(C) Using the Law of Cosines in $\Delta ABC$:$\cos B = \frac{c^2 + a^2 - b^2}{2ac}$Given that $c^2 + a^2 - b^2 = ac$,we substitute this into the formu

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(C) Using the Law of Cosines in $\Delta ABC$:$\cos B = \frac{c^2 + a^2 - b^2}{2ac}$Given that $c^2 + a^2 - b^2 = ac$,we substitute this into the formula:$\cos B = \frac{ac}{2ac} = \frac{1}{2}$Since $\cos B = \frac{1}{2}$,we have $B = \frac{\pi}{3}$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

Medium

In a $\Delta ABC$,if ${c^2} + {a^2} - {b^2} = ac$,then $\angle B = $

A
$\frac{\pi}{6}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{3}$
D
None of these

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

If in a $\triangle ABC$,$\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$,then $\angle C$ is equal to (in $^{\circ}$)

DifficultAP EAMCET 2013

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The number of triangles $ABC$ that can be formed with $a = 3, b = 8$ and $\sin A = \frac{5}{13}$ is

Medium

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In $\triangle ABC$,if $A, B, C$ are in arithmetic progression,then $\sqrt{a^2-ac+c^2} \cdot \cos \left(\frac{A-C}{2}\right) =$

MediumAP EAMCET 2025

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In a $\triangle ABC$,if $3a = b + c$,then $\cot \frac{B}{2} \cot \frac{C}{2}$ is equal to :

DifficultAP EAMCET 2003

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If the altitudes of a triangle are in arithmetic progression,then the sides of the triangle are in

DifficultAP EAMCET 2002

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In a $\Delta ABC$,if ${c^2} + {a^2} - {b^2} = ac$,then $\angle B = $

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