In a triangle [PQR,;; R = 2.,,If,, ,( P 2 )] and [ ( Q 2 )] are the roots of [ax^2 + bx + …

Question

In a triangle $$PQR,\;\;\angle R = {\pi  \over 2}.\,\,If\,\,\tan \,\left( {{P \over 2}} \right)$$ and $$ \tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0,\,\,a \ne 0$$  then

Accepted Answer

$$\angle $$R = 90o $$ \therefore $$ $$\angle $$P + $$\angle $$Q = 90o

$$ \Rightarrow $$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$o

$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $a{x^2} + bx + c = 0$

$$ \therefore $$ $$\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) =  - {b \over a},\,\,$$

and $$\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}$$

$${{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}$$

$$ = \tan \left( {{P \over 2} + {Q \over 2}} \right) $$= tan 45o = 1

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

$$ \Rightarrow  - {b \over a} = {a \over a} - {c \over a}$$

$$ \Rightarrow  - b = a - c$$ or $c = a + b.$

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