Formula for angle between faces based on angles of corners

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:

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

Tags:

3D

Trigonometry

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