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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

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

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$x^2-2\sqrt 3x+2=0$
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
$a+b=2\sqrt 3$
$\Rightarrow 2ab.\large\frac{1}{2}$$=a^2+b^2-c^2$
$(2\sqrt 3)^2-2\times 2-c^2=2$
$4\times 3-4-c^2=2$
$c=\sqrt 6$
Perimeter =$a+b+c$
$\qquad\;\;\;\;\;=2\sqrt 3+\sqrt 6$
Hence (b) is the correct answer.
answered Nov 18, 2013 by sreemathi.v

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