An equilateral triangular prism cuts $x-y$ plane forming triangle $ABC$ whose vertices $ A(\alpha_1,\alpha_2),B(\beta_1,\beta_2),C(\gamma_1,\gamma_2) $ make angle pairs on prism sides as indicated in parantheses.
Find a trigonometric relation between the six angles.
EDIT1:
Self Answer
A development of the prism is added below.
$\alpha_1+\beta_1 =\pi,$ by Euclid's parallel lines (cut by transversal, internal angles sum $\pi$) thm. Considering all three faces
$$ \alpha_1+\beta_1+\alpha_2+\beta_2+\alpha_3+\beta_3 =3\pi $$
Alternately, considering sum of all external angles in a convex polygon $ABCA=2 \pi$ to which, due to one re-entrant vertex, we need to add another $\pi$ so that this angle sum $ \Sigma \psi_i= 3 \pi$ in the common tangent plane.
One can use the spherical law of cosines to calculate the dihedral angles of the prism (all of which are $\pi/3$) in terms of the plane angles at $A$, $B$, $C$:
$$\frac12=\cos\frac{\pi}{3} = \frac{\cos A-\cos\alpha_1\cos\alpha_2}{\sin\alpha_1\sin\alpha_2} = \frac{\cos B-\cos\beta_1\cos\beta_2}{\sin\beta_1\sin\beta_2} = \frac{\cos C-\cos\gamma_1\cos\gamma_2}{\sin\gamma_1\sin\gamma_2} \tag{1}$$ so that $$\cos A = \frac12\sin\alpha_1 \sin\alpha_2 + \cos\alpha_1 \cos\alpha_2, \quad\text{etc} \tag{2}$$
Since $A+B+C=\pi$, we know $$1 - \cos^2 A - \cos^2B - \cos^2C - 2 \cos A \cos B \cos C = 0 \tag{3}$$
Substituting from $(2)$ into $(3)$ gives a relation in the $\alpha_i$, $\beta_i$, $\gamma_i$. I haven't found a "simple" representation of that relation, however, so I'll leave that as an exercise to the reader. $\square$