With the usual notation, in $∆$ A B C , if $⌊\mathrm{A}+⌊\mathrm{B}={120}^{\circ },\mathrm{a}=\sqrt{3}+1$and $\mathrm{b}=\sqrt{3}-1$, then the ratio $\lfloor \mathrm{A}:\lfloor \mathrm{B}$ is

For a $∆$ABC , it is given that $\mathrm{a}=\sqrt{3}+1,\mathrm{b}=\sqrt{3}-1$ and $⌊\mathrm{A}+\mathrm{B}̸={120}^{\circ }$

Clearly, $⌊\mathrm{c}={60}^{\circ }\left[\because ⌊\mathrm{A}+\cancel{\mathrm{B}}+\left.⌊\mathrm{C}={180}^{\circ }\right]\right.$

Now, by tangent law, we have $\tan \frac{\mathrm{A}-\mathrm{B}}{2}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\cot \frac{\mathrm{C}}{2}$

$=\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)}\cot \left(\frac{{60}^0}{2}\right)=\frac{2}{2\sqrt{3}}\cot \left({30}^{\circ }\right)=\frac{1}{\sqrt{3}}\times \sqrt{3}=1$

$\Rightarrow \tan \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=1=\tan {45}^{\circ }\Rightarrow \frac{\mathrm{A}-\mathrm{B}}{2}={45}^{\circ }\Rightarrow ⌊\mathrm{A}-⌊\mathrm{B}={90}^{\circ }$

On solving $⌊\mathrm{A}-⌊\mathrm{B}={90}^{\circ }\mathrm{and} ⌊\mathrm{A}+⌊\mathrm{B}={120}^{\circ }$, 

We get $⌊\mathrm{A}={105}^{\circ }$ and $⌊\mathrm{B}={15}^{\circ }$So, $\lfloor \mathrm{A}:\lfloor \mathrm{B}=7:1$