With usual notations, in triangle ABC, if cos | vedclass

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(C) Using the Sine Rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, we have $\sin A = \frac{a}{2R}$ and $\sin B = \frac{b}{2R}$.Su

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$b = c$

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(C) Using the Sine Rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, we have $\sin A = \frac{a}{2R}$ and $\sin B = \frac{b}{2R}$.Substituting these into the given equation: $\cos C = \frac{a/2R}{2(b/2R)} = \frac{a}{2b}$.Using the Cosine Rule, $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Equating the two expressions for $\cos C$: $\frac{a^2 + b^2 - c^2}{2ab} = \frac{a}{2b}$.Multiplying both sides by $2ab$: $a^2 + b^2 - c^2 = a^2$.Simplifying gives $b^2 - c^2 = 0$, which implies $b^2 = c^2$, so $b = c$.

With usual notations, in $\triangle ABC$, if $\cos C = \frac{\sin A}{2 \sin B}$, then which of the following is true?

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