In a triangle $PQR,\mathrm{\angle }R=\pi /4.\text{ If }\tan (P/3)$ and $\tan (Q/3)$ are the roots of the equation $a{x}^2+bx+c=0$, then

$\begin{array}{r}\tan \left(\frac{P}{3}\right)+\tan \left(\frac{Q}{3}\right)=-\frac{b}{a}\\ \tan \left(\frac{P}{3}\right)\tan \left(\frac{Q}{3}\right)=-\frac{c}{a}\end{array}$

Now, $\mathrm{\angle }R=\pi /4\Rightarrow P+Q=3\pi /4$

$\begin{array}{l}\therefore 1=\tan \left(\frac{\pi }{4}\right)=\frac{\tan (P/3)+\tan (Q/3)}{1-\tan (P/3)(Q/3)}\\ =\frac{-b/a}{1-c/a}=\frac{-b}{a-c}\end{array}$

$\Rightarrow a+b=c.$