For any regular polyhedron (tetrahedron, cube, octahedron, dodecahedron, icosahedron, ...), how do I calculate an angle from the center of the regular polyhedron to the center of each face? Obviously, the angle decreases with the increasing number of faces. What is the formula, when the number of faces n is the variable?
For cube, n= 6, the angle is 90 degrees:
On
Théophile's answer is probably the more practical one, but here is a nice technique that works for all regular polytopes in all dimension:
If $P$ is a regular polytope, let $G^\circ$ be the edge-graph of its polar dual (e.g., if $P$ is the cube, then $G^\circ$ is the edge-graph of the octahedron). Alternatively, $G^\circ$ is the graph in which the vertices are the facets of $P$, and two are adjacent in $G^\circ$ if they are incident.
Let $\theta_2$ be the second-largest eigenvalue of the adjacency matrix of $G^\circ$, then the desired angle $\alpha$ is
$$\alpha=\cos^{-1}\!\Big(\frac{\theta_2}{\mathrm{deg}(G^\circ)}\Big),$$
where $\mathrm{deg}(G^\circ)$ is the vertex-degree of $G^\circ$.
On
A full outline on corner angles of polygons, dihedral angles of polyhedra, and dihedral angles of polychora can be found here: https://bendwavy.org/klitzing/explain/dihedral.htm
It also includes the asked for supplements as well as an explicite listing of all relevant cases, even the acrons.
--- rk
What you're describing is the supplementary angle of the dihedral angle of the solid. This latter angle $\theta$ is the interior angle between two connected faces, with
$$\sin\frac\theta2 = \frac{\cos\frac\pi q}{\sin\frac\pi p}$$
for a solid with Schläfli symbol $\{p,q\}$. See the full chart in the Wikipedia article on Platonic solids.
Therefore, taking the supplement, you're looking for $$2\pi - \theta.$$