I know the dimensions of the base of a tetrahedron and the angles between the non base sides at the apex. I want to know the lengths of the three non base sides. Let the base's corner points be $A, B,$ and $C, $ and the apex be $O$. I have used the law of cosines to create three equations: $$|AB|^2=|AO|^2+|BO|^2-2\cdot |AO|\cdot |BO|\cdot \cos(\angle AOB)$$ $$|BC|^2=|BO|^2+|CO|^2-2\cdot |BO|\cdot |CO|\cdot \cos(\angle BOC)$$ $$|CA|^2=|CO|^2+|AO|^2-2\cdot |CO|\cdot |AO|\cdot \cos(\angle COA)$$ Let: $$X=|BC|^2,Y=|CA|^2,Z=|AB|^2,a=|AO|,b=|BO|,c=|CO|,$$ $$\alpha=2\cdot \cos(\angle BOC),\beta=2\cdot \cos(\angle COA),\gamma=2\cdot \cos(\angle AOB)$$ Now the equations can be written: $$0=b^2+c^2-\alpha b c-X$$ $$0=c^2+a^2-\beta c a-Y$$ $$0=a^2+b^2-\gamma a b-Z$$ I tried to solve this system using the quadratic formula, but the equations became far too large by the time I had it in terms of a and there were three groups of $\pm$.
My main question: Is there a better way to solve this system than the quadratic formula? Also, is there a better way to solve the tetrahedron than my law of cosines method?