Question

In $\Delta ABC$, right angled at $A$, the circumradius, inradius and radius of the excircle opposite to $A$ are respectively in the ratio $2:5:\lambda$, then the roots of the equation $x^2-(\lambda -5)x+(\lambda -6)=0$ are

• A. 3,4
• B. 5,13
• C. 1,3
• D. 8,13

Answer 1

Answer: (C)

$\because A=\frac{\pi }{2}$
$\therefore B+C=\frac{\pi }{2}$
and $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$\Rightarrow \frac{r}{R}=4\frac{1}{\sqrt{2}}\sin \frac{B}{2}\sin \left(\frac{\pi }{4}-\frac{B}{2}\right)$
$\Rightarrow \frac{5}{2}=2\sqrt{2}\sin \frac{B}{2}\left[\frac{1}{\sqrt{2}}\cos \frac{B}{2}-\frac{1}{\sqrt{2}}\sin \frac{B}{2}\right]$
$\Rightarrow \frac{5}{4}=\sin \frac{B}{2}\cos \frac{B}{2}-{\sin }^2\frac{B}{2}....(i)$
and $r_1=4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
$\Rightarrow \frac{r_1}{R}=4\times \frac{1}{\sqrt{2}}\cos \frac{B}{2}\cos \left(\frac{\pi }{4}-\frac{B}{2}\right)$
$\Rightarrow \frac{\lambda }{2}=2\sqrt{2}\cos \frac{B}{2}\left[\frac{1}{\sqrt{2}}\cos \frac{B}{2}+\frac{1}{\sqrt{2}}\sin \frac{B}{2}\right]$
$\Rightarrow \frac{\lambda }{4}={\cos }^2\frac{B}{2}+\cos \frac{B}{2}\sin \frac{B}{2}....(ii)$
From Eqs. (i) and (ii),
$\frac{\lambda }{4}-\frac{5}{4}=1$
$\Rightarrow \lambda -5=4$
$\Rightarrow \lambda =9$
So, the roots of quadratic equation
$x^2-4x+3=0$ are $1,3$