Given that the angles between the consecutive lateral edges AB, AC & AD meeting at the vertex A of a tetrahedron ABCD are $ α, β, γ$ (as shown in the diagram below). Is there any set-formula to find out the solid angle subtended by the tetrahedron at the same vertex?
Note: A tetrahedron is a solid having 4 triangular faces, 6 edges & 4 vertices. Three triangular faces meet together at each of four vertices & each of six edges is shared (common) by two adjacent triangular faces.
Denote the solid angle by $\omega$ and let $v_1,v_2,v_3$ be the vectors from vertex $A$ along the edges $AB, AD, AC$. Then we have (using the usual cross product, dot product, and Euclidean norm):
$$(4 \pi)\omega + \pi = \cos ^{-1} \left( \frac{ (v_1 \times v_2) \cdot (v_1 \times v_3)}{||v_1 \times v_2|| ||v_1 \times v_3||}\right) + \cos ^{-1} \left( \frac{ (v_2 \times v_1) \cdot (v_2 \times v_3)}{||v_2 \times v_1|| ||v_2 \times v_3||}\right) + \cos ^{-1} \left( \frac{ (v_3 \times v_1) \cdot (v_3 \times v_2)}{||v_3 \times v_1|| ||v_3 \times v_2||}\right)$$