Question

If $△ABC$, if $a^2−c^2=b(b−c),\sqrt{2}a=2b−c$ 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
$a^2−c^2=b(b−c),\sqrt{2}a=2b−c$ and $R=\frac{1}{\sqrt{3}}$
$⇒a^2−c^2=−b^2−bc$
$⇒b^2+c^2−a^2=bc$
$⇒\frac{b^2+c^2−a^2}{2bc}=\frac{1}{2}$
$⇒\cos A=\frac{1}{2}⇒A=60^{∘}$
Since, $\frac{a}{\sin A}=2R⇒\frac{a}{\sin 60^{∘}}=\frac{2}{\sqrt{3}}⇒a=1$
$\sqrt{2}a=2b−c⇒2a^2=4b^2+c^2−4bc$
$⇒2a^2=4\left(b^2−bc\right)+c^2$
$⇒2=4\left(a^2−c^2\right)+c^2$
$⇒2=4−3c^2⇒c^2=\frac{2}{3}$
$⇒c=\frac{\sqrt{2}}{\sqrt{3}}$
Now, $2b=\sqrt{2}a+c$
$⇒2b=\sqrt{2}+\frac{\sqrt{2}}{\sqrt{3}}$
$⇒b=\frac{\sqrt{2}(\sqrt{3}+1)}{2\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{6}}$