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

The correct option is D: ${120}^{\circ }$

$\because (\sqrt{3}-1)a=2b$

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

Applying sine Rule, we get $\frac{a}{\text{sin}A}=\frac{b}{\text{sin}B}=\frac{c}{\text{sin}C}$

$\therefore \frac{a}{b}=\frac{\text{sin}A}{\text{sin}B}$

$\frac{\text{sin~A}}{\text{sin~B}}=\frac{2}{\sqrt{3}-1}=\frac{2}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\sqrt{3}+1$

$\text{Given,}A=3B$

$\therefore \frac{\text{sin}3B}{\text{sin}B}=\sqrt{3}+1$

$\Rightarrow \frac{\text{3sinB}-4{\text{sin}}^{3}B}{\text{sin}B}=\sqrt{3}+1$

$\Rightarrow 3-4{\text{sin}}^{2}B=\sqrt{3}+1$

$\Rightarrow 3-\sqrt{3}-1=4{\text{sin}}^{2}B$

$\Rightarrow \frac{2-\sqrt{3}}{4}={\text{sin}}^{2}B$

$\Rightarrow {\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}^2={\text{sin}}^{2}B$

$\Rightarrow \text{sin}B=\frac{\sqrt{3}-1}{2\sqrt{2}}$

Since, $\text{sin}{15}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\Rightarrow B={15}^{\circ }$

and $A=3B=3\times {15}^{\circ }={45}^{\circ }$

$\therefore C={180}^{\circ }-A-B={180}^{\circ }-{60}^{\circ }={120}^{\circ }$