Помогите решить 2 вариант все задания

По теореме косинусов:

$\displaystyle\boxed{1}\\\\\tt AC=\sqrt{4^2+(3\sqrt{2})^2-2\cdot4\cdot3\sqrt{2}\cdot \frac{\sqrt{2}}{2}} =\sqrt{16+18-24} =\sqrt{10} \\\\\\ \boxed{2}\\\\\tt AC=\sqrt{(2\sqrt{3})^2+3^2-2\cdot2\sqrt{3}\cdot3\cdot\bigg(-\frac{\sqrt{3}}{2}\bigg)} =\sqrt{12+9+18} =\sqrt{39}$

По теореме синусов:

$\displaystyle\boxed{3}\\\\\tt \frac{AB}{\sin C}= \frac{AC}{\sin B} \ \ \Rightarrow \ \ AC=\frac{AB\cdot\sin B}{\sin C}=\frac{4\sqrt{2}\cdot\frac{\sqrt{3}}{2}}{ \frac{\sqrt{2}}{2}}=4\sqrt{3}\\\\\\ \boxed{4}\\\\ \angle{C}=180-(135+15)=30^o\\\\ AC=\frac{AB\cdot\sin B}{\sin C}=\frac{4\cdot\frac{\sqrt{2}}{2}}{ \frac{1}{2}}=4\sqrt{2}$