In ABC- ab2-c2-cb2-a2-0 then B-

The correct option is D π3 answer is D ...

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

Vector Addition

In ABC, ab2...

Question

In $△ A B C, \frac{a}{b^{2} - c^{2}} + \frac{c}{b^{2} - a^{2}} = 0$ then $B$ =

\frac{π}{2}

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

\frac{π}{4}

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

\frac{2 π}{3}

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

\frac{π}{3}

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

Open in App

Solution

Suggest Corrections

Similar questions

△ A B C

∠ C = \frac{π}{2}

sin (A - B) = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}

△ A B C,

(\frac{a^{2} + b^{2}}{a^{2} - b^{2}}) sin (A - B) = 1

C

cos (A - B) = tan (\frac{C}{2} - \frac{π}{4})

{sin}^{- 1} (a - \frac{a^{2}}{3} + \frac{a^{3}}{9} + . . .) + {cos}^{- 1} (1 + b + b^{2} + . . .) = \frac{π}{2}

s i n^{- 1} (a - \frac{a^{2}}{3} + \frac{a^{3}}{9} - . . . . . . \infty) + c o s^{- 1} (1 + b + b^{2} + . . . . \infty) = \frac{π}{2}

s i n^{- 1} (a - \frac{a^{2}}{3} + \frac{a^{3}}{9} - . . . . . . \infty) + c o s^{- 1} (1 + b + b^{2} + . . . . \infty) = \frac{π}{2}

Join BYJU'S Learning Program