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

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(B) Using the sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,which implies $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2

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an equilateral triangle

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(B) Using the sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,which implies $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Given the condition $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,we substitute the values of $a, b, c$: $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$. This simplifies to $\cot A = \cot B = \cot C$. Since $A, B, C$ are angles of a triangle,$\cot A = \cot B = \cot C$ implies $A = B = C$. Therefore,the triangle is an equilateral triangle.

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

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