What equation calculates degrees at a point on a sphere where a tetrahedral triangle touches from the inside?

Question

Below is an answer received previously, from a similar question.

It is arcsin(1/3). The center of a tetrahedron is 1/4 of the way along any of its altitudes. Therefore, the radius of the circumscribing sphere is the remaining 3/4. Since the base of the tetrahedron extends 1/4 of the altitude below the center, the sine of the latitude must be 1/4 divided by 3/4, or 1/3. That gives about 19.5 degrees.

Incidentally, the Great Red Spot is at 22 degrees south latitude, not 19.5. It's probably just coincidence, since there are not two other similar storms separated by 120 degrees of longitude from the GRS.

Answer 1

In a symmetric  tetrahedron (with 4 triangles) if the side is '1',

diameter of 'circumsphere' is sqrt(3/2). If one vertex is pole the other three vertices will be 109d 28' 16.29" away (=> 19d 28' 16.29" S, a distance of 118385 feet on the arc of great circle).

The equation is simple

a^2 = 2{r^2[1 - cos A]}

where

a=side of tetrahedron,

A=angle at center of tetrahedron =109d 28' 16.29"

r=radius of circumshere= sqrt[(3/2)r] /2.

If it is Jovian great red spot you're talking about, I hope the flattening at poles is considered to arrive at 22deg.