Question

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