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

Start with Sine Law since you are given the quotient of sides.
Make a rough sketch of segment containing an angle $60$, to note that $A$ must be > 90 an obtuse angle anyhow, B is noted to be a narrow angle, should be an acute angle. 
Arguments in degrees to avoid Latex. (I prefer such practice only when there is no degree/radian conflict anyway).
$\dfrac{\sin A}{\sin (A + 60)} =  \dfrac{2 + \sqrt 3}{1}$, recognize the right hand side as $ \tan 75 . $ 
So, using sine supplementary angle identiity $ \sin\theta = \sin ( 180 - \theta) $:
$$\dfrac{\sin A}{\sin (A + 60)}  = \dfrac{\sin 75}{\cos 75}  = \dfrac{\sin 75}{\sin 15} =  \dfrac{\sin 105}{ \sin 165} = \dfrac{\sin (105)}{\sin (  105 + 60) }$$ 
Comparing arguments of first and last fractions we straightaway note that  $ A = 105 $, $$ (A,B,C)= (105,15,60).$$
