Question

In a triangle ABC, $\angle A={60}^{\circ }$ and $b:c=\sqrt{3}+1:2$. Then the value of $(\angle B-\angle C)$ is

• A. ${30}^{\circ }$
• B. ${15}^{\circ }$
• C. ${75}^{\circ }$
• D. $37\frac{1}{2}$${}^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$

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

Now using $\tan \frac{B-C}{2}=\frac{b-c}{b+c}\cot \frac{A}{2}$, we get

$\tan \frac{B-C}{2}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\cdot \frac{\sqrt{3}}{\sqrt{3}}=2-\sqrt{3}=\tan {15}^{\circ }$

$\Rightarrow \frac{B-C}{2}={15}^{\circ }$

$\therefore B-C={30}^{\circ }$