In triangle ABC with usual notation,frac{cos | vedclass

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

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$\frac{1}{8 \sqrt{3}}$ sq. units.

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(D) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these,we get $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$,which implies $\cot A = \cot B = \cot C$.Since $A, B, C$ are angles of a triangle,$A = B = C = 60^\circ$,so the triangle is equilateral.Given $a = \frac{1}{\sqrt{6}}$,the area of an equilateral triangle is $\frac{\sqrt{3}}{4} a^2$.Area $= \frac{\sqrt{3}}{4} \left( \frac{1}{\sqrt{6}} ight)^2 = \frac{\sqrt{3}}{4} 	imes \frac{1}{6} = \frac{\sqrt{3}}{24} = \frac{1}{8 \sqrt{3}}$ sq. units.

In $\triangle ABC$ with usual notation,$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=\frac{1}{\sqrt{6}}$,then the area of the triangle is

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