设 $\triangle A B C$ 的内角 $A, B, C$ 的对边分别为 $a, b, c$, 已知 $b=4, B=30^{\circ}, C=45^{\circ}$, 则 $\triangle A B C$的面积为
$: \because \mathrm{b}=4, \mathrm{~B}=30^{\circ}, \mathrm{C}=45^{\circ}$,

$\therefore$ 由正弦定理: $\frac{b}{\sin B}=\frac{c}{\sin C}$, 可得: $\mathrm{c}=\frac{b \cdot \sin C}{\sin B}=\frac{4 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}}=4 \sqrt{2}$,

$\therefore \mathrm{S}_{\triangle \mathrm{ABC}}=\frac{1}{2} \mathrm{bcsinA}=\frac{1}{2} \times 4 \times 4 \sqrt{2} \times \sin \left(180^{\circ}-30^{\circ}-45^{\circ}\right)=8 \sqrt{2} \times \frac{\sqrt{6}+\sqrt{2}}{4}=4 \sqrt{3}+4$.