In triangle ABC,if a^2-c^2=b(b-c),sqrt{2}a=2b | vedclass

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(C) Given: $a^2-c^2=b^2-bc$ $\Rightarrow a^2-c^2=b^2-bc$ $\Rightarrow b^2+c^2-a^2=bc$. Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac

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$\frac{\sqrt{3}+1}{\sqrt{6}}$

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(C) Given: $a^2-c^2=b^2-bc$ $\Rightarrow a^2-c^2=b^2-bc$ $\Rightarrow b^2+c^2-a^2=bc$. Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{bc}{2bc} = \frac{1}{2}$. Thus,$A = 60^{\circ}$. Using the sine rule,$\frac{a}{\sin A} = 2R$ $\Rightarrow \frac{a}{\sin 60^{\circ}} = 2 	imes \frac{1}{\sqrt{3}}$ $\Rightarrow a = \frac{2}{\sqrt{3}} 	imes \frac{\sqrt{3}}{2} = 1$. Given $\sqrt{2}a = 2b-c$,substitute $a=1$: $\sqrt{2} = 2b-c \Rightarrow c = 2b-\sqrt{2}$. Substitute $c$ into $a^2-c^2=b^2-bc$: $1^2 - (2b-\sqrt{2})^2 = b^2 - b(2b-\sqrt{2})$. $1 - (4b^2 - 4\sqrt{2}b + 2) = b^2 - 2b^2 + \sqrt{2}b$. $1 - 4b^2 + 4\sqrt{2}b - 2 = -b^2 + \sqrt{2}b$. $-3b^2 + 3\sqrt{2}b - 1 = 0 \Rightarrow 3b^2 - 3\sqrt{2}b + 1 = 0$. Using the quadratic formula $b = \frac{-(-3\sqrt{2}) \pm \sqrt{(-3\sqrt{2})^2 - 4(3)(1)}}{2(3)} = \frac{3\sqrt{2} \pm \sqrt{18-12}}{6} = \frac{3\sqrt{2} \pm \sqrt{6}}{6} = \frac{\sqrt{2}}{2} \pm \frac{\sqrt{6}}{6} = \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{6} = \frac{3\sqrt{2}+\sqrt{6}}{6} = \frac{\sqrt{3}+1}{\sqrt{6}}$.

In $\triangle ABC$,if $a^2-c^2=b(b-c)$,$\sqrt{2}a=2b-c$ and $R=\frac{1}{\sqrt{3}}$,then $b=$

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