In a triangle $\mathrm{ABC}$ with usual notations, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$,…

In a triangle $\mathrm{ABC}$ with usual notations, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$, then area of
  1. $\frac{\sqrt{3}}{2}$ sq. units
  2. $\frac{3 \sqrt{3}}{2}$ sq. units
  3. $\frac{2}{\sqrt{3}}$ sq. units
  4. $\frac{5 \sqrt{3}}{2}$ sq. units

Solution

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 $=\frac{\sqrt{3}}{4} \mathrm{a}^{2}=\frac{\sqrt{3}}{4}(\sqrt{6})^{2}=\frac{\sqrt{3}}{4} \times 6=\frac{3 \sqrt{3}}{2}$

Asked in: MHT CET 2020 (16 Oct Shift 1)

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