In a triangle ABC,if A = 2B and the sides opp | vedclass

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(C) From the sine rule: $\frac{\alpha+1}{\sin A} = \frac{\alpha-1}{\sin B} = \frac{\alpha}{\sin C}$.Given $A = 2B$,so $\frac{\alpha+1}{\sin 2B} = \fra

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$5$

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(C) From the sine rule: $\frac{\alpha+1}{\sin A} = \frac{\alpha-1}{\sin B} = \frac{\alpha}{\sin C}$.Given $A = 2B$,so $\frac{\alpha+1}{\sin 2B} = \frac{\alpha-1}{\sin B}$.Using $\sin 2B = 2 \sin B \cos B$,we get $\frac{\alpha+1}{2 \sin B \cos B} = \frac{\alpha-1}{\sin B}$,which simplifies to $\cos B = \frac{\alpha+1}{2(\alpha-1)} \dots(1)$.From the cosine rule: $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(\alpha+1)^2 + \alpha^2 - (\alpha-1)^2}{2(\alpha+1)(\alpha)}$.Simplifying the numerator: $(\alpha^2 + 2\alpha + 1) + \alpha^2 - (\alpha^2 - 2\alpha + 1) = \alpha^2 + 4\alpha$.So,$\cos B = \frac{\alpha^2 + 4\alpha}{2\alpha(\alpha+1)} = \frac{\alpha+4}{2(\alpha+1)} \dots(2)$.Equating $(1)$ and $(2)$: $\frac{\alpha+1}{2(\alpha-1)} = \frac{\alpha+4}{2(\alpha+1)}$.$(\alpha+1)^2 = (\alpha-1)(\alpha+4) \Rightarrow \alpha^2 + 2\alpha + 1 = \alpha^2 + 3\alpha - 4$.Solving for $\alpha$,we get $\alpha = 5$.

In a triangle $ABC$,if $A = 2B$ and the sides opposite to the angles $A, B, C$ are $\alpha + 1, \alpha - 1$ and $\alpha$ respectively,then $\alpha =$

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