One observation to be made is a tetrahedron does not necessarily exist with an arbitrary set of dihedral angles. As the exercise shows below, we must have **the dihedral angles at each vertex sum to at least $180°$.** They must also satisfy the triangle inequality-like relation: **at each vertex the sum of any two dihedral angles must be equal to or less than the third plus $180°$.** Equality with these criteria is acceptable if we allow degenerate cases.

We think of the tetrahedron, which may be non-Euclidean, as being formed from a space enclosed by four planes or four congruent spheres. For such an enclosure to occur there must be a correct combination of dihedral angles.

Consider the case of all dihedral angles being equal. We model this case as four congruent spheres whose radius is taken as one unit and whose centers are at the vertices of a Euclidean regular tetrahedron. If we place the spheres so they just touch each other, a distance of two units between sphere centers, that corresponds to zero dihedral angles ... but the tetrahedron does not form. There are triangular gaps through which a small bug initially trapped between the spheres can crawl or fly out.

To really trap the bug and enclose a tetrahedron, we must move the centers closer allowing the spheres to overlap. We finally close off the gaps when the center-to-center distance is reduced to $\sqrt3$ units. If we then consider the dihedral angles formed locally along the edge of any small circle where two spheres intersect, we discover that this minimal dihedral angle for a tetrahedron to close is $60°$ or $\pi/6$ radians. (We note that each vertex has three of these dihedral angles coming together and these add up to $180°$, a sum we shall encounter later on.) Clearly the faces of this tetrahedron curve away from each other and the angles of the triangular faces are decreased as if in a hyperbolic space; this is a **hyperbolic** tetrahedron.

If we continue to push the sphere centers closer together increasing the dihedral angles beyond $60°$, the curvature of the faces and decrement of the vertex angles become less; the tetrahedron tends towards Euclidean. The Euclidean shape is obtained as a limit when the enclosed region shrinks towards a point (the overall center of the set of spheres); we may recover the **Euclidean** tetrahedron with dihedral angles approximately $70°32'$ (actually, $\cos^{-1}(1/3)$) by regarding the spheres as infinitesimally removed from the overall center and looking close-up at the resultant limiting shape.

If we go on so that now the spheres overlap and the enclosed region now curves positively around the collective center, then we have an **elliptic** tetrahedron, in this case with equal dihedral angles exceeding the $70°32'$ figure cited above. We maintain an enclosed tetrahedron all the way until the centers of the individual spheres coincide, in which case the dihedral angles hit their maximum of $180°$ and the tetrahedron is now the whole sphere (which may be considered a degenerate case). We conclude that with equal dihedral angles, a tetrahedron exists only for angle values between $60°$ and $180°$, including endpoints if we allow degenerate cases.

How do the planar angles at vertices on the faces correlate with the dihedral angles and the hyperbolic/Euclidean/elliptic character of the tetrahedron as cited above? Let's examine a hyperbolic case from above. We set the dihedral angles all to $65°$ and then imagine each vertex as the center of a small sphere. This small sphere will intersect the tetrahedron to form a spherical triangle, whose angles are the given dihedral angles around the vertex and whose arcs measure the planar angles at the vertices.

Before solving, we first note that the dihedral angles, each rendered as an angle of a spherical triangle, must add up to more than $180°$ (exactly $80°$ would he acceptable in the bounding, degenerate case). That checks. We also note that each dihedral angle of $65°$ plus $180°$ is properly greater than the third such angle plus ($2×65°=130°$). So we have a soluble case at every vertex according to spherical geometry.

With all dihedral angles set to $65°$ in this regular example, we apply the Spherical Law of Cosines and infer that the arcs of the small spherical triangle around each vertex measures approximately $42°57'$. Therefore the planar angkes at the vertices are also $\approx42°57'$. Since this applies at all three corners of any face, the angles of the triangles sum to less than $180°$, as in hyperbolic geometry. To get the sum up to $180°$ (three $60°$ arcs in the small sphere surrounding each vertex) and realize a Euclidean tetrahedron by this method, we find that the dihedral angles must be bumped up to $\cos^{-1}(1/3)\approx70°32'$, corresponding exactly to the enclosed region becoming infinitesimal and its faces simultaneously Euclidean-planarized. Once beyond this point the sums of angles of all triangular faces simultaneously go above $180°$ as the entire tetrahedron goes over to elliptic. For instance, in the regular case if all dihedral angles are $90°$ we find the same is true of the planar angles at the vertices, so the sum of these angles at every face is $270°>180°$.

Let us look more briefly at a slightly more complicated example. The dihedral angles at $\overline{AB},\overline{AC},\overline{AD}$ are fixed at $90°$ and the remaining (basal) dihedral angles are equal. The spherical model for this contains three fixed unit spheres with center-to-center distances $\sqrt2$ forming the fixed $90°$ dihedral angles with their pair wise intersections; the fourth sphere moves in with its center along thectgreefold axis. The tetrahedron first closes when the basal dihedral angles reach $45°$, where all vertices initially satisfy the $180°$ sum requirement. It becomes infinitesimal and Euclidean, with all triangular faces simultaneously achiving planar-angle suns of $180°$, at basal dihedral angles of $\cos^{-1}(1/\sqrt3)\approx54°44'$. It then becomes elliptic for larger basal dihedral angles ... up to $135°$ where the basal vertices have two dihedral angles adding up to the third one plus $180°$ ($135°+135°=90°+180°$). If we try to increase the dihedral angke further, the tetrahedron breaks; the enclosed region suddenly needs a fifth face. Thereby a $C_{3v}$ tetrahedron with $90°$ lateral dihedral angles exists for basal angles between $45°$ (where the basal vertices vertices first satisfy the $\ge180°$ sum requirement) and $135°$ (where the basal vertices hit the limit of two dihedral angkes summing to no more than the third plus $180°$).