Triangle related question

Posted July 15th, 2009 by alpha

In a triangle PQR , angle R=pi/2 .It tan(P/2)and tan(Q/2) are the roots of the equation $ax^2+bx+c=0,(a\ne0)$ , then
a)a+b=c
b)b+c=a
c)a+c=b
d)b=c

Comments

On July 15th, 2009 Structure says:

$\frac{P}{2}=u;\:\frac{Q}{2}=v$

$u+v=\frac{\pi}{4}$

$\tan(u+v)=\frac{\tan u+\tan v}{1-\tan u\tan v}=1$

$\tan u+\tan v=1-\tan u\tan v$

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

$a+b=c$