Question

If $\triangle ABC$, if $a^{2}-c^{2}=b(b-c), \sqrt{2} a=2 b-c$ and $R=\frac{1}{\sqrt{3}}$ then $b=$

• A. $\frac{\sqrt{2}}{\sqrt{3}}$
• B. $\frac{\sqrt{3}-1}{\sqrt{6}}$
• C. $\frac{\sqrt{3}+1}{\sqrt{6}}$
• D. $\frac{\sqrt{3}}{\sqrt{2}}$

Answer 1

Answer: (C)

We have
$a^{2}-c^{2}=b(b-c), \sqrt{2} a=2 b-c$ and $ R=\frac{1}{\sqrt{3}} $
$\Rightarrow a^{2}-c^{2}=-b^{2}-b c $
$\Rightarrow b^{2}+c^{2}-a^{2}=b c$
$\Rightarrow \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{1}{2}$
$ \Rightarrow \cos A=\frac{1}{2} \Rightarrow A=60^{\circ}$
Since, $\frac{a}{\sin A}=2 R \Rightarrow \frac{a}{\sin 60^{\circ}}=\frac{2}{\sqrt{3}} \Rightarrow a=1 $
$\sqrt{2} a=2 b-c \Rightarrow 2 a^{2}=4 b^{2}+c^{2}-4 b c $
$\Rightarrow 2 a^{2}=4\left(b^{2}-b c\right)+c^{2} $
$\Rightarrow 2=4\left(a^{2}-c^{2}\right)+c^{2} $
$\Rightarrow 2=4-3 c^{2} \Rightarrow c^{2}=\frac{2}{3} $
$\Rightarrow c=\frac{\sqrt{2}}{\sqrt{3}} $
Now, $2 b=\sqrt{2} a+c $
$\Rightarrow 2 b=\sqrt{2}+\frac{\sqrt{2}}{\sqrt{3}}$
$ \Rightarrow b=\frac{\sqrt{2}(\sqrt{3}+1)}{2 \sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{6}}$