11. 在 $\triangle A B C$ 中, 已知 $a=2 \sqrt{2}, A=30^{\circ}, B=45^{\circ}$, 解三角形.
$\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$,

$\therefore b=\frac{a \sin B}{\sin A}=\frac{2 \sqrt{2} \sin 45^{\circ}}{\sin 30^{\circ}}=\frac{2 \sqrt{2} \times \frac{\sqrt{2}}{2}}{\frac{1}{2}}=4$.
$\because C=180^{\circ}-(A+B)=180^{\circ}-\left(30^{\circ}+45^{\circ}\right)=105^{\circ}$,

$\therefore c=\frac{a \sin C}{\sin A}=\frac{2 \sqrt{2} \sin 105^{\circ}}{\sin 30^{\circ}}=\frac{2 \sqrt{2} \sin 75^{\circ}}{\frac{1}{2}}=2+2 \sqrt{3}$.