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

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(B) Given,in $	riangle ABC$,$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c} \dots (i)$ From the sine rule,$\frac{\sin A}{a}=\frac{\sin B}{b}=\fra

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$\sqrt{3}$

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(B) Given,in $	riangle ABC$,$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c} \dots (i)$ From the sine rule,$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \dots (ii)$ Dividing Eq. $(ii)$ by Eq. $(i)$,we get $	an A = 	an B = 	an C$. Since $A, B, C$ are angles of a triangle,$A = B = C$. Thus,$	riangle ABC$ is an equilateral triangle. Since $A+B+C = 180^{\circ}$,we have $3A = 180^{\circ}$,so $A = B = C = 60^{\circ}$. The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} a^2$. Given $a = 2$,the area is $\frac{\sqrt{3}}{4} (2)^2 = \frac{\sqrt{3}}{4} 	imes 4 = \sqrt{3}$.

In a $\Delta ABC$,if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=2$,then its area is

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