( I ) 求 $A$ 的大小;
(II) 若 $2 \sqrt{2} \sin B=3 \sin C$, 再从下列条件(1), 条件(2)中任选一个作为已知, 求 $\triangle A B C$ 的面积. 条件(1): $a \sin C=2$; 条件(2) $a c=2 \sqrt{10}$.

【答案】 解: ( I ) $\because \frac{b}{a}=\sin C+\cos C$,

\begin{aligned} & \therefore \sin B=\sin (A+C)=\sin A \cos C+\cos A \sin C=\sin A \sin C+\sin A \cos C . \\ & \therefore \cos A \sin C=\sin A \sin C . \because C \in(0, \pi), \therefore \sin C \neq 0 . \\ & \therefore \sin A=\cos A . \\ & \because A \in(0, \pi), \therefore A=\frac{\pi}{4} . \end{aligned}

(II) 若选择条件(1), 由正弦定理 $\frac{a}{\sin A}=\frac{c}{\sin C}$, 得 $a \sin C=c \sin A=\frac{\sqrt{2}}{2} c=2$. $\therefore c=2 \sqrt{2}$.

\begin{aligned} & \therefore b=3 . \\ & \therefore S_{\triangle A B C}=\frac{1}{2} b c \sin A=\frac{1}{2} \times 3 \times 2 \sqrt{2} \sin \frac{\pi}{4}=3 . \end{aligned}

\begin{aligned} & \therefore a=\sqrt{5}, b=3, c=2 \sqrt{2} . \\ & \therefore S_{\triangle A B C}=\frac{1}{2} b c \sin A=\frac{1}{2} \times 3 \times 2 \sqrt{2} \sin \frac{\pi}{4}=3 . \end{aligned}