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?


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

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


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



(Theorem 1).

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.