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

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(D) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substitutin

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$\sqrt{3}$

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(D) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these,we get $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$.This simplifies to $\cot A = \cot B = \cot C$.Since $A, B, C$ are angles of a triangle,$A = B = C = 60^\circ$.Thus,$\Delta ABC$ is an equilateral triangle.The area of an equilateral triangle is given by $	ext{Area} = \frac{\sqrt{3}}{4} a^2$.Given $a = 2$,$	ext{Area} = \frac{\sqrt{3}}{4} (2)^2 = \sqrt{3}$.

In a $\Delta ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$ and the side $a = 2$,then the area of the triangle is:

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