<s>[INST] Dodecahedron, angle between edge and face. In an effort to build a dodecahedron frame in Fusion360 I need to know some of the angles. Looking around I found out that the angle between an edge and a face on a regular dodecahedron is $121.7^\circ$ but I couldn't find the mathematical formula nor the way to calculate this angle. The formula is needed so the exact angle can be used so the simulation is precise. Can anyone help?

 [/INST] This angle can be derived from the dihedral angle $\delta = \arccos (-1/\sqrt{5})$ of the dodecahedron using a vector method. (The dihedral angle $\delta$ is derived during construction of the dodecahedron - eg see this answer to How does this proof of the regular dodecahedron's existence fail?).
Consider a group of three adjacent pentagon faces as shown in Fig 1, in which pentagon $A$ is in the plane of the page, and $B$ and $C$ are tilted forwards to meet $A$ at the dihedral angle $\delta$. Each pentagon makes dihedral angle $\delta$ with the other two. The angle required is $\alpha$ between edge $e$ and the plane of pentagon $A$. If $\beta$ is the angle between the unit normal $\underline{\mathbf{u}}_A = \underline{\mathbf{k}}$ to $A$ and the unit vector $\underline{\mathbf{e}}$ along $e$, then $\alpha = \beta + 90$. Since edge $e$ is the intersection line of the planes $B$ and $C$, then if $\underline{\mathbf{u}}_B$ is a unit normal to $B$ and $\underline{\mathbf{u}}_C$ is a unit normal to $C$ then noting that the angle between $\underline{\mathbf{u}}_B$ and $\underline{\mathbf{u}}_C$ is the dihedral angle $\delta$, we have :
$$\underline{\mathbf{e}} = \frac{1}{|\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C|} \;  (\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C) = \frac{1}{\sin \delta} \; (\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C).
$$





Setting up an $xyz$-axes system as shown in Figs 1 and 2 we have :
\begin{eqnarray*}
\underline{\mathbf{u}}_B & = & -\sin \epsilon\; \underline{\mathbf{i}} + \cos \epsilon\; \underline{\mathbf{k}} \\
\Rightarrow \hspace{1em} \underline{\mathbf{u}}_B & = & -\sin \delta\; \underline{\mathbf{i}} - \cos \delta\; \underline{\mathbf{k}}
\end{eqnarray*}
Since pentagon $C$ is obtained from pentagon $B$ by rotating $B$ by $+72^\circ$ about the $z$-axis, we can obtain $\underline{\mathbf{u}}_C$ from $\underline{\mathbf{u}}_B$ by rotating $\underline{\mathbf{u}}_B$ by $72^\circ$ about the $z$-axis. (Note we are rotating vectors, which possess only direction and magnitude - they have no starting or ending point, in contrast to directed line segments, but can be 'placed' anywhere and will always point in the same direction). This rotation is a 2D rotation in the $xy$-plane and leaves the $\underline{\mathbf{k}}$ component unchanged, resulting in :
$$
\underline{\mathbf{u}}_C = -\sin \delta \cos 72\; \underline{\mathbf{i}} - \sin \delta \sin 72\; \underline{\mathbf{j}} - \cos \delta\; \underline{\mathbf{k}}
$$
Then :
$$
\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C = \sin^2 \delta \sin 72\; \underline{\mathbf{k}} -\sin \delta \cos \delta\; \underline{\mathbf{j}} + \sin \delta \cos \delta \cos 72\; \underline{\mathbf{j}} - \sin \delta \cos \delta \sin 72\; \underline{\mathbf{i}}
$$
and so :
$$
\cos \beta = \underline{\mathbf{k}} \cdot \underline{\mathbf{e}} = \frac{1}{\sin \delta} \; (\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C) \cdot \underline{\mathbf{k}} = \sin \delta \sin 72
$$
and the desired angle $\alpha$ is given by :
$$
\cos \alpha = \cos(\beta + 90) = -\sin \beta = -\sqrt{1 - \cos^2 \beta}.
$$
The 18-72-90$^\circ$ triangle (which comes from the diagonal of a regular pentagon) tells us :
$$
\sin 72 = \frac{\sqrt{4\Phi + 3}}{2\Phi}
$$
where $\Phi$ is the golden ratio $\frac{1 + \sqrt5}{2}$.
The dihedral angle $\delta$ satisfies $\sin^2 \delta = 4/5$ so making use of the identity $\Phi^2 = \Phi + 1$ we now have:
\begin{eqnarray*}
\cos^2 \beta & = & \sin^2 \delta \sin^2 72 = \frac{4}{5} \cdot \frac{4\Phi + 3}{4\Phi^2} = \frac{4\Phi + 3}{5\Phi + 5} \\
\Rightarrow 1 - \cos^2 \beta & = & \frac{\Phi + 2}{5\Phi^2} \\
\Rightarrow \cos \alpha & = & -\frac{1}{\Phi \sqrt{5}} \sqrt{\Phi + 2} = -\frac{1}{\Phi \sqrt{5}} \sqrt{\frac{5 + \sqrt{5}}{2}} = -\frac{2}{1 + \sqrt{5}} \sqrt{\frac{5 + \sqrt{5}}{10}} \\
& = & -\frac{1}{2} \sqrt{\frac{(\sqrt{5} - 1)^2 (5 + \sqrt{5})}{10}} = -\frac{1}{2} \sqrt{\frac{(6 - 2\sqrt{5}) (5 + \sqrt{5})}{10}} \\
& = & -\frac{1}{2} \sqrt{\frac{20 - 4\sqrt{5}}{10}} = -\sqrt{\frac{5 - \sqrt{5}}{10}}.
\end{eqnarray*}
Thus :
$$
\alpha = \arccos \left\{ -\sqrt{\frac{5 - \sqrt{5}}{10}} \right\}.
$$
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