[OT]: looking for 3D geometry help!

[ijabbott]

So it's basically part of a regular, spherical octahedron?

I think the angle you are looking for is the dihedral angle between the base of an equilateral square pyramid (the base bisects a regular octahedron) and one of the sides, which is \( \tan^{-1}(\sqrt 2) \).

Alternatively, the dihedral angle between faces of a regular octahedron is \( \cos^{-1}(-\frac{1}{3}) \), and the angle you want is half of that: \( \frac{\cos^{-1}(-\frac{1}{3})}{2} \).

At least I assume that's the angle you want because it is approx 54.74° whereas its complement is approx 35.26°.