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(C) Let the sides be $a = \alpha - \beta$,$b = \alpha + \beta$,and $c = \sqrt{3\alpha^2 + \beta^2}$.Since $\alpha > \beta > 0$,we observe that $c^2 =

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$\frac{2\pi}{3}$

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(C) Let the sides be $a = \alpha - \beta$,$b = \alpha + \beta$,and $c = \sqrt{3\alpha^2 + \beta^2}$.Since $\alpha > \beta > 0$,we observe that $c^2 = 3\alpha^2 + \beta^2$ is the largest square,as $a^2 + b^2 = 2\alpha^2 + 2\beta^2  \beta^2$).Thus,$c$ is the longest side,and the angle $C$ opposite to $c$ is the largest angle.Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the values: $\cos C = \frac{(\alpha - \beta)^2 + (\alpha + \beta)^2 - (3\alpha^2 + \beta^2)}{2(\alpha - \beta)(\alpha + \beta)}$.$\cos C = \frac{(\alpha^2 - 2\alpha\beta + \beta^2) + (\alpha^2 + 2\alpha\beta + \beta^2) - 3\alpha^2 - \beta^2}{2(\alpha^2 - \beta^2)}$.$\cos C = \frac{2\alpha^2 + 2\beta^2 - 3\alpha^2 - \beta^2}{2(\alpha^2 - \beta^2)} = \frac{\beta^2 - \alpha^2}{2(\alpha^2 - \beta^2)}$.$\cos C = \frac{-(\alpha^2 - \beta^2)}{2(\alpha^2 - \beta^2)} = -\frac{1}{2}$.Since $\cos C = -\frac{1}{2}$,the angle $C = \frac{2\pi}{3}$.

The lengths of the sides of a triangle are $\alpha - \beta$,$\alpha + \beta$,and $\sqrt{3\alpha^2 + \beta^2}$,where $\alpha > \beta > 0$. Its largest angle is:

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