If in a Delta ABC,frac{cos A}{a} = frac{cos B | vedclass

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(C) Using the sine rule,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.Substituting these into the given eq

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equilateral

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(C) Using the sine rule,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.Substituting these into the given equation: $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$.This simplifies to $\cot A = \cot B = \cot C$.Since $A, B, C$ are angles of a triangle,this implies $A = B = C$.Therefore,the triangle is equilateral.

If in a $\Delta ABC$,$\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then the triangle is

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