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# Two sides of a $\Delta$le are given by the roots of the equation $x^2-2\sqrt 3x+2=0$,The angle between the sides is $\large\frac{\pi}{3}$.The perimeter of the triangle is

$\begin{array}{1 1}(a)\;6+\sqrt 3&(b)\;2\sqrt 3+\sqrt 6\\(c)\;2\sqrt 3+\sqrt{10}&(d)\;None\;of\;these\end{array}$

$x^2-2\sqrt 3x+2=0$
Angle =$\large\frac{\pi}{3}$
We know that
$\cos c=\large\frac{a^2+b^2-c^2}{2ab}$
$\cos 60^{\large\circ}=\large\frac{a^2+b^2-c^2}{2ab}$
From the given equation we have
$a+b=2\sqrt 3$
$ab=2$
$\large\frac{1}{2}=\frac{a^2+b^2-c^2}{2ab}$