Question

In a $△ABC,$, if $A={30}^0,b=2,c=\sqrt{3}+1$, then $\frac{C-B}{2}$ is equal to

• A. ${15}^0$
• B. ${30}^0$
• C. ${45}^0$
• D. None of these

Answer 1

The correct option is B ${30}^0$

Given, $A={30}^0,b=2,c=\sqrt{3}+1,\frac{C-B}{2}=$

$\tan \left(\frac{C-B}{2}\right)=\frac{c-b}{c+b}\cot \frac{A}{2}=\frac{\sqrt{3}-1}{\sqrt{3}+3}\cot {15}^0$

$=\frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)}\left[\frac{\cot {30}^0+1}{\cot {30}^0-1}\right]$

$=\frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)}\left[\frac{\sqrt{3}+1}{\sqrt{3}-1}\right]=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{C-B}{2}=\frac{\pi }{6}={30}^0$