CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In a triangle $$PQR,\ if\ \angle R=\dfrac{\pi}{2}$$ and  $$\displaystyle \tan(\frac{P}{2})$$ , $$\displaystyle \tan(\frac{Q}{2})$$ are the roots of the equation $$ax^{2}+bx+c=0(a\neq 0)$$ , then


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

Solution

The correct option is A $$a+b=c$$
Given In $$\triangle PQR\ \angle R=90^\circ,\tan\dfrac{ P}{2}  , \tan \dfrac{Q}{2}  $$ are roots of the equation $$ax^2+bx+c=0(a\not =0)$$

$$P+Q+R=\pi\Rightarrow P+Q+\dfrac{\pi}{2}=\pi$$

$$P+Q = \dfrac {\pi}{2} $$ ,  sum of roots$$=  \tan\dfrac{ P}{2}  +  \tan \dfrac{Q}{2} =  \dfrac{-b}{a}  $$

$$ \dfrac{P+Q}{2}  = \cfrac{\pi}{4} $$,   Products of roots$$= \tan \dfrac{P}{2} . \tan\dfrac{ Q}{2} = \dfrac{c}{a}$$ 

$$\tan  \big(\dfrac{P+Q}{2}\big)  = 1$$

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

$$\Rightarrow  \cfrac{\dfrac{-b}{a}}{1-\dfrac{c}{a}}  = 1$$

$$\Rightarrow  \cfrac{b}{c-a} = 1$$

$$\Rightarrow  a+b = c$$

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image