Chamfered Tetrahedron Question

Feb 2020
1
0
San Diego
Does anybody know the internal angles of the irregular (bi-mirror-symmetric) hexagons in the chamfered tetrahedron also referred to as the alternate truncated cube? I have searched the web far and wide!
I do know that the internal angles of a hexagon, whether regular or irregular add up to 720 and that these ‘bi-mirror-symmetric’ irregular are elongated in the process of chamfering a tetrahedron or likewise alternate truncating a cube, although I cannot find what the lengths of the edges are in order to calculate the internal angles of the irregular hexagons!
Please excuse my ignorance, as I am new to this and may have exact wording wrong!
Or if someone can direct me to a forum or a Goldberg Polyhedral wiz I could find out!
Thank you!