In triangle ABC , if $A=45^{\circ}, C=75^{\circ}$ and $R=\sqrt{2}$, then $r=$

• A

  $\frac{3+\sqrt{3}}{\sqrt{3}+\sqrt{2}+1}$
• Correct answer
  B

  $\frac{3+\sqrt{3}}{\sqrt{3}+\sqrt{2}+1}$
• C

  $\frac{\sqrt{3}+1}{\sqrt{6}+\sqrt{3}+3}$
• D

  $\frac{\sqrt{3}+1}{\sqrt{3}+\sqrt{2}}$

Solution

In $\triangle A B C, A=45^{\circ}, C=75^{\circ}, B=60^{\circ}, R=\sqrt{2}$
$\begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 \mathbb{R}=2 \sqrt{2} \\ & a=2, b=\sqrt{6}, c=\sqrt{3}+1 \\ & \text { We know } S=\frac{2+\sqrt{6}+\sqrt{3}+1}{2} \\ & \Rightarrow S=\frac{3+\sqrt{3}+\sqrt{6}}{2}\end{aligned}$ $\therefore r=\frac{a b c}{4 R S} \Rightarrow r=\frac{2 \times \sqrt{6} \times(\sqrt{3}+1)}{4 \times \sqrt{2} \times \frac{3+\sqrt{3}+\sqrt{6}}{2}}=\frac{\sqrt{3}+1}{1+\sqrt{2}+\sqrt{3}}$.

Hence, the correct answer is option: 'B'.