The two adjacent sides of a  cyclic quadrilateral are 2 and 5 and the angle between them is 60° . If the area of the quadrilateral is $4\sqrt{3}$, then the perimeter of the quadrilateral is

Here $\cos \theta =\frac{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}{2\mathrm{ab}}\text{ and }\theta ={60}^{\circ }$

$\Rightarrow \cos {60}^{\circ }=\frac{4+25-c^2}{22.5}\Rightarrow 10-29-c^2\Rightarrow c=\sqrt{19}\text{ Also, }$

$\cos \theta =\frac{a^2+b^2-19}{2ab}\text{ and }\theta ={120}^{\circ }$

$\Rightarrow -\frac{1}{2}=\frac{a^2+b^2-19}{2ab}\Rightarrow a^2+b^2-19=-ab\text{ Aera }=$   $=\frac{1}{2}\times 2\times 5\times \cos 60+\frac{1}{2}\mathrm{absin}{120}^{\circ }=4\sqrt{3}$

$\Rightarrow \frac{5\sqrt{3}}{2}+\frac{\mathrm{ab}\sqrt{3}}{4}=4\sqrt{3}=\frac{\mathrm{ab}}{4}=4-\frac{5}{2}=\frac{3}{2}$

$\Rightarrow a^2+b^2+ab=19$

$ab=6$

$a^2+b^2=13\Rightarrow a=2,b=3$

$\text{ perimeter }=\text{ sum of allsides }=2+5+2+3=12$