Question

Two sides of triangle are given by the roots of the equation $x^2-5x-6=0$, the angle between the sides is $\pi /3$. Then find the perimeter of the triangle.

• A. $5+\sqrt{2}$
• B. $5+\sqrt{3}$
• C. $5+\sqrt{5}$
• D. $5+\sqrt{7}$

Answer 1

Answer: (A)

$5+\sqrt{2}$

Two sides of triangle are given by the roots of the equation $x^2-2\sqrt{3x}+2=0$, the angle between the sides is $\pi /3$
Let $a,b$ be the roots of above equation.
$\Rightarrow a+b=2\sqrt{3}$ and $ab=2$
$\angle C={60}^{\circ }$
$\cos C=\frac{a^2+b^2-c^2}{2ab}$

$\frac{1}{2}=\frac{a^2+b^2-c^2}{2ab}$
$\Rightarrow (a+b{)}^2-2ab-c^2=ab$
$\Rightarrow c^2=6$
$\therefore$ Perimeter$=a+b+c=2\sqrt{3}+\sqrt{6}$