse, $c = \sqrt{a^2 + b^2}$ (just for fun, the plot is here:
http://www.wolframalpha.com/input/?i=plot+c+%3D+sqrt+%28a%5E2+%2B+b%5E2%29 )
but how do I calculate the exact solution? using the spherical law of cosines seems to be the way. https://en.wikipedia.org/wiki/Spherical_law_of_cosines
$$\cos(c) = \cos(a)\cos(b) + \sin(a)\sin(b)\cos(\gamma)$$
but $\gamma$ is $\pi/2$, so this simplifies to
$$\cos(c) = \cos(a)\cos(b)$$
and this simplifies to
$$c = \arccos(\cos(a)\cos(b))$$
but this plot is WAY different
http://www.wolframalpha.com/input/?i=plot+c+%3D+cos%5E-1%28cos%28a%29cos%28b%29%29
Where did I screw up?
Second part of question:
Now I realize that this law only works for spherical triangles, meaning that the circle defined by each limb of the triangle has to pass through the sphere's center. (the drawing does not make that clear, on purpose because the next question is how to calculate c if the triangle is not a spherical triangle, i.e., one of the limbs of the triangle lies not on a meridian but a line of latitude that is not the equator)