Prolly just tetrahedral (not sure how to prove this). So $\cos \frac{\alpha}{2}=\frac{1}{3}$. This means $\cos\alpha=-\frac{7}{9}$

the result is

Oh I see, it appears I did my tetrahedron geometry wrong. It still remains true that they are tetrahedral angles. Let's say we have a tetrahedron of inradius $1$ and circumradius $3$. Then the circumradius of one of the faces is $\sqrt{8}$ and the inradius is $\sqrt{2}$. So the distance from the center of the tetrahedron to the midpoint of an edge is $\sqrt{3}$. Then $\cos\frac{\alpha}{2}=\frac{\sqrt{3}}{3}$ and therefore $\cos\alpha=-\frac{1}{3}$