Question

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

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

Answer 1

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

Given In $△PQR\angle R={90}^{\circ },\tan \frac{P}{2},\tan \frac{Q}{2}$ are roots of the equation $a{x}^2+bx+c=0(a\ne 0)$

$P+Q+R=\pi \Rightarrow P+Q+\frac{\pi }{2}=\pi$

$P+Q=\frac{\pi }{2}$ , sum of roots$=\tan \frac{P}{2}+\tan \frac{Q}{2}=\frac{-b}{a}$

$\frac{P+Q}{2}=\frac{\pi }{4}$, Products of roots$=\tan \frac{P}{2}.\tan \frac{Q}{2}=\frac{c}{a}$

$\tan (\frac{P+Q}{2})=1$

$\Rightarrow \frac{\tan P/2+\tan Q/2}{1-\tan P/2\tan Q/2}=1$

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

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

$\Rightarrow a+b=c$