sin ∠ B=sin120 ° =sqrt(3)/2
sin ∠ C=sin45 ° =sqrt(2)/2
AB=c=?
По теореме синусов:
[m]\frac{b}{sin\angle B}=\frac{c}{sin\angle C}[/m]
[m]\frac{\sqrt{6}}{\frac{\sqrt{3}}{2}}=\frac{c}{\frac{\sqrt{2}}{2}}[/m]
[m]c=\frac{\sqrt{6}\cdot \sqrt{2}}{\sqrt{3}}=2[/m]
Ответ: 2