Q73 / 160 · TS EAMCET 2005 · Mathematics
Two sides of a triangle are given by the roots of the equation $x^2-5 x+6=0$ and the angle between the sides is $\frac{\pi}{3}$. Then, the perimeter of the triangle is
- A$5+\sqrt{2}$
- B$5+\sqrt{3}$
- C$5+\sqrt{5}$
- Correct answerD$5+\sqrt{7}$
Solution
Given equation is
$\begin{aligned}
x^2-5 x+6 & =0 \\
\Rightarrow(x-3)(x-2)=0 \Rightarrow x & =3,2
\end{aligned}$
These are the sides of a triangle
Let $a=3, b=2, \angle C=\frac{\pi}{3}$
$\begin{aligned}
& \therefore \quad \cos C=\frac{a^2+b^2-c^2}{2 a b} \\
& \Rightarrow \quad \cos \left(\frac{\pi}{3}\right)=\frac{3^2+2^2-c^2}{2 \cdot 3 \cdot 2} \\
& \Rightarrow \quad \frac{1}{2}=\frac{13-c^2}{12} \\
& \Rightarrow \quad c^2=13-6=7 \\
& \Rightarrow \quad c= \pm \sqrt{7} \\
& \Rightarrow \quad c=\sqrt{7} \quad \text { (neglect }- \text { ve sign) } \\
& \text { Perimeter of a triangle }=a+b+c \\
& =3+2+\sqrt{7} \\
& =5+\sqrt{7} \\
&
\end{aligned}$
Hence, the correct answer is option: 'D'.