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In a triangle $\mathrm{ABC}$ with usual notations, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$, then area of
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The correct answer is:
$\frac{3 \sqrt{3}}{2}$ sq. units
(C)
We know, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$...(1)
Given : $\frac{\cos \mathrm{A}}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}=\frac{\cos \mathrm{C}}{\mathrm{c}}$...(2)
Divide (1) by (2)
$\tan \mathrm{A}=\tan \mathrm{B}=\tan \mathrm{C} \Rightarrow \Delta \mathrm{ABC}$ is equilateral
Area $=\frac{\sqrt{3}}{4} \mathrm{a}^{2}=\frac{\sqrt{3}}{4}(\sqrt{6})^{2}=\frac{\sqrt{3}}{4} \times 6=\frac{3 \sqrt{3}}{2}$
We know, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$...(1)
Given : $\frac{\cos \mathrm{A}}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}=\frac{\cos \mathrm{C}}{\mathrm{c}}$...(2)
Divide (1) by (2)
$\tan \mathrm{A}=\tan \mathrm{B}=\tan \mathrm{C} \Rightarrow \Delta \mathrm{ABC}$ is equilateral
Area $=\frac{\sqrt{3}}{4} \mathrm{a}^{2}=\frac{\sqrt{3}}{4}(\sqrt{6})^{2}=\frac{\sqrt{3}}{4} \times 6=\frac{3 \sqrt{3}}{2}$
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