△ABC的内角A、B、C的对边分别为abc．已知asinA+C2＝bsinA．1求B；2若△ABC为锐角三角形且c＝1求△ABC面积的取值范围．

（1）由正弦定理可得：

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$

其中$R$为三角形的外接圆半径。将$\sin B$代入已知条件得：

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$

$$\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin A\cos C+\cos A\sin C}=\frac{1}{\sin C}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin A}{\cos A\sin C+\sin A\cos C}=\frac{\sin A}{\sin (A+C)}=\frac{\sin B}{\sin C}$$

因为$\triangle ABC$为三角形，所以$\angle A+\angle B+\angle C=180^\circ$，即$A+B+C=180^\circ$，所以$\sin C=\sin (180^\circ-A-B)=\sin (A+B)$。代入上式得：

$$\frac{a}{b}=\frac{\sin B}{\sin (A+B)}$$

$$\Rightarrow a\sin (A+B)=b\sin B$$

$$\Rightarrow a(\sin A\cos B+\cos A\sin B)+a\cos A\cos B=b\sin B$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\cos A\cos B+\sin A\sin B}=\frac{\sin B}{\cos (A-B)}$$

又因为$\triangle ABC$为三角形，所以$\angle A$和$\angle B$都是锐角或直角或钝角。若$\angle A$和$\angle B$都是锐角或直角，则$\cos (A-B)>0$，所以$\dfrac{a}{b}>0$。若$\angle A$和$\angle B$都是钝角，则$\cos (A-B)<0$，所以$\dfrac{a}{b}<0$。因为$\dfrac{a}{b}=\dfrac{\sin B}{\cos (A-B)}$，所以$\sin B$和$\cos (A-B)$的符号相同，即$\angle B$和$\angle A-B$的符号相同。因此，如果$\angle A$和$\angle B$都是锐角或直角，则$B>0$；如果$\angle A$和$\angle B$都是钝角，则$B<0$。综上所述，$B>0$。

将$\dfrac{a}{b}=\dfrac{\sin B}{\cos (A-B)}$代入已知条件得：

$$\frac{a}{b}=\frac{\sin A\sin C}{\sin B\sin (A+C)/2}=\frac{2\sin A\sin C}{\sin B\cos (A-C)/2}$$

$$\Rightarrow \frac{\sin A\sin C}{\sin B}=\frac{2a}{b}\cos \frac{A-C}{2}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin (180^\circ-A-B)}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin (B+C)}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\cos A\sin B\sin C+\sin A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin (A+B)\sin C}\cdot \frac{1}{2\cos \frac{A-C}{2}}$$

$$\Rightarrow \frac{a}{b}=\frac{\sin B}{\sin A\sin B\cos C+\cos A\cos B\sin C}\cdot \frac{1}{2\cos \fra