In a triangle PQR - R -2 If tan P -2 and tan Q -2 are the roots of the equation -ax - 2 - bx - c -0 a -0 - then-A- a-b-cB- b-c-aC- b - cD- a-c-b

The correct option is A a+b=c ∠ P + ∠ Q = π/2 P/2 + Q/2 = π/4 tan ( P/2 + Q/2) = tan π/4 ⇒ 1 = tan P/2 + tan Q/2/1 tan P/2 tan Q/2 tan P/2 + tan Q/2 = b ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Graph of Trigonometric Ratios

In a triangle...

Question

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

a+b=c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

b+c=a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a+c=b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

b=c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
a+b=c

$∠ P + ∠ Q = \frac{π}{2}$

$\frac{P}{2} + \frac{Q}{2} = \frac{π}{4}$

$t a n (\frac{P}{2} + \frac{Q}{2}) = t a n \frac{π}{4}$

$\Rightarrow$ 1 = $\frac{t a n \frac{P}{2} + t a n \frac{Q}{2}}{1 - t a n \frac{P}{2} t a n \frac{Q}{2}}$

$t a n \frac{P}{2} + t a n \frac{Q}{2} = - \frac{b}{a}$

$t a n \frac{P}{2} t a n \frac{Q}{2} = \frac{c}{a}$

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

Suggest Corrections

Similar questions

Q. In a triangle

P Q R, ∠ R = \frac{π}{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

Q. In a triangle

P Q R, i f ∠ R = \frac{π}{2}

and

tan (\frac{P}{2})

tan (\frac{Q}{2})

are the roots of the equation

a x^{2} + b x + c = 0 (a \neq 0)

, then

Q. In a triangle

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, a \neq 0

, then

Q. In a

△ P Q R, ∠ R = \frac{π}{2} .

tan \frac{P}{2}

and

tan \frac{Q}{2}

are the roots of the equation

a x^{2} + b x + c = 0

then

Q. In a

△ P Q R

, if

∠ R = \frac{π}{2}

tan (\frac{P}{2})

and

tan (\frac{Q}{2})

are the roots of

a x^{2} + b x + c = 0, a \neq 0

then

Join BYJU'S Learning Program

In a triangle PQR, R = π2 If tan P2and tanQ2are the roots of the equation \( ax^2\) +bx+c=0 (a≠0), then:

The correct option is A a+b=c ∠P+∠Q=π2 P2+Q2=π4 tan(P2+Q2)=tanπ4 ⇒ 1 = tanP2+tanQ21−tanP2tanQ2 tanP2+tanQ2=−ba tanP2tanQ2=ca ⇒ 1 = −ba1−ca=−ba−c

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