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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?

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    $\begingroup$ You are using $A$ to mean both a point and a number. Please do NOT! $\endgroup$
    – sds
    Commented Jul 29, 2014 at 16:47
  • $\begingroup$ I don't really know how to do vectors, but I'm not sure this would work because I know the angles between the rising sides and each other, not the rising sides and the base. $\endgroup$
    – hacatu
    Commented Jul 29, 2014 at 16:50
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    $\begingroup$ Just to clarify things, you have the angles between the edges at the apex, not the dihedral angle between the faces, right? If you look at the inscribed angle, then you can see that the locus of all points $O$ which satisfy one of the angle conditions would be a circle (or arc) in the plane, but actually a self-intersecting spindle torus in 3D. So you are intersecting three such objects, for which a quadratic approach looks perfectly reasonable. $\endgroup$
    – MvG
    Commented Jul 29, 2014 at 20:36
  • $\begingroup$ Yes, the angles are between the edges at the apex, not the faces (not dihedral). $\endgroup$
    – hacatu
    Commented Jul 29, 2014 at 22:22
  • $\begingroup$ By the Spherical Law of Cosines, if you know the face angles at the apex, then you can find the dihedral angles at the apex, and vice-versa. $\endgroup$
    – Blue
    Commented Jul 29, 2014 at 23:54

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