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

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

Aicharya · Accepted Answer

Given, $\quad R=\frac{\pi}{4}$ Also, $P+Q+R=\pi$ $\begin{array}{ll}\Rightarrow & P+Q+\frac{\pi}{4}=\pi \ \Rightarrow & P+Q=\frac{3 \pi}{4} \ \Rightarrow & \frac{P}{3}+\frac{Q}{3}=\frac{\pi}{4} \ \Rightarrow \quad & 	an \left(\frac{P}{3}+\frac{Q}{3}ight)=	an \left(\frac{\pi}{4}ight) \ \Rightarrow \quad & \frac{	an \frac{P}{3}+	an \frac{Q}{3}}{1-	an \frac{P}{3} 	an \frac{Q}{3}}=1\end{array}$ Since, $	an \frac{P}{3}$ and $	an \frac{Q}{3}$ are the roots of the equation $a x^2+b x+c=0$. $	herefore \quad 	an \frac{P}{3}+	an \frac{Q}{3}=-\frac{b}{a}$ and $	an \frac{P}{3} \cdot 	an \frac{Q}{3}=\frac{c}{a}$ $	herefore$ From Eq. (i), $\frac{-\frac{b}{a}}{1-\frac{c}{a}}=1$ $\begin{array}{ll}\Rightarrow & \frac{-b}{a-c}=1 \ \Rightarrow & a+b=c\end{array}$

In $\triangle P Q R, \angle R=\frac{\pi}{4}, \tan \left(\frac{P}{3}\right), \tan \left(\frac{Q}{3}\right)$…

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