Mathematics | TS EAMCET 2018 | Discussion Page

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 $\alpha$=

A) 3

B) 4

C) 5

D) 6

Option C

According to sine law,

$\frac{ \sin A}{ \alpha +1}= \frac{ \sin B}{\alpha-1}$

$\Rightarrow$ $\frac{2 \sin B \cos B}{\alpha+1}=\frac{\sin B}{\alpha-1}$ [$\because$ A=2B]

$\Rightarrow$ $ \cos B=\frac{\alpha+1}{2(\alpha-1)}$

$\Rightarrow$ $\frac{(\alpha+1)^{2}+\alpha^{2}-(\alpha-1)^{2}}{2\alpha(\alpha+1)}=\frac{\alpha+1}{2(\alpha-1)}$

$\Rightarrow$ $\frac{4\alpha+\alpha^{2}}{2\alpha(\alpha+1)}=\frac{\alpha+1}{2(\alpha-1)}$

$\Rightarrow$ $(4+\alpha)(\alpha-1)=(\alpha+1)^{2}$

$\Rightarrow$ $\alpha^{2}+3\alpha-4=\alpha^{2}+2\alpha+1$

$\Rightarrow$ $\alpha$=5

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