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

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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None of these

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Solution

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

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