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

We have a triangle $$\triangle ABC$$ with the usual notation: side $$a$$ is opposite angle $$A$$ and side $$b$$ is opposite angle $$B$$. The given data are

$$\angle A+\angle B = 120^{\circ}, \qquad a = \sqrt{3}+1, \qquad b = \sqrt{3}-1.$$

From the Law of Sines, which states $$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C},$$ we obtain the proportionality

$$\dfrac{a}{b}=\dfrac{\sin A}{\sin B}. \quad -(1)$$

Substituting the given side lengths into (1) gives

$$\dfrac{\sin A}{\sin B}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}.$$

To simplify the right-hand side we rationalise the denominator:

$$\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\times\dfrac{\sqrt{3}+1}{\sqrt{3}+1} =\dfrac{(\sqrt{3}+1)^{2}}{(\sqrt{3})^{2}-1^{2}} =\dfrac{3+1+2\sqrt{3}}{3-1} =\dfrac{4+2\sqrt{3}}{2} =2+\sqrt{3}.$$

Hence

$$\dfrac{\sin A}{\sin B}=2+\sqrt{3}. \quad -(2)$$

Now let us set $$\angle A = x^{\circ}.$$ Because $$\angle A+\angle B=120^{\circ},$$ we can write $$\angle B = 120^{\circ}-x^{\circ}.$$ Substituting these into equation (2) yields

$$\dfrac{\sin x}{\sin(120^{\circ}-x)} = 2+\sqrt{3}. \quad -(3)$$

We next expand the denominator using the sine subtraction formula $$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.$$ Taking $$\alpha=120^{\circ}$$ and $$\beta=x$$ gives

$$\sin(120^{\circ}-x)=\sin120^{\circ}\cos x-\cos120^{\circ}\sin x.$$

Because $$\sin120^{\circ}=\dfrac{\sqrt{3}}{2}$$ and $$\cos120^{\circ}=-\dfrac12,$$ we have

$$\sin(120^{\circ}-x)=\dfrac{\sqrt{3}}{2}\cos x-\left(-\dfrac12\right)\sin x =\dfrac{\sqrt{3}}{2}\cos x+\dfrac12\sin x =\dfrac{\sqrt{3}\cos x+\sin x}{2}.$$

Substituting this expression back into (3) we get

$$\dfrac{\sin x}{\dfrac{\sqrt{3}\cos x+\sin x}{2}}=2+\sqrt{3}.$$ Multiplying numerator and denominator by $$2$$ yields

$$\dfrac{2\sin x}{\sqrt{3}\cos x+\sin x}=2+\sqrt{3}.$$

Cross-multiplying, we obtain

$$2\sin x=(2+\sqrt{3})(\sqrt{3}\cos x+\sin x).$$

Expanding the right-hand side gives

$$2\sin x=(2+\sqrt{3})\sqrt{3}\cos x+(2+\sqrt{3})\sin x =(2\sqrt{3}+3)\cos x+(2+\sqrt{3})\sin x.$$

Bringing all terms to the left and combining like terms:

$$2\sin x-(2+\sqrt{3})\sin x-(2\sqrt{3}+3)\cos x=0,$$ $$\bigl[2-(2+\sqrt{3})\bigr]\sin x-(2\sqrt{3}+3)\cos x=0,$$ $$-\sqrt{3}\sin x-(2\sqrt{3}+3)\cos x=0.$$

Multiplying by $$-1$$ simplifies this to

$$\sqrt{3}\sin x+(2\sqrt{3}+3)\cos x=0.$$

Dividing throughout by $$\cos x$$ (which is non-zero for the angle we will obtain) gives an equation in $$\tan x$$:

$$\sqrt{3}\tan x+(2\sqrt{3}+3)=0,$$ $$\tan x=-\dfrac{2\sqrt{3}+3}{\sqrt{3}}.$$

Splitting the numerator and simplifying:

$$-\dfrac{2\sqrt{3}}{\sqrt{3}}-\dfrac{3}{\sqrt{3}} =-(2+\sqrt{3}).$$

Thus

$$\tan x=-(2+\sqrt{3}). \quad -(4)$$

We recognise that $$\tan75^{\circ}=2+\sqrt{3},$$ so equation (4) becomes

$$\tan x=-\tan75^{\circ}.$$

The principal value that satisfies this while lying between $$0^{\circ}$$ and $$120^{\circ}$$ is

$$x=105^{\circ},$$ because $$\tan105^{\circ}=\tan(180^{\circ}-75^{\circ})=-\tan75^{\circ}.$$

Therefore

$$\angle A = 105^{\circ}, \qquad \angle B = 120^{\circ}-105^{\circ}=15^{\circ}.$$

The required ratio is then

$$\angle A:\angle B = 105^{\circ}:15^{\circ}=7:1.$$

Hence, the correct answer is Option A.