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?
On
The faces are regular pentagons. That means at each vertex, the external angle is $1/5$ of the full circle, thus $360^\circ/5=72^\circ$. Therefore the internal angle must be $180^\circ-72^\circ=108^\circ$.
Three edges are vectors emanating from that vertex. Think of that vertex as the origin in a three-dimensional coordinate system. Let the $z$-axis go from there toward the center of the sphere, so the center is at $(0,0,R)$, and so $R$ is the radius. Let the $x$-axis go in the direction of one of those three edges, so that that vector is $(1,0,a)$, and we must find $a$. The other two edges must then be $\left(\frac{-1}{2}, \pm\frac{\sqrt{3}}{2}, a\right)$. The cosine of the angle between those vectors is their dot product divided by the product of their lengths, and their lengths are $\sqrt{1+a^2}$. So $$ \cos108^\circ = \frac{a^2-\frac12}{a^2+1}. $$ This gives us $$ a=\sqrt{\frac{1+2\cos180^\circ}{2(1-\cos108^\circ)}}. $$
The distance from the center $(0,0,R)$ to $(1,0,a)$ must be $R$, so $$ R^2 = 1+(R-a)^2. $$ Solve that equation for $R$, and use the fact that you know $a$. The equation is not truly quadratic, since the $R^2$ terms cancel.
Now you have the coordinates of the center and of two adjacent vertices, so you can find the cosine of the angle between the lines from the center to those vertices.
On
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, $$(1,1,1)\cdot(-1,1,1)=\sqrt{3}\sqrt{3}\cos\theta\implies\theta=\arccos(\frac{1}{3})\approx 70.53^{\circ}.$$
As noted in Wikipedia's "Dodecahedron" entry, if $s$ is the length of an edge of a dodecahedron, and $r$ the radius of its circumsphere, then
$$r = s \frac{\sqrt{3}}{4}\left( 1 + \sqrt{5} \right)$$
So, if two points $A$ and $B$ are joined by and edge, and $O$ is the center of the dodecahedron, then $\triangle AOB$ is isosceles with legs $r$ and base $s$; applying the Law of Cosines to its vertex angle, we have ...
$$s^2 = r^2 + r^2 - 2 r\cdot r\cos\angle AOB = 2 r^2 \left( 1 - \cos\angle AOB \right) = 4 r^2 \sin^2\frac{1}{2}\angle AOB$$
so that
$$\sin\frac{1}{2}\angle AOB = \frac{s}{2r} = \frac{2s}{s\sqrt{3}\left(1+\sqrt{5}\right)} = \frac{\sqrt{3}\left(\sqrt{5}-1\right)}{6}$$
whence
$$\angle AOB = 2 \arcsin \frac{\sqrt{3}\left(\sqrt{5}-1\right)}{6} = 41.8103\dots^\circ$$
If $A$ and $C$ are non-adjacent vertices on a face, then $d := |AC|$ is a diagonal of a regular pentagon with side length $s$. Thus,
$$d = \frac{s}{2}\left( 1 + \sqrt{5} \right)$$
Just as above, we can compute
$$\sin\frac{1}{2}\angle AOC = \frac{d}{2r} = \frac{s\left(1+\sqrt{5}\right)}{s\sqrt{3}\left(1+\sqrt{5}\right)} = \frac{\sqrt{3}}{3}$$
whence
$$\angle AOC = 2 \arcsin \frac{\sqrt{3}}{3} = 70.5288\dots^\circ$$
(You may recognize this as the central angle between adjacent vertices of a cube. It's often helpful to realize that a dodecahedron's face diagonals form the edges of a family of cubes, as shown in the Wikipedia entry. Moreover, one can think of constructing a dodecahedron by taking a cube and pitching a pup-tent on each face, where a triangular tent face and a quadrilateral tent face form a regular pentagon.)