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Finding dihedral angles between tetrahedron faces from triangles' angles at the tip has been answered in:

Dihedral angles between tetrahedron faces from triangles' angles at the tip

My questions is the converse: at the tip of a tetrahedron, if I know the three dihedral angles, can I find the three triangles' angles?

Blue
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joan doe
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  • Yes. The formulas are nearly identical. They're usually presented as the "Laws of Cosines for Spherical Triangles", but don't let that intimidate you. (The interpretation comes from positioning a unit sphere at the vertex of the tetrahedron. The planes of the adjacent faces determine a curved triangle on the surface of the sphere. The "angles" of the triangle exactly match the dihedral angles of the tetrahedron; the lengths of the "sides" actually match the tetrahedron's face angles, because of how arc length works.) – Blue Apr 06 '23 at 23:52
  • See, for instance, this answer of mine. – Blue Apr 06 '23 at 23:55

1 Answers1

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This vertex angle can be computed easily using Girard's formula for the area of a spherical triangle, defined as the intersection of three hemispherical regions on the unit sphere.

To see how these two problems are related, suppose that the three faces of the tetrahedron that meet at your vertex have their outer normals listed as $\vec n_1, \vec n_2, \vec n_3$. Then the 3D angle at your vertex corresponds to the portion of the unit sphere consisting of all direction vectors $\vec u$ that satisfy three simultaneous inequalities: $\vec u\cdot n_1 \leq 0, \vec u\cdot \vec n_2 \leq 0, \vec u\cdot \vec n_3 \leq0$. (The size of that 3D angle is basically defined as the area of this region on the unit sphere.) But these inequalities also carve out three hemispheres on the unit sphere! Thus they define a spherical triangle $\Delta$ whose area we wish to determine.

Girard's formula $$ A(\Delta)= \alpha+ \beta +\gamma- \pi$$ tells you that area directly in terms of data you know: the angles between the normal vectors, which after a bit of thought you may recognize are also the dihedral angles formed by the faces of your tetrahedron. They are also the angles that one measures at the vertices of $\Delta$.

Recapitulation. Note that there is lots of duality occurring in this discussion, and you must convert back and forth between the original 3D tetrahedron and the associated spherical triangle. At any vertex of the tetrahedron, the faces of the tetrahedron determine points (normal vectors) on the unit sphere, and their three associated orthogonal great circles carve out a triangle $\Delta$ on the sphere. These great circles intersect pairwise at three vertices of a spherical triangle $\Delta$, meeting at three angles $\alpha, \beta, \gamma$ which are also the dihedral angles of your original tetrahedron.

P.S. Girard's formula is a special instance of the more general Gauss-Bonnet formula for geodesic triangles on any surface. Some would just call it Gauss-Bonnet on the sphere.

MathFont
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