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?
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.)
There are 3 angles to consider in a polygon involving the origin point. the circumsphere radius (vertices) what you are seeking, the midsphere (center of edges, aka. 1/2 dihedral angle), and the insphere (right angle perpendicular to the face).

Formulas for finding these three radii are:

C = a/4 * √3 * (1 + √5)

M = a/4 * (3 + √5)

I = a/2 * √([25 + {11*√5}]/10)

Also useful to know the radii formulas for the pentagon.

c = a/10 * √(50 + [10*√5]) 

i = a/10 * √(25 + [10*√5])

Then you just solve for the right triangle OIC, where you have the three sides and the right angle, height=I, base=c, and hypotenuse=C, just have to find the other two angles.