With the usual notation in ∆ ABC , if ∠A + ∠B = 120 ° , a = 3 + 1...

Let angle $B$ is $x$.

Using sine rule, we get

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

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

$⇒ \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = \dfrac{\sin 120 ° \cos x - \cos 120 ° \sin x}{\sin x}$

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

$⇒ \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} - \dfrac{1}{2} = \dfrac{\sqrt{3}}{2}\cot x$

$⇒ \dfrac{3 + 2\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}\cot x$

$⇒ \cot x = \sqrt{3} + 2$

$⇒ \tan x = 2 - \sqrt{3}$

$⇒ x = ∠ B = 15 °$

$⇒ 120 - x = ∠ A = 105 °$

$⇒ ∠ A: ∠ B = 7: 1$