In a triangle $PQR,\ \angle R = \frac{\pi}{2}$. If $ an\left( \frac{P}{2} ight)$ and $ an\left( \frac{\mathcal{Q}}{2} ight)$ are the roots of $ax^{2} + bx + c = 0,\ a
eq 0$, then

Question

In a triangle $PQR,\ \angle R = \frac{\pi}{2}$. If $	an\left( \frac{P}{2} ight)$ and $	an\left( \frac{\mathcal{Q}}{2} ight)$ are the roots of $ax^{2} + bx + c = 0,\ a 
eq 0$, then

Dr. Abubacker · Accepted Answer

(c)
Here, $	an{\frac{P}{2} + 	an{\frac{\mathcal{Q}}{2} = - \frac{b}{a}}}$
And $	an{\frac{P}{2}	an{\frac{\mathcal{Q}}{2} = \frac{c}{a}}\ \ \ \ \ \ \ \ldots(i)}$
Also, $\frac{P}{2} + \frac{\mathcal{Q}}{2} + \frac{R}{2} = \frac{\pi}{2}\ \ \ \ \ \ \lbrack\because P + \mathcal{Q +}R = \pibrack$
$\Rightarrow \frac{P + \mathcal{Q}}{2} = \frac{\pi}{4}\ \ \ \ \ \ \left\lbrack \because\angle R = \frac{\pi}{2},\ given ightbrack$
$	an\left( \frac{P}{2} + \frac{\mathcal{Q}}{2} ight) = 1 \Rightarrow \frac{	an{\frac{P}{2} + 	an\frac{\mathcal{Q}}{2}}}{1 - 	an{\frac{P}{2}	an\frac{\mathcal{Q}}{2}}} = 1$
$\Rightarrow \frac{- \frac{b}{a}}{1 - \frac{c}{a}} = 1\ \  \Rightarrow \ \  - \frac{b}{a} = 1 - \frac{c}{a}$ [from Eq. (i)]
$\Rightarrow c = a + b$

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

Explanation

Reviewed by Dr. Abubacker

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