In a $$\Delta ABC$$, $$\frac{a}{b} = 2 + \sqrt{3}$$, and $$\angle C = 60^\circ$$. Then the ordered pair $$(\angle A, \angle B)$$ is equal to:

We have a triangle $$\triangle ABC$$ with side lengths $$a,\,b,\,c$$ opposite to angles $$A,\,B,\,C$$ respectively. It is given that $$\frac{a}{b}=2+\sqrt{3}$$ and that $$\angle C = 60^\circ.$$

First we recall the Sine Rule, which states

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.$$

Using the first two fractions of the rule we can write

$$\frac{a}{b}=\frac{\sin A}{\sin B}.$$

Substituting the given ratio of the sides we obtain

$$\frac{\sin A}{\sin B}=2+\sqrt{3}.$$ Hence $$\sin A=(2+\sqrt{3})\sin B.$$

Because the three angles of a triangle sum to $$180^\circ,$$ and we already know $$C=60^\circ,$$ we may write

$$A+B+60^\circ = 180^\circ \quad\Longrightarrow\quad A+B = 120^\circ.$$ So $$A = 120^\circ - B.$$

Now we substitute this value of $$A$$ into the earlier sine equation:

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

We expand $$\sin(120^\circ - B)$$ using the identity $$\sin(x-y)=\sin x \cos y - \cos x \sin y.$$ Here $$x=120^\circ$$ and $$y=B,$$ so

$$\sin(120^\circ - B) = \sin 120^\circ \cos B - \cos 120^\circ \sin B.$$

We know the exact trigonometric values

$$\sin 120^\circ = \sin(180^\circ-60^\circ)=\sin 60^\circ=\frac{\sqrt{3}}{2},$$ $$\cos 120^\circ = -\cos 60^\circ = -\frac12.$$

Therefore

$$\sin(120^\circ - B)=\frac{\sqrt{3}}{2}\cos B - \left(-\frac12\right)\sin B =\frac{\sqrt{3}}{2}\cos B +\frac12 \sin B.$$

Equating this to $$(2+\sqrt{3})\sin B$$ we get

$$\frac{\sqrt{3}}{2}\cos B + \frac12\sin B = (2+\sqrt{3})\sin B.$$

Multiplying both sides by $$2$$ to clear denominators gives

$$\sqrt{3}\cos B + \sin B = (4+2\sqrt{3})\sin B.$$

Now we bring the terms involving $$\sin B$$ to one side:

$$\sqrt{3}\cos B + \sin B - (4+2\sqrt{3})\sin B = 0,$$ $$\sqrt{3}\cos B + \left[1-(4+2\sqrt{3})\right]\sin B = 0,$$ $$\sqrt{3}\cos B + (-3-2\sqrt{3})\sin B = 0.$$

Rewriting, we have

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

Dividing both sides by $$\cos B$$ (which is non-zero for a triangle angle) we obtain

$$\sqrt{3} = (3+2\sqrt{3})\tan B,$$ so $$\tan B = \frac{\sqrt{3}}{3+2\sqrt{3}}.$$

To simplify this fraction we rationalise the denominator by multiplying numerator and denominator by the conjugate $$3-2\sqrt{3}:$$

$$\tan B = \frac{\sqrt{3}(3-2\sqrt{3})}{(3+2\sqrt{3})(3-2\sqrt{3})}.$$

The denominator becomes a difference of squares:

$$(3+2\sqrt{3})(3-2\sqrt{3}) = 3^2-(2\sqrt{3})^2 = 9-4\cdot3 = 9-12=-3.$$

Hence

$$\tan B = \frac{\sqrt{3}(3-2\sqrt{3})}{-3} = -\frac{\sqrt{3}}{3}(3-2\sqrt{3}).$$

Distributing the factor $$-\frac{\sqrt{3}}{3}$$ inside the bracket gives

$$\tan B = -\sqrt{3} + \frac{2\sqrt{3}\cdot\sqrt{3}}{3} = -\sqrt{3} + \frac{2\cdot3}{3} = -\sqrt{3} + 2.$$ So $$\tan B = 2-\sqrt{3}.$$

The exact value $$\tan 15^\circ$$ is known to be $$2-\sqrt{3},$$ therefore

$$B = 15^\circ.$$

Finally we find $$A$$ using $$A+B=120^\circ:$$

$$A = 120^\circ - B = 120^\circ - 15^\circ = 105^\circ.$$

Thus the ordered pair $$(\angle A,\;\angle B)$$ equals $$(105^\circ,\,15^\circ).$$

Hence, the correct answer is Option A.