Given the angles between the edges of a triangular pyramid, what is the angle between any one edge and it's opposing side? Rendition in paint
You have a triangular pyramid with the edges a, b, and c.
You know the angles between those edges - alpha (between a & c), beta (between b & c), and gamma (between a & b).
I hope I am also right in assuming that a plane can be formed by any of the two edges.
So what I'm looking for is the angle between an edge and the plane that is formed by the other edges.
As far as I can see, it's all pretty basic geometry, but after multiple hours and a few sheets of paper, I am probably approaching it the wrong way, and making some mistakes somewhere.
I don't think it would be of any help, but I can include a long description of one of my example solutions, which yielded an incorrect answer.


Draw a triangular pyramid. $ABCD$ such that $\angle{BAC}=\beta,\angle CAD=\alpha$ and $\angle BAD=\gamma$. Drop a perpendicular from $C$ to $BD$ at point $E$. 

First check  that $\angle CAE$ is your required angle.


Without loss of generality, we can make $AB=AC=AD=1$ unit. 
Applying sine rule:$${BC\over \sin \beta}={1\over\cos(\beta/2)}$$So you know $BC$. Similarly you also know $CD\ \&\ BD$$${CD\over \sin\alpha}={1\over\cos(\alpha/2)}\ \ \ \ \&\ \ \ {BD\over\sin\gamma}={1\over \cos(\gamma/2)}$$
Now in $\Delta BCD,$ $$BC^2-BE^2=CD^2-(BD-BE)^2$$ You get $BE$, and hence $CE$.

Check that the only thing to find is $AE$, because once you get that , you get $AE,AC\ \&\ CE$ and hence the entire triangle $ACE$ and hence $\angle CAE$.

Finding $AE$ is easy:
In triangle $ABD$, $\angle BAD=\gamma,\angle ABD=90^\circ -\gamma/2$. 
As we know $BE\ \&\ DE$ , let $\angle BAE=x$ . Applying sine rule ,$${BE\over \sin x}={1\over \sin\angle AEB}={1\over \sin(180-\angle AEB)}={1\over \sin \angle AED}={ED\over \sin(\gamma-x)}$$
Find out $x$ and then apply $${BE\over \sin x}={AE\over \cos(\gamma/2)}$$
BINGO!!
