In triangle ABC,if a=2, b=sqrt{6},and c=sqrt{ | vedclass

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(C) Given $a=2, b=\sqrt{6}, c=\sqrt{3}+1$. Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{2^2 + (\sqrt{6})^2 - (\sqrt{3}+1)^2}{2(

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$\frac{\sqrt{3}}{4}$

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(C) Given $a=2, b=\sqrt{6}, c=\sqrt{3}+1$. Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{2^2 + (\sqrt{6})^2 - (\sqrt{3}+1)^2}{2(2)(\sqrt{6})} = \frac{4+6-(3+1+2\sqrt{3})}{4\sqrt{6}} = \frac{6-2\sqrt{3}}{4\sqrt{6}} = \frac{2\sqrt{3}(\sqrt{3}-1)}{4\sqrt{6}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$. Then $\sin^2 C = 1 - \cos^2 C = 1 - \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2 = 1 - \frac{3+1-2\sqrt{3}}{8} = 1 - \frac{4-2\sqrt{3}}{8} = 1 - \frac{2-\sqrt{3}}{4} = \frac{4-2+\sqrt{3}}{4} = \frac{2+\sqrt{3}}{4}$. Using the Law of Cosines for $A$: $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{6 + (\sqrt{3}+1)^2 - 2^2}{2(\sqrt{6})(\sqrt{3}+1)} = \frac{6 + 4 + 2\sqrt{3} - 4}{2\sqrt{6}(\sqrt{3}+1)} = \frac{6+2\sqrt{3}}{2\sqrt{6}(\sqrt{3}+1)} = \frac{2\sqrt{3}(\sqrt{3}+1)}{2\sqrt{6}(\sqrt{3}+1)} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$. Then $\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{1}{\sqrt{2}}\right)^2 = 1 - \frac{1}{2} = \frac{1}{2}$. Therefore,$\sin^2 C - \sin^2 A = \frac{2+\sqrt{3}}{4} - \frac{1}{2} = \frac{2+\sqrt{3}-2}{4} = \frac{\sqrt{3}}{4}$.

In $\triangle ABC$,if $a=2, b=\sqrt{6}$,and $c=\sqrt{3}+1$,then $\sin^2 C - \sin^2 A =$

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