In a triangle ABC,if frac{cos A}{a} = frac{co | vedclass

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(C) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.By the sine rule,we know that $a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,wher

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equilateral

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(C) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.By the sine rule,we know that $a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,where $k$ is a constant.Substituting these values into the given equation:$\frac{\cos A}{k \sin A} = \frac{\cos B}{k \sin B} = \frac{\cos C}{k \sin C}$$\Rightarrow \cot A = \cot B = \cot C$Since $A, B, C$ are angles of a triangle,$A = B = C$.Since all angles are equal,the triangle is an equilateral triangle.

In a $\triangle ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then $\triangle ABC$ is

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