Quadratic Equation

Geometry Level 3

In a triangle $PQR$ , $\angle R = 90^\circ$ .

If $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are the roots of $ax^2 + bx + c = 0$ , where $a\ne 0$ , then which of the following is true?

b = a + c

b = c

c = a + b

a = b + c

3 solutions

Rishabh Jain
Jan 16, 2016

$\frac{P+Q}{2}=\frac{180-R}{2}=45$ $\Rightarrow \tan(\frac{P+Q}{2})=\tan45=1$ $\Rightarrow \frac{\tan\frac{P}{2}+\tan\frac{Q}{2}}{1-(\tan\frac{P}{2})(\tan{Q}{2})}=1$ $\Rightarrow \frac{\color{#3D99F6}{\frac{-b}{a}}}{1-\color{#3D99F6}{\frac{c}{a}}}=1 \Rightarrow \color{#D61F06}{c=a+b}$

Gogul Raman Thirunathan
Feb 19, 2016

Ved Sharda
Jan 16, 2016

Let 'x' = tan(P/2) and 'y' = tan(Q/2)

P + Q = 90 degree

(P/2) + (Q/2) = 45 degree

tan[P/2 + Q/2] = 1 {as tan45 = 1}

we know that, tan(P/2)=x and tan(Q/2)=y

so tan[P/2 + Q/2] = (x+y)/1-xy

x + y = 1 - xy {x+y=sum of roots and xy=product of roots}

-b/a = 1 - (c/a)

(c-b)/a = 1

c-b = a

c = a + b

Quadratic Equation

3 solutions

0 pending reports