In a triangle PQR, R =π4 and tan (P/3) and tan (Q/3) are the roots of the equation ax2+bx+c=0, then

In Δle PQRP+Q+R=1800⇒ P+Q+450=1800⇒ P+Q=1350⇒ P3+Q3=450−−(1)and Tan(P3)+Tan(Q3)=−baTan(P3).Tan(Q3)=cafrom (1) Tan (P3+Q3)=Tan450⇒ Tan(P3)+Tan(Q3)1− Tan(P3).Tan(Q3)=1⇒ −ba1−ca=1⇒ −ba=1−ca⇒ −b=a−c ⇒ a+b=c

In a triangle PQR, R =π4 and tan (P/3) and tan (Q/3) are the roots of the equation ax2+bx+c=0, then

In Δle PQRP+Q+R=1800⇒ P+Q+450=1800⇒ P+Q=1350⇒ P3+Q3=450−−(1)and Tan(P3)+Tan(Q3)=−baTan(P3).Tan(Q3)=cafrom (1) Tan (P3+Q3)=Tan450⇒ Tan(P3)+Tan(Q3)1− Tan(P3).Tan(Q3)=1⇒ −ba1−ca=1⇒ −ba=1−ca⇒ −b=a−c ⇒ a+b=c

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$I n a t r i a n g l e P Q R, R = \frac{π}{4} a n d \tan (P / 3) a n d \tan (Q / 3) a r e t h e r o o t s o f t h e e q u a t i o n a x^{2} + b x + c = 0, t h e n$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$a + b = c$

$b + c = a$

$a + c = b$

$b = c$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} I n Δ^{l e} P Q R \\ P + Q + R = 180^{0} \\ \Rightarrow P + Q + 45^{0} = 180^{0} \\ \Rightarrow P + Q = 135^{0} \\ \Rightarrow \frac{P}{3} + \frac{Q}{3} = 45^{0} - - (1) \\ a n d T a n (\frac{P}{3}) + T a n (\frac{Q}{3}) = \frac{- b}{a} \\ T a n (\frac{P}{3}) . T a n (\frac{Q}{3}) = \frac{c}{a} \\ f r o m (1) T a n (\frac{P}{3} + \frac{Q}{3}) = T a n 45^{0} \\ \Rightarrow \frac{T a n (\frac{P}{3}) + T a n (\frac{Q}{3})}{1 - T a n (\frac{P}{3}) . T a n (\frac{Q}{3})} = 1 \\ \Rightarrow \frac{\frac{- b}{a}}{1 - \frac{c}{a}} = 1 \\ \Rightarrow \frac{- b}{a} = 1 - \frac{c}{a} \\ \Rightarrow - b = a - c \\ \Rightarrow a + b = c \end{array}$