The Wikipedia page for the snub dodecahedron provides explicit coordinates for its vertices, from which one can of course by brute force calculate that there are only two dihedral angles. Can anyone give a (hopefully insightful/more intuitive) shorter proof that there must be only two such angles? The symmetries of the object immediately imply that there can be at most three such angles: the triangle-pentagon one, the angle between two triangles each of which is adjacent to a pentagon, and the angle between a triangle not adjacent to any pentagon and one that is. But is there any "quick" way to see that the latter two angles must be equal?
Why only two dihedral angles for a snub dodecahedron
anglegeometrypolyhedra
Related Solutions
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.)
This is really just a comment that's too long. Hopefully, once I understand fully what you're asking and we have uniform terminology, someone will be able to answer your (at least specific) question.
The dihedral angle is angle at which two planes meet; for polyhedra, it's the angle at which two faces meet. All polyhedra, convex or not, have dihedral angles.
However, due to the way it's measured, the dihedral angle can tell convex from non-convex polyhedra: We always measure the dihedral angle as the angle inside the polygon.
In the picture above, imagine the lines are altitudes of triangle faces (and we can't see the faces; they're perpendicular to our line of sight). Then the angle $\varphi$ corresponds to a dihedral angle of a convex portion of our polyhedron; the angles all "bend toward" the interior of the polyhedron. This would be a convex meeting of faces, as we have $\varphi < \pi$.
However, for a non-convex polyhedron, the angles may "bend away" from the interior of the polytope. In this case, it would actually be $2\pi - \varphi$ that's the interior dihedral angle, and we'd have $\varphi > \pi$.
Thus, we can characterize convex polytopes based on their interior dihedral angles: They're all less than $\pi$.
There's another angle measure for polyhedra that can (only sometimes) detect a lack of convexity: The angular defect. To calculate the angular defect at a vertex, we add up all the angles of faces that make up the vertex, and subtract this sum from $2\pi$. For example, in a regular tetrahedron, three faces meet at each vertex, and the angles of each face are $\pi/3$; thus our angular defect is $2\pi - 3(\pi/3) = \pi$. In a regular dodecahedron, the angular defect is $2\pi - 3(3\pi/5) = 2\pi - 9\pi/5 = \pi/5$.
I only bring 'angular defect' up because I can't quite parse your phrase that "the dihedral angles are uniform at each vertex." To my knowledge, in $3$-space, dihedral angles really only apply to edges (where just two faces meet), and not to vertices.
For the Catalan PD, apparently the dihedral angle is constant for all edges. I suspect that the angular defect is not constant, and can be one of two different things, but I haven't done any calculations.
Best Answer
Focus on a pair of adjacent triangles $T_1$ and $T_2$. Their vertices lie on the circumscribing sphere $S$. Their dihedral angle is supplementary to the angle $\angle P_1OP_2$ where $O$ is the centre of $S$ and $P_i$ is the centroid of $T_i$. But this angle only depends on the radius of the sphere and the edge lengths of the equilateral triangles $T_i$. (There's a rotation of the sphere taking $T_1$ and $T_2$ to any configuration of two adjacent triangles with the same sidelengths with vertices on $S$.)