# Is there a formula for the solid angle at each vertex of tetrahedron?

A tetrahedron has four vertices as much as a triangle has three vertices. A tetrahedron therefore can have four solid angles as much as a triangle can have three angles.

I am wondering:

Is there a general formula for the solid angle at each vertex of tetrahedron, if the length of each of the six edges are given? As much as one can use the law of cosines to determine the angle at each vertex of a triangle, as long as the lengths of each sides of triangle is given?

Further question: what kind of area in mathematics does such a theorem belong to? What is the most general textbook for such mathematics?

• If you have all of the edge lengths, you can compute the corner angles for each face. The solid angle at each vertex is then the area of a spherical triangle whose sides are the corner angles of the three faces that meet. – hmakholm left over Monica Oct 14 '11 at 12:16
• On that note, there's l'Huilier's theorem... see also this paper. – J. M. is a poor mathematician Oct 14 '11 at 12:43
• – joriki Oct 14 '11 at 13:15

I realize that this post is years old, but the only documentation of an answer that I know of is younger. There is a formula. For dihedral angles $\theta_{ij}$ and solid angle $\Omega_i$,

$$\Omega_i = \theta_{ij} + \theta_{ik} + \theta_{il} - \pi$$ (Lemma 1).

If you have the six edge lengths of the tetrahedron, you can obtain the dihedral angles. For edges $e_{ij}$ with lengths denoted $d_{ij}$ and face area $F_l$ (the face opposite vertex $V_l$), the dihedral angle $\theta_{ij}$ is given by

$$\cos \theta_{ij}=\frac{D_{ij}}{\sqrt{D_{ijk}D_{ijl}}},$$

where

$$D_{ij}=-d_{ij}^4+ (d_{ik}^2+d_{il}^2+d_{jk}^2+d_{jl}^2-2d_{kl}^2)d_{ij}^2 +(d_{ik}^2-d_{jk}^2)(d_{jl}^2-d_{il}^2)$$

and

$$D_{ijk}=-16{F_l}^2$$

(Theorem 1).

Wirth and Dreiding (2014), Relations between edge lengths, dihedral and solid angles in tetrahedra