Given a tetrahedron ABCD with known sides $a, b, c - AB, AC, BC$ of the base and the angles $\alpha, \beta, \gamma - \angle{ADB}, \angle{ADC}, \angle{BDC}$ of the side faces opposite to these sides. Find angle $\phi$ (side edge or height)
I have some ideas on how to solve this with python scipy, but it іs too complicated and has some edge case problems.
$BD = \frac{a\sin\alpha}{\sin\phi}$, $AD = \frac{a\sin(\alpha+\phi)}{\sin\alpha},$ $$\angle{ACD}=\arcsin\left(\frac{a\sin(\alpha+\phi)\sin\beta}{b\sin\alpha}\right),$$ $CD=\frac{b\sin(\beta+\angle{ACD})}{\sin\beta}=\frac{b\sin\left(\beta+\arcsin\left(\frac{a\sin(\alpha+\phi)\sin\beta}{b\sin\alpha}\right)\right)}{\sin\beta},$ $$\frac{CD^2+BD^2-c^2}{2*BD*CD}-\cos\gamma=0,$$ $$\frac{\left(\frac{b\sin\left(\beta+\arcsin\left(\frac{a\sin(\alpha+\phi)\sin\beta}{b\sin\alpha}\right)\right)}{\sin\beta}\right)^2+\left(\frac{a\sin\alpha}{\sin\phi}\right)^2-c^2}{2\left(\frac{a\sin\alpha}{\sin\phi}\right)\left(\frac{b\sin\left(\beta+\arcsin\left(\frac{a\sin(\alpha+\phi)\sin\beta}{b\sin\alpha}\right)\right)}{\sin\beta}\right)}-\cos\gamma=0,$$ And then I solve this equasion in terms of $\phi$ using python scipy.
So... Are there any other ways? any ideas are welcome.