In $\triangle ABC$, if $\;$ $\dfrac{\cos A}{a} = \dfrac{\cos B}{b} = \dfrac{\cos C}{c}$ $\;$ and $\;$ $a = 2$, find the area of $\triangle ABC$.
By cosine rule,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
Given: $\;$ $\dfrac{\cos A}{a} = \dfrac{\cos B}{b} = \dfrac{\cos C}{c}$
i.e. $\;$ $\dfrac{b^2 + c^2 - a^2 / 2 b c}{a} = \dfrac{c^2 + a^2 - b^2 / 2 c a}{b} = \dfrac{a^2 + b^2 - c^2 / 2 a b}{c}$ $\;\;$ [by cosine rule]
i.e. $\;$ $\dfrac{b^2 + c^2 - a^2}{2 a b c} = \dfrac{c^2 + a^2 - b^2}{2 a b c} = \dfrac{a^2 + b^2 - c^2}{2 a b c}$
i.e. $\;$ $b^2 + c^2 - a^2 = c^2 + a^2 - b^2 = a^2 + b^2 - c^2$
Now, $\;$ $b^2 + c^2 - a^2 = c^2 + a^2 - b^2$ $\implies$ $2 a^2 = 2 b^2$ $\implies$ $a = b$
and $\;$ $b^2 + c^2 - a^2 = a^2 + b^2 - c^2$ $\implies$ $2 a^2 = 2 c^2$ $\implies$ $a = c$
$\therefore \;$ We have $\;$ $a = b = c$
i.e. $\;$ $\triangle ABC$ is an equilateral triangle.
$\therefore \;$ $A = B = C = \dfrac{\pi}{3}$
Area of $\triangle ABC$ $= \Delta = \dfrac{1}{2} a b \sin C$
Here, $\;$ $\Delta = \dfrac{1}{2} a^2 \sin \left(\dfrac{\pi}{3}\right)$
i.e. $\;$ $\Delta = \dfrac{1}{2} \times \left(2\right)^2 \times \dfrac{\sqrt{3}}{2}$ $\;\;$ [Given: $\; a = 2$]
i.e. $\;$ Area of $\triangle ABC$ $= \Delta = \sqrt{3}$