In Triangle ABC,
CosA /a = CosB / b = Cos C / c
and the side a = 2. find the area of the triangle.

Hi Anushka,
In any triangle, we have
a/SinA = b/SinB = c/SinC (this is the Sin rule applicable for any triangle).
And here CosA/a = CosB/b = CosC/c
So CosA/SinA = CosB/SinB = CosC/SinC
or CotA = CotB = CotC
In a triangle this is possible only when A = B = C = 60 degrees.
So the trianle is equilateral.
And hence a = b = c = 2
Or area = [root(3)/4] * (2^2) = root(3). { Area of an equilateral triangle is [root(3)/4]*(a^2) }

Hope that helps.
Best Regards,
Ashwin (IIT Madras).

Using th SIne law
Sin A /a = Sin B / b = Sin C / c  ------(1)
CosA /a = CosB / b = Cos C / c ------(2)
Solving Both the the Equations we get
tan A =tan B= tan C
This can Be possible only when  A = B = C =60
This means the triangle is equilateral.
thus the area is ((3)1/2 /4)(a)2 = ((3)1/2 /4)(2)2= (3)1/2

In ∆ABC , THE sides b and angle B are fixed and remaining elements undergo small changes . Show that A∆c + cos C ∆a = 0