In a $△$ ABC, if $c^{2} + a^{2} - b^{2} = a c,$ then $∠ B =$
[MP PET 1983, 89, 90]

\frac{π}{6}

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{π}{3}

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

None of these

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

Open in App

Solution

The correct option is C $\frac{π}{3}$
cos B $= \frac{c^{2} + a^{2} - b^{2}}{2 a c} \Rightarrow c o s B = \frac{1}{2} i . e ., B = \frac{π}{3} .$

Suggest Corrections

Similar questions

△

c^{2} + a^{2} - b^{2} = a c,

∠ B =

△

c^{2} + a^{2} - b^{2} = a c,

∠ B =

△ A B C, \frac{s i n (A - B)}{s i n (A + B)} =

△ A B C,

2 △^{2} = \frac{a^{2} b^{2} c^{2}}{a^{2} + b^{2} + c^{2}}

In the adjoining figure, if $B C = a, A C = b, A B = c$ and $∠ C A B = 120^{\circ}$ , then the correct relation is

Join BYJU'S Learning Program