<s>[INST] 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
 [/INST] sine rule $$\frac{sin A}{a}=\frac{sin B}{b}=\frac{sin C}{C}\\\frac{a}{b}=\frac{sin A}{sin B}=2+\sqrt{3}\\A+B=180-C=120\\B=120-A\\\frac{sin A}{sin B}=2+\sqrt{3}\\\frac{sin A}{sin (120-A)}=2+\sqrt{3}\\$$now solve for A $$\frac{sin A}{sin 120 cos A -cos 120 sin A}=2+\sqrt{3}$$but it seem to solve for B is easier !$$\frac{sin(120-B)}{sin B}=2+\sqrt{3}\\\frac{\frac{\sqrt{3}}{2}cos B-\frac{1}{2}sin B}{sin B}=2+\sqrt{3}\\\frac{\sqrt{3}}{2}cot B -\frac{1}{2}=2+\sqrt{3}$$
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