Let $\Delta$ denote the area of $\triangle{ABC}$. Using $\Delta=\frac{abc}{4R}=rs$ we get $b+c=\frac{bc-2rR}{2\sqrt{rR}}\quad (\diamond)$ after some simple manipulation. Now by the Cosine rule, $b^2+c^2-2bc\cos \alpha = a^2$ or $(b+c)^2-2bc(1+\cos \alpha)=rR$. Plugging in our equation for $b+c$ and letting $t:=bc$ we get $\left(\frac{t-2rR}{2\sqrt{rR}}\right)^2-2t(1+\cos \alpha)-rR=0$ or $\frac{t^2}{4rR}+rR-t-2t(1+\cos \alpha)-rR=0$ or $t^2-4rR(3+2\cos \alpha)t=0$. Since $t=bc\ne 0$ we get divide by it obtaining $t=4rR(3+2\cos\alpha)$. Finally, feeding this and $c=\frac{4rR}{b}(3+2\cos\alpha)$ into $(\diamond)$ we can get a quadratic in $b$, then we solve for $b$ and make a back-substitution to get $c$. (The last part is really bashy, best avoid it.)