Angles between two vertices on a dodecahedron Say $20$ points are placed across a spherical planet, and they are all spaced evenly, forming the vertices of a dodecahedron. I would like to calculate the distances between the points, but that requires me to find out the angles between the vertices.
From the origin of the dodecahedron, how would I find the angle between two adjacent vertices on the same face, and the angle between two vertices on the same face but not connected by an edge?

It is maybe not a good habit to use short-cut formulas/constructions from Wikipedia,  since a real mathematician should construct his geometric or analytic geometric figures herself/himself, however, I believe that it is necessary when you have not enough time.
Wikipedia already constructed a dodec with 20 vertices: $(\pm 1,\pm 1, \pm 1), (0,\pm\phi,\pm\phi^{-1}), (\pm\phi^{-1},0,\pm\phi), (\pm\phi,\pm\phi^{-1},0).$
If we take two of them which are vertices of an edge we can compute the first angle asked. For example, $(1,1,1)$ and $(\phi,\phi^{-1},0)$ are such vertices! By using the identities $\phi^{-2}=2-\phi$ and $\phi^2=\phi+1$, we compute the angle which sees an edge from the origin: $$(1,1,1)\cdot(\phi,\phi^{-1},0)=\sqrt{1^2+1^2+1^2}\sqrt{\phi^2+\phi^{-2}}\cos\theta\implies\theta=\arccos\frac{\sqrt{5}}{3}\approx 41.81^{\circ}$$
For the angle which sees a diagonal of a face, we take $(1,1,1)$ and $(-1,1,1)$. Hence,
$$(1,1,1)\cdot(-1,1,1)=\sqrt{3}\sqrt{3}\cos\theta\implies\theta=\arccos(\frac{1}{3})\approx 70.53^{\circ}.$$
