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$\triangle \mathrm{ABC}$, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and side $a=2$, then area of the $\triangle \mathrm{ABC}$ (in sq. units) is
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Verified Answer
The correct answer is:
$\sqrt{3}$
$\begin{aligned} & \frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}, a=2 \\ \Rightarrow & \frac{b^2+c^2-a^2}{2 a b c}=\frac{a^2+c^2-b^2}{2 a b c}=\frac{a^2+b^2-c^2}{2 a b c} \\ \therefore \quad & b^2+c^2-a^2=a^2+c^2-b^2=a^2+b^2-c^2\end{aligned}$
$$
\begin{array}{ll}
\because \quad b^2+c^2-a^2=a^2+c^2-b^2 \\
& b^2=a^2
\end{array}
$$
Similarly, $a^2+c^2-b^2=a^2+b^2-c^2$
$$
\begin{aligned}
& b^2=c^2 \\
& \therefore \quad a^2=b^2=c^2 \Rightarrow a=b=c \\
& \therefore \quad \operatorname{Ar}(\triangle A B C)=\frac{\sqrt{3}}{4}(a)^2=\sqrt{3} .
\end{aligned}
$$
$$
\begin{array}{ll}
\because \quad b^2+c^2-a^2=a^2+c^2-b^2 \\
& b^2=a^2
\end{array}
$$
Similarly, $a^2+c^2-b^2=a^2+b^2-c^2$
$$
\begin{aligned}
& b^2=c^2 \\
& \therefore \quad a^2=b^2=c^2 \Rightarrow a=b=c \\
& \therefore \quad \operatorname{Ar}(\triangle A B C)=\frac{\sqrt{3}}{4}(a)^2=\sqrt{3} .
\end{aligned}
$$
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