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In $\triangle \mathrm{ABC}$ with usual notation, $\frac{\cos \mathrm{A}}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}=\frac{\cos \mathrm{C}}{\mathrm{c}}$ and $\mathrm{a}=\frac{1}{\sqrt{6}}$, then the area of triangle is
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The correct answer is:
$\frac{1}{8 \sqrt{3}}$
If $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$, then the triangle is equilateral.
$$
\begin{aligned}
\therefore \quad \mathrm{A}(\triangle \mathrm{ABC}) & =\frac{\sqrt{3}}{4} \mathrm{a}^2 \\
& =\frac{\sqrt{3}}{4}\left(\frac{1}{\sqrt{6}}\right)^2 \\
& =\frac{\sqrt{3}}{24}=\frac{1}{8 \sqrt{3}} \text { sq. units }
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad \mathrm{A}(\triangle \mathrm{ABC}) & =\frac{\sqrt{3}}{4} \mathrm{a}^2 \\
& =\frac{\sqrt{3}}{4}\left(\frac{1}{\sqrt{6}}\right)^2 \\
& =\frac{\sqrt{3}}{24}=\frac{1}{8 \sqrt{3}} \text { sq. units }
\end{aligned}
$$
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