In ΔPQR, R=π4, TanP3, TanQ3 are the roots of the equation ax2+bx+c=0 then

P + Q + R = π ⇒ P + Q = π -π4=3π4⇒ P3+Q3=π4 ⇒ TanP3+TanQ31-TanP3TanQ3=1⇒ -ba-c=1 ⇒ a+b = c

In ΔPQR, R=π4, TanP3, TanQ3 are the roots of the equation ax2+bx+c=0 then

P + Q + R = π ⇒ P + Q = π -π4=3π4⇒ P3+Q3=π4 ⇒ TanP3+TanQ31-TanP3TanQ3=1⇒ -ba-c=1 ⇒ a+b = c

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In $ΔPQR, R = \frac{π}{4}, Tan \frac{P}{3}, Tan \frac{Q}{3}$ are the roots of the equation ${ax}^{2} + bx + c = 0$ then

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$a + b = c$

$b + c = 0$

$a + c = 0$

$b = c$

answer is A.

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$P + Q + R = π \Rightarrow P + Q = π - \frac{π}{4} = \frac{3π}{4}$

$\Rightarrow \frac{P}{3} + \frac{Q}{3} = \frac{π}{4} \Rightarrow \frac{Tan \frac{P}{3} +Tan \frac{Q}{3}}{1-Tan \frac{P}{3} Tan \frac{Q}{3}} =1$

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

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In ΔPQR, R=π4, TanP3, TanQ3 are the roots of the equation ax2+bx+c=0 then

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In $ΔPQR, R = \frac{π}{4}, Tan \frac{P}{3}, Tan \frac{Q}{3}$ are the roots of the equation ${ax}^{2} + bx + c = 0$ then