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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 $42°57'$. 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 $270°>180°$.

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 $270°>180°$.

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 kess than the third plus $180°$. Equality with these criteria is acceptable if we allow degenerate cases.

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 at basal dihedral angles if $\cos^{-1}(1/\sqrt3)\approx54°44'$, then elliptic for larger basal dihedral angkes ... 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 $180°$).

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.

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 $180°$).

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 at basal dihedral angles if $\cos^{-1}(1/\sqrt3)\approx54°44'$, then elliptic for larger basal dihedral angkes ... 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 $180°$).