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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.

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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:

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

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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

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

according to the link above

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