Question

In a $△ABC$, if $A={30}^o,b=2,c=\sqrt{3}+1$, then $\frac{C-B}{2}=$

• A. ${15}^o$
• B. ${30}^o$
• C. ${45}^o$
• D. $noneofthese$

Answer 1

The correct option is B ${30}^o$

Given In $△ABC,A={30}^{\circ }⟹B+C={150}^{\circ },b=2,c=\sqrt{3}+1$

$\frac{b+c}{c-b}=\frac{\tan \frac{B+C}{2}}{\tan \frac{C-B}{2}}⟹\frac{\sqrt{3}+3}{\sqrt{3}-1}=\frac{\tan {75}^{\circ }}{\tan \frac{C-B}{2}}$

$\tan \frac{C-B}{2}=\frac{\sqrt{3}-1}{\sqrt{3}+3}\times \tan {75}^{\circ }=\frac{1}{\sqrt{3}}⟹\frac{C-B}{2}={30}^{\circ }$