《2025年高中必修课程复习与检测数学》

23. 在△ABC中，已知$c = \sqrt{6}$，$A = 45^{\circ}$，$a = 2$，解此三角形.

答案： 解：$\because\frac{a}{\sin A}=\frac{c}{\sin C}$，
$\therefore\sin C=\frac{c\sin A}{a}=\frac{\sqrt{6}\sin45^{\circ}}{2}=\frac{\sqrt{3}}{2}$.
$\because C\in(0^{\circ},180^{\circ})$，$\therefore C = 60^{\circ}$或$C = 120^{\circ}$.
当$C = 60^{\circ}$时，$B = 75^{\circ}$，$b=\frac{c\sin B}{\sin C}=\frac{\sqrt{6}\sin75^{\circ}}{\sin60^{\circ}}=\sqrt{3}+1$；
当$C = 120^{\circ}$时，$B = 15^{\circ}$，$b=\frac{c\sin B}{\sin C}=\frac{\sqrt{6}\sin15^{\circ}}{\sin120^{\circ}}=\sqrt{3}-1$.
$\therefore b=\sqrt{3}+1$，$B = 75^{\circ}$，$C = 60^{\circ}$或$b=\sqrt{3}-1$，$B = 15^{\circ}$，$C = 120^{\circ}$.