ABC

S

\angle BAC=\alpha

AC=b

BC

\sqrt{\frac{4S^{2}}{b^{2}\sin^{2}\alpha}+b^{2}-4S\ctg\alpha}

S=\frac{1}{2}AB\cdot AC\sin\alpha

AB=\frac{2S}{b\sin\alpha}

BC^{2}=AB^{2}+AC^{2}-2AB\cdot AC\cos\alpha=

=\frac{4S^{2}}{b^{2}\sin^{2}\alpha}+b^{2}-2b\cos\alpha\cdot\frac{2S}{b\sin\alpha}=\frac{4S^{2}}{b^{2}\sin^{2}\alpha}+b^{2}-4S\ctg\alpha.