Question

In a triangle ABC, if $(\sqrt{3}-1)a,A=3B$ ,then $\angle$C is

• A. ${60}^0$
• B. ${120}^0$
• C. ${30}^0$
• D. ${45}^0$

Answer 1

The correct option is B ${120}^0$

Using sine rule

$2sinB=(3–\surd -1)sin3B=(...)\left(sinB\right)(3-4si{n}^{2}B)$

As $sinB>0,$

$si{n}^{2}B=\frac{2-\sqrt{3}}{4}$

$cos2B=1-\frac{2-\sqrt{3}}{2}=cos{30}^{\circ }$

$0<2B<360,2B=360n\pm 30$

for some integer $n$

$⟹2B=30\circ ,B={15}^{\circ }$

$⟹A=3B=3\times 15={45}^{\circ }$

$\angle C=180-(A+B)=180-(45+15)=180-60={120}^{\circ }$