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.

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.
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?
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Step 1
There is a formula. For dihedral angles ${\theta }_{ij}$ and solid angle ${\mathrm{\Omega }}_{i}$,
${\mathrm{\Omega }}_{i}={\theta }_{ij}+{\theta }_{ik}+{\theta }_{il}-\pi$
(Lemma 1).
Step 2
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 $\mathrm{cos}{\theta }_{ij}=\frac{{D}_{ij}}{\sqrt{{D}_{ijk}{D}_{ijl}}},$
where ${D}_{ij}=-{d}_{ij}^{4}+\left({d}_{ik}^{2}+{d}_{il}^{2}+{d}_{jk}^{2}+{d}_{jl}^{2}-2{d}_{kl}^{2}\right){d}_{ij}^{2}+\left({d}_{ik}^{2}-{d}_{jk}^{2}\right)\left({d}_{jl}^{2}-{d}_{il}^{2}\right)$ and ${D}_{ijk}=-16{{F}_{l}}^{2}$