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?


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 :
$$
\alpha = \arccos \left\{ -\sqrt{\frac{5 - \sqrt{5}}{10}} \right\}.
$$
