Triangle with given data constructible? So, this is a problem in abstract algebra, not some elementary geometry: Is it possible to construct a triangle $ABC$ given $a$, which is side $BC$, $b$ which is side $AC$ and angle $\beta-\gamma$, where $\beta$ is the angle at point $B$ while $\gamma$ is the angle at point $C$? Please help, I have no idea how I could even start, I just know that I need some polynomial on something inside triangle which would not be constructible....

\begin{eqnarray*}
\sin \beta = \frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{2ca} \\
\sin \gamma = \frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{2ab} \\
\cos \beta = \frac{c^2+a^2-b^2}{2ca} \\
\cos \gamma = \frac{a^2+b^2-c^2}{2ab} \\
\end{eqnarray*}
Quick sanity check ...
\begin{eqnarray*}
\cos \alpha = -\cos \beta \cos \gamma + \sin \beta \sin \gamma = \cdots = \frac{b^2+c^2-a^2}{2bc}.
\end{eqnarray*}
Now to the problem in hand ... 
\begin{eqnarray*}
\cos (\beta- \gamma)= \cos \beta \cos \gamma + \sin \beta \sin \gamma = \cdots = \frac{a^2 b^2+a^2c^2-b^4+2b^2c^2-c^4}{2a^2bc}
\end{eqnarray*}
Now a little bit of algeabra & we have the following quartic for $c$
\begin{eqnarray*}
c^4-c^2(2b^2+a^2)+2a^2bc \cos (\beta - \gamma)+b^4-a^2b^2=0.
\end{eqnarray*}
