If $\triangle ABC\;,\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^0,$ Then find ordered pairs $\left(\angle A,\angle B\right)$ 
In a $\displaystyle \triangle ABC\;,\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^0,$ Then find the ordered pairs $\left(\angle A,\angle B\right)=$
$\bf{Options}::$ $(i)\; \left(45^0,75^0\right)\;\;\;\; (ii)\; \left(105^0,15^0\right)\;\;\;\; (ii) \; \left(15^0,105^0\right)\;\;\;\; (ii)\;\;\;\; (75^0\;,45^0)$

$\bf{My\; Solution::}$ Using Cosine Rule, $$\displaystyle \cos (60^0)=\frac{a^2+b^2-c^2}{2ab}\Rightarrow a^2+b^2-c^2=ab.$$
Now Divide both side by $b^2\;,$ We get $$\displaystyle \left(\frac{a}{b}\right)^2+1-\left(\frac{c}{b}\right)^2=\left(\frac{a}{b}\right)\;,$$ Where $b\neq 0$
Now put value of  $$\displaystyle \frac{a}{b}=2+\sqrt{3}\;,$$ We get $$\displaystyle 7+4\sqrt{3}+1-\left(\frac{c}{b}\right)^2=2+\sqrt{3}.$$
So we get $$\displaystyle \left(\frac{c}{b}\right)^2=6+3\sqrt{3}\Rightarrow \frac{c}{b}=\sqrt{6+3\sqrt{3}}$$
Now How can i solve it, Help required, Thanks

Instead of cosine law, you can try sine law.
Starting from $\frac{\sin A}{\sin (A + 60^0)} =  \frac{2 + \sqrt 3}{1}$, and $\sin (A + 60^0) = \sin A \cos 60^0 + \cos A sin 60^0$
Then, $\sin A = [2 + \sqrt (3)][\sin A .(\frac {1}{2}) + \cos A.(\frac {\sqrt 3}{2})$
∴ $\tan A = -\frac {2 \sqrt 3 + 3}{\sqrt 3}$
∴ $\angle A = 105^0$ (but not using special angle values).
