How to find out the solid angle subtended by a tetrahedron at its vertex? 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.   

 A: 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):
$$(4 \pi)\omega + \pi = \cos ^{-1} \left( \frac{ (v_1 \times v_2) \cdot (v_1 \times v_3)}{||v_1 \times v_2|| ||v_1 \times v_3||}\right) + \cos ^{-1} \left( \frac{ (v_2 \times v_1) \cdot (v_2 \times v_3)}{||v_2 \times v_1|| ||v_2 \times v_3||}\right) + \cos ^{-1} \left( \frac{ (v_3 \times v_1) \cdot (v_3 \times v_2)}{||v_3 \times v_1|| ||v_3 \times v_2||}\right)$$
A: The solid angle $\omega$ subtended at a vertex by any tetrahedron having (vertex) angles $\alpha, \beta$ & $\gamma$ between consecutive lateral edges meeting at the same vertex, is given by the following HCR's Generalized Formula
$$\omega=\cos^{-1}\left(\frac{\cos\alpha-\cos\beta\cos\gamma}{\sin\beta\sin\gamma}\right)-\sin^{-1}\left(\frac{\cos\beta-\cos\alpha\cos\gamma}{\sin\alpha\sin\gamma}\right)-\sin^{-1}\left(\frac{\cos\gamma-\cos\alpha\cos\beta}{\sin\alpha\sin\beta}\right)$$
OR
$$\omega=\frac{\pi}{2}-\sin^{-1}\left(\frac{\cos\alpha-\cos\beta\cos\gamma}{\sin\beta\sin\gamma}\right)-\sin^{-1}\left(\frac{\cos\beta-\cos\alpha\cos\gamma}{\sin\alpha\sin\gamma}\right)-\sin^{-1}\left(\frac{\cos\gamma-\cos\alpha\cos\beta}{\sin\alpha\sin\beta}\right)$$
OR
$$\omega=\cos^{-1}\left(\frac{\cos\alpha-\cos\beta\cos\gamma}{\sin\beta\sin\gamma}\right)+\cos^{-1}\left(\frac{\cos\beta-\cos\alpha\cos\gamma}{\sin\alpha\sin\gamma}\right)+\cos^{-1}\left(\frac{\cos\gamma-\cos\alpha\cos\beta}{\sin\alpha\sin\beta}\right)-\pi$$
It is worth noticing that the above formula has internal symmetry i.e. the vertex-angles $\alpha, \beta$ & $\gamma$ can be taken in any order/sequence but the result obtained in each case remains the same.
