Question

In $\Delta A B C$, 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 $r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$\Rightarrow \frac{r}{R}=4 \frac{1}{\sqrt{2}} \sin \frac{B}{2} \sin \left(\frac{\pi}{4}-\frac{B}{2}\right)$
$\Rightarrow \frac{5}{2}=2 \sqrt{2} \sin \frac{B}{2}\left[\frac{1}{\sqrt{2}} \cos \frac{B}{2}-\frac{1}{\sqrt{2}} \sin \frac{B}{2}\right]$
$\Rightarrow \frac{5}{4}=\sin \frac{B}{2} \cos \frac{B}{2}-\sin ^{2} \frac{B}{2}\,....(i)$
and $r_{1}=4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
$\Rightarrow \frac{r_{1}}{R}=4 \times \frac{1}{\sqrt{2}} \cos \frac{B}{2} \cos \left(\frac{\pi}{4}-\frac{B}{2}\right)$
$\Rightarrow \frac{\lambda}{2}=2 \sqrt{2} \cos \frac{B}{2}\left[\frac{1}{\sqrt{2}} \cos \frac{B}{2}+\frac{1}{\sqrt{2}} \sin \frac{B}{2}\right]$
$\Rightarrow \frac{\lambda}{4}=\cos ^{2} \frac{B}{2}+\cos \frac{B}{2} \sin \frac{B}{2}\,....(ii)$
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
$x^{2}-4 x+3=0$ are $1,3$