Question

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

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

Answer 1

The correct option is A $a+b=c$

$a{x}^2+bx+c=0,\tan \frac{P}{3},\tan \frac{Q}{3},$ and
$\angle R=\frac{\pi }{4}$

$\angle P+\angle Q+\angle R=\pi$

$\Rightarrow \angle P+\angle Q=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

$\frac{\angle P}{3}+\frac{\angle Q}{3}=\frac{\pi }{4}$

$\tan (\frac{\angle P}{3}+\frac{\angle Q}{3})=1=\frac{\tan \frac{\angle P}{3}+\tan \frac{\angle Q}{3}}{1-\tan \frac{\angle p}{3}\tan \frac{Q}{3}}$

$1=\frac{-\frac{b}{a}}{1-\frac{c}{a}}\Rightarrow 1=\frac{-\frac{b}{a}}{\frac{a-c}{a}}$

$\Rightarrow a-c=-b$

$\Rightarrow a+b=c$