In a P Q R , R = 2 . If tan ( P 2 ) and tan ( Q 2 ) are the roots of a x 2 + b x + c = 0 , where a 0 , then

Question

In a $∆ P Q R,$ $∠ R = \frac{π}{2} .$ If $\tan (\frac{P}{2})$ and $\tan (\frac{Q}{2})$ are the roots of $a x^{2} + b x + c = 0,$ where $a \neq 0,$ then

Smart Achievers · Accepted Answer

(3)

Since, $\tan \frac{P}{2} + \tan \frac{Q}{2} = \frac{- b}{a}$

and $\tan \frac{P}{2} \tan \frac{Q}{2} = \frac{c}{a}$

Also, $\frac{P}{2} + \frac{Q}{2} + \frac{R}{2} = \frac{π}{2}$

$\Rightarrow \frac{P + Q}{2} = \frac{π}{4}$ $[∵ ∠ R = \frac{π}{2}]$

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

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

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

$\Rightarrow$ $c = a + b$

Solution

Q.
In a $∆ P Q R,$ $∠ R = \frac{π}{2} .$ If $\tan (\frac{P}{2})$ and $\tan (\frac{Q}{2})$ are the roots of $a x^{2} + b x + c = 0,$ where $a \neq 0,$ then

1 $b = a + c$

2 $b = c$

3 $c = a + b$

4 $a = b + c$

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Solution

Q. In a ∆PQR, ∠R=π2. If tan(P2) and tan(Q2) are the roots of ax2+bx+c=0, where a≠0, then

1 b=a+c

2 b=c

3 c=a+b

4 a=b+c