In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$

Question

In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$

My Attempt $$ b=\frac{\sqrt{3}-1}{2}a\quad\& \quad \frac{A-B}{2}=B\quad\&\quad\frac{A+B}{2}=2B\\\frac{a-b}{a+b}=\frac{\tan\frac{A-B}{2}}{\tan\frac{A+B}{2}}\implies \frac{3-\sqrt{3}}{\sqrt{3}+1}=\frac{\tan B}{\tan 2B}=\frac{1-\tan^2B}{2} $$

Answer 1

In the standard notation by law of sines we obtain: $$\frac{\sin\beta}{\sin3\beta}=\frac{\sqrt3-1}{2}$$ or $$\frac{1}{3-4\sin^2\beta}=\frac{\sqrt3-1}{2}$$ or $$3-4\sin^2\beta=\sqrt3+1$$ or $$8\sin^2\beta=(\sqrt3-1)^2$$ or $$\sin\beta=\frac{\sqrt3-1}{2\sqrt2},$$ which gives $$\beta=15^{\circ}.$$ Can you end it now?

Answer 2

$$2\sin B=(\sqrt3-1)\sin3B=(...)(\sin B)(3-4\sin^2B)$$

As $\sin B>0,$ $$\sin^2B=\dfrac{2-\sqrt3}4$$

$$\cos2B=1-\dfrac{2-\sqrt3}2=\cos30^\circ$$

$0<2B<360^\circ,2B=360^\circ n\pm30^\circ$ for some integer $n$

$\implies2B=30^\circ$