Question

In a triangle ABC, if $\frac{cosA}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=\frac{1}{\sqrt{6}}$,

then the area of the triangle in square units is equal to?

• A. $\frac{1}{24}$
• B. $\frac{1}{8}$
• C. $\frac{1}{8\sqrt{3}}$
• D. $\frac{1}{24\sqrt{3}}$

Answer 1

The correct option is C $\frac{1}{8\sqrt{3}}$

Given, In a triangle ABC,

$\frac{cosA}{a}=\frac{cosB}{b}=\frac{cosC}{c}$

$a=\frac{1}{\sqrt{6}}$

Formuls used-

$cosA=\frac{b^2+c^2-a^2}{2bc}$

$cosB=\frac{a^2+c^2-b^2}{2ac}$

$cosC=\frac{a^2+b^2-c^2}{2ab}$

$\text{Area of equilateral triangle with side 'a' is}$ $\frac{\sqrt{3}}{4}\times a^2$

Since,

$\frac{cosA}{a}=\frac{cosB}{b}=\frac{cosC}{c}$

$\Rightarrow \frac{b^2+c^2-a^2}{2abc}=\frac{c^2+a^2-b^2}{2abc}=\frac{a^2+b^2-c^2}{2abc}$

$\Rightarrow b^2+c^2-a^2=c^2+a^2-b^2=a^2+b^2-c^2$

$\Rightarrow a^2=b^2=c^2$

$\Rightarrow a=b=c$

Thus $\Delta$ ABC is an equilateral triangle

So, Area of $\Delta$ ABC = $\frac{\sqrt{3}}{4}\times a^2$

$=\frac{\sqrt{3}}{4}\times (\frac{1}{\sqrt{6}}{)}^2$

$=\frac{1}{8\times \sqrt{3}}$

Hence, option (C) is correct.