In triangle 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 cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,so $\frac{\cos A}{a} = \frac{b

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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 cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,so $\frac{\cos A}{a} = \frac{b^2+c^2-a^2}{2abc}$.Thus,$\frac{b^2+c^2-a^2}{2abc} = \frac{a^2+c^2-b^2}{2abc} = \frac{a^2+b^2-c^2}{2abc}$.This implies $b^2+c^2-a^2 = a^2+c^2-b^2 = a^2+b^2-c^2$.From $b^2+c^2-a^2 = a^2+c^2-b^2$,we get $2b^2 = 2a^2$,so $a=b$.From $a^2+c^2-b^2 = a^2+b^2-c^2$,we get $2c^2 = 2b^2$,so $b=c$.Therefore,$a=b=c$,which means $	riangle ABC$ is an equilateral triangle.Given $a=2$,the area is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (2)^2 = \sqrt{3}$ sq. units.

In $\triangle ABC$,if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and side $a=2$,then the area of the $\triangle ABC$ (in sq. units) is

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