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!