Properties of Triangles 1 Question 9
9. In a $\triangle P Q R, \angle R=\frac{\pi}{2}$, if $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are the roots of the equation $a x^{2}+b x+c=0(a \neq 0)$, then
(a) $a+b=c$
(b) $b+c=a$
(c) $a+c=b$
(d) $b=c$
(1999, 2M)
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Answer:
Correct Answer: 9. (a)
Solution:
- It is given that, $\tan (P / 2)$ and $\tan (Q / 2)$ are the roots of the quadratic equation $a x^{2}+b x+c=0$
$$ \text { and } \quad \angle R=\pi / 2 $$
$\therefore \quad \tan (P / 2)+\tan (Q / 2)=-b / a$
and $\tan (P / 2) \tan (Q / 2)=c / a$
Since, $\quad P+Q+R=180^{\circ}$
$$ \Rightarrow \quad P+Q=90^{\circ} $$
$$ \begin{aligned} & \Rightarrow \quad \frac{P+Q}{2}=45^{\circ} \\ & \Rightarrow \quad \tan \frac{P+Q}{2}=\tan 45^{\circ} \end{aligned} $$
$$ \begin{array}{lc} \Rightarrow & \frac{\tan (P / 2)+\tan (Q / 2)}{1-\tan (P / 2) \tan (Q / 2)}=1 \\ \Rightarrow & \frac{-b / a}{1-c / a}=1 \\ \Rightarrow & \frac{-b / a}{\frac{a-c}{a}}=1 \\ \Rightarrow & \frac{-b}{a-c}=1 \\ \Rightarrow & -b=a-c \\ \Rightarrow & a+b=c \end{array} $$
