In a triangle PQR, R=(π)/(2) . If ( P 2 ) and ( Q 2 ) are the roots of a x^2+b x+c=0, a ≠ 0 then

Correct answer: B. c=a+b ( P 2 ), ( Q 2 ) are the roots of ax^2+bx+c=0 ( P 2 )+ ( Q 2 )=- b a ( P 2 ) ( Q 2 )= c a ( P 2 )+ ( Q 2 ) 1- ( P 2 ) ( Q 2 ) = ( P 2 + Q 2 )=1 - b a 1- c a =1 - b a = a a - c a - b=a -c c=a+b .

JEE Main 2005

Exam: JEE Main
Year: 2005
Subject: Mathematics
Chapter: Properties of Triangles

JEE Main 2005 · Mathematics · Properties of Triangles

In a triangle PQR, R=(π)/(2) . If ( P 2 ) and ( Q 2 ) are the roots...

In a triangle

\mathrm{PQR}, \angle \mathrm{R}=\frac{\pi}{2}

. If

\tan \left(\frac{\mathrm{P}}{2}\right)

and

\tan \left(\frac{\mathrm{Q}}{2}\right)

are the roots of

a x^2+b x+c=0, a \neq 0

then

Options

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

Correct answer

B. $c=a+b$

Step-by-step solution

\tan \left(\frac{\mathrm{P}}{2}\right), \tan \left(\frac{\mathrm{Q}}{2}\right)

are the roots of

\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0

\tan \left(\frac{\mathrm{P}}{2}\right)+\tan \left(\frac{\mathrm{Q}}{2}\right)=-\frac{\mathrm{b}}{\mathrm{a}}

\tan \left(\frac{\mathrm{P}}{2}\right) \tan \left(\frac{\mathrm{Q}}{2}\right)=\frac{\mathrm{c}}{\mathrm{a}}

\frac{\tan \left(\frac{\mathrm{P}}{2}\right)+\tan \left(\frac{\mathrm{Q}}{2}\right)}{1-\tan \left(\frac{\mathrm{P}}{2}\right) \tan \left(\frac{\mathrm{Q}}{2}\right)}=\tan \left(\frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}\right)=1

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

\mathrm{b}=\mathrm{a}

-\mathrm{c}

\mathrm{c}=\mathrm{a}+\mathrm{b}

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