Question

In $\Delta P Q R, \angle R=\frac{\pi}{4}, \tan \left(\frac{P}{3}\right), \tan \left(\frac{Q}{3}\right)$ are the roots of the equation $a x^{2}+b x+c=0$, then

• A. a + b = c
• B. b + c = 0
• C. a + c = 0
• D. b = c

Answer 1

Answer: (A)

Given, $R=\frac{\pi}{4}$
Also, $P+Q+R=\pi$
$\Rightarrow P+Q+\frac{\pi}{4}=\pi$
$\Rightarrow P+Q=\frac{3 \pi}{4}$
$\Rightarrow \frac{P}{3}+\frac{Q}{3}=\frac{\pi}{4} $
$ \Rightarrow \tan \left(\frac{P}{3}+\frac{Q}{3}\right)=\tan \left(\frac{\pi}{4}\right)$
$\Rightarrow \frac{\tan \frac{P}{3}+\tan \frac{Q}{3}}{1-\tan \frac{P}{3} \tan \frac{Q}{3}}=1$....(i)
Since, $\tan \frac{P}{3}$ and $\tan \frac{Q}{3}$ are the
roots of the equation $a x^{2}+b x+c=0$
$\therefore \tan \frac{P}{3}+\tan \frac{Q}{3}=-\frac{b}{a}$
and $\tan \frac{P}{3} \cdot \tan \frac{Q}{3}=\frac{c}{a}$
$\therefore $ From Eq. (i),
$\frac{-\frac{b}{a}}{1-\frac{c}{a}}=1$
$\Rightarrow \frac{-b}{a-c}=1$
$ \Rightarrow a+b=c$