Question

If $a,b,c$ are the sides opposite to angles $A,B,C$ of a triangle $ABC,$ respectively and $\angle A=\frac{\pi }{3},b:c=\sqrt{3}+1:2,$ then the value of $\angle B-\angle C$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{12}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B $\frac{\pi }{6}$

$\frac{b}{c}=\frac{\sqrt{3}+1}{2}$
$\Rightarrow \frac{b-c}{b+c}=\frac{\sqrt{3}+1-2}{\sqrt{3}+1+2}$
$\Rightarrow \frac{b-c}{b+c}=\frac{\sqrt{3}-1}{\sqrt{3}+3}$

Now, $\tan \frac{B-C}{2}=\frac{b-c}{b+c}\cot \frac{A}{2}$
$\Rightarrow \tan \frac{B-C}{2}=\frac{\sqrt{3}-1}{\sqrt{3}+3}\cdot \sqrt{3}$
$\Rightarrow \tan \frac{B-C}{2}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$
$\Rightarrow \frac{B-C}{2}=\frac{\pi }{12}$
$\Rightarrow B-C=\frac{\pi }{6}$