With the usual notation in $∆ABC$, if $\angle A+\angle B=120°, a=\sqrt{3}+1$ units and $b=\sqrt{3}-1$ units, then the ratio $\angle A:\angle B$ is

Let angle $B$ is $x$.

Using sine rule, we get

$\frac{\sqrt{3}+1}{\sin \left(120°-x\right)}=\frac{\sqrt{3}-1}{\sin x}$.

$\Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\sin \left(120°-x\right)}{\sin x}$

$\Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\sin 120°\cos x-\cos 120°\sin x}{\sin x}$

$\Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1}=\sin 120°\cot x-\cos 120°$

$\Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1}-\frac{1}{2}=\frac{\sqrt{3}}{2}\cot x$

$\Rightarrow \frac{3+2\sqrt{3}}{2}=\frac{\sqrt{3}}{2}\cot x$

$\Rightarrow \cot x=\sqrt{3}+2$

$\Rightarrow \tan x=2-\sqrt{3}$

$\Rightarrow x=\angle B=15°$

$\Rightarrow 120-x=\angle A=105°$

$\Rightarrow \angle A:\angle B=7:1$