Discussion Forum

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 $\alpha$= 

Answer:

Explanation:

 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