Dihedral angles between tetrahedron faces from triangles' angles at the tip

This is to address one of the comments in how to derive the formula for the dihedral angles.

Given a tip of a tetrahedron, label the 3 edges attached to it as $1, 2, 3$ in such a way when you view the tip within the tetrahedron, the edges $1, 2, 3$ are arranged in counterclockwise manner.

Let $\vec{v}_1$, $\vec{v}_2$, $\vec{v}_3$ be the 3 unit vectors pointing away from the tip along the direction of edge $1, 2, 3$ respectively. Let us look at one of edge, say edge $2$. It is in contact with two faces. One bounded by vectors $\vec{v}_1$ and $\vec{v}_2$. Another bounded by vectors $\vec{v}_2$ and $\vec{v}_3$. The inward normal vectors of these two faces are given by:

$$ \vec{n}_{12} = \frac{\vec{v}_1 \times \vec{v}_2}{|\vec{v}_1 \times \vec{v}_2|} \,\,\,\text{ and }\,\,\, \vec{n}_{23} = \frac{\vec{v}_2 \times \vec{v}_3}{|\vec{v}_2 \times \vec{v}_3|} $$ In terms of the normal vectors, the dihedral angle at edge $2$, $\phi_2$, satisfies:

$$\begin{align} \cos(\phi_2) &= -\vec{n}_{12}\cdot\vec{n}_{23}\\ &= -\frac{( \vec{v}_1 \times \vec{v}_2 )\cdot(\vec{v}_2 \times \vec{v}_3)}{ |\vec{v}_1 \times \vec{v}_2||\vec{v}_2 \times \vec{v}_3|}\\ &= -\frac{(\vec{v}_1\cdot\vec{v}_2)(\vec{v}_2\cdot\vec{v}_3) - (\vec{v}_1\cdot\vec{v}_3)|\vec{v}_2|^2}{|\vec{v}_1 \times \vec{v}_2||\vec{v}_2 \times \vec{v}_3|}\\ &= \frac{\vec{v}_1\cdot\vec{v}_3 - (\vec{v}_1\cdot\vec{v}_2)(\vec{v}_2\cdot\vec{v}_3)}{|\vec{v}_1 \times \vec{v}_2||\vec{v}_2 \times \vec{v}_3|}\tag{*} \end{align}$$ Let $\theta_{ij}$ be the vertex angle between edge $i$ and $j$. We can simplify R.H.S of (*) to get: $$\cos(\phi_2) = \frac{\cos(\theta_{13}) - \cos(\theta_{12})\cos(\theta_{23})}{\sin(\theta_{12})\sin(\theta_{23})}$$

The formula of other dihedral angles can be derived in same manner.

With the tetrahedron having three triangular faces $A$, $B$, and $C$ with vertex angles $a$, $b$, and $c$, respectively, and dihedral angles between each face $\angle AB$, $\angle BC$, and $\angle AC$, the dihedral angles are given by:

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

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

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

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