For a (possibly irregular) Tetrahedron $ABCD$, if I know the lengths of $AB$, $BC$, and $CA$ and the measure of $\angle ADB$, $\angle ADC$ and $\angle BDC$, is there a way to find the lengths of sides $AD$, $BD$ and $CD$.
Using the law of cosines, I think this is equivalent to solving the fallowing system $AD^2+BD^2-2(AD)(BD)cos(\angle ADB)=AB^2$, $BD^2+CD^2-2(BD)(CD)cos(\angle BDC)=BC^2$, $CD^2+AD^2-2(CD)(AD)cos(\angle CDA)=CA^2$, but I'm not sure how to solve the system.