In a triangle ABC with usual notations,if fra | vedclass

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(B) We know from the Sine Rule that $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$. Given that $\frac{\cos A}{a} = \frac{\cos B}{b} = \f

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$\frac{3 \sqrt{3}}{2}$ sq. units

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(B) We know from the Sine Rule that $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$. Given that $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$. Dividing the Sine Rule equation by the given equation,we get $	an A = 	an B = 	an C$. Since $A, B, C$ are angles of a triangle,this implies $A = B = C = 60^{\circ}$,so the triangle is equilateral. The area of an equilateral triangle with side $a$ is given by $	ext{Area} = \frac{\sqrt{3}}{4} a^2$. Given $a = \sqrt{6}$,the area is $\frac{\sqrt{3}}{4} (\sqrt{6})^2 = \frac{\sqrt{3}}{4} 	imes 6 = \frac{3 \sqrt{3}}{2}$ sq. units.

In a triangle $ABC$ with usual notations,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then the triangle is equilateral. If the side length is $a = \sqrt{6}$,find the area of the triangle.

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