Question

In a triangle $PQR,\angle R=\frac{\pi }{2}$. If $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are the roots of $a{x}^2+bx+c=0,a\ne 0$, then

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

Answer 1

Answer: (D)

$c=a+b$

$\Rightarrow R={90}^o\Rightarrow P+Q={90}^o$
$\therefore \frac{P}{2}+\frac{Q}{2}={45}^o$
$\frac{S}{1-P}=1\Rightarrow S=1-P$
$-\frac{b}{a}=1-\frac{c}{a}or-b=a-c\therefore c=a+b$