# Formula for angle between faces based on angles of corners

I've got an irregular solid with a vertex at which three faces (and thus also 3 edges) come together. None of the angles involved are right angles. I know the angles of the corners, at that vertex, of the three faces. I need to find the angle between two of the faces (in a plane perpendicular to their shared edge; the plane of the third face isn't perpendicular, so I can't just use its corner angle).

What is the formula to relate the three face-corner angles to the angle between two of the faces?

Edit: I want this because I'm trying to fabricate an irregularly-shaped box to fit into a constrained space. I have college mathematics, including geometry, and calculus-based geometry, but excluding whatever spherical geometry is.

The answer comes from spherical trigonometry, but don't let the "spherical" part throw you.

Let $$a$$, $$b$$, $$c$$ be the "face-corner" angles (bounded by pairs of edges), and let $$A$$, $$B$$, $$C$$ be the "dihedral angles" (bounded by pairs of faces), with $$a$$ opposite $$A$$, $$b$$ opposite $$B$$, and $$c$$ opposite $$C$$.

The two Spherical Laws of Cosines relates these angles thusly:

\begin{align} \cos c &= \phantom{-}\cos a\cos b+\sin a \sin b \cos C \tag{1}\\[4pt] \cos C &= -\cos A\cos B+\sin A\sin B \cos c \tag{2} \end{align}

Note: If the plane containing face-angle $$c$$ is perpendicular to the edge along dihedral angle $$C$$, then $$a=b=90^\circ$$ so that $$(1)$$ reduces to $$\cos c=\cos C$$, so that $$c=C$$. This is consistent with OP's parenthetical about those angles matching in this situation.

FYI: There's also the Spherical Law of Sines:

$$\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C} \tag{3}$$

http://whistleralley.com/polyhedra/derivations.htm

3 faces A, B and C intersect at a point with corner angles a, b and c respectively

Angle between B and C

$$= \cos^{-1}\frac{ (\cos a - \cos b \cos c)}{(\sin b \sin c) }$$