Question

If two sides of a triangle are roots of the equation $x^2-7x+8=0$ and the angle between these sides is ${60}^{\circ },$ then the product of in-radius and circum-radius of triangle is

• A. $\frac{8}{7}$
• B. $\frac{5}{3}$
• C. $\frac{5\sqrt{2}}{3}$
• D. $8$

Answer 1

The correct option is B $\frac{5}{3}$

Let $a$ and $b$ be the roots of $x^2-7x+8=0.$ Then $a+b=7,ab=8$
Also, $C={60}^{\circ }$
Now, $\frac{1}{2}=\frac{a^2+b^2-c^2}{2ab}$
$\Rightarrow ab=(a+b{)}^2-2ab-c^2\therefore c^2=(a+b{)}^2-3ab=49-24=25\Rightarrow c=5$
Thus, $r\cdot R=\frac{abc}{2(a+b+c)}=\frac{8\times 5}{2(7+5)}=\frac{5}{3}$