In a triangle ABC, if `(cosA)/a=(cosB)/b=(cosC)/c` and the side `a =2`, then area of triangle is

Explanation

To solve the problem, we need to find the area of triangle ABC given that \(\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}\) and that side \(a = 2\).

### Step 1: Understand the given condition
We are given that:
\[ \frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c} \]
This implies that the ratios of the cosines of the angles to their respective opposite sides are equal.

### Step 2: Use the relationship between angles and sides
From the given condition, we can express this as:
\[ \frac{\cos A}{a} = k, \quad \frac{\cos B}{b} = k, \quad \frac{\cos C}{c} = k \]
where \(k\) is some constant. Rearranging gives us:
\[ \cos A = ka, \quad \cos B = kb, \quad \cos C = kc \]

### Step 3: Substitute the value of side \(a\)
Since we know \(a = 2\), we can write:
\[ \cos A = 2k \]

### Step 4: Apply the Law of Cosines
Using the Law of Cosines, we have:
\[ c^2 = a^2 + b^2 - 2ab \cos C \]
Substituting \(a = 2\) into the equation, we can express the other angles in terms of \(k\) and the sides.

### Step 5: Establish the equality of angles
Given that \(\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}\), we can derive that:
\[ \tan A = \tan B = \tan C \]
This implies that:
\[ A = B = C \]
Thus, triangle ABC is an equilateral triangle.

### Step 6: Calculate the area of the equilateral triangle
The area \(A\) of an equilateral triangle with side length \(a\) is given by:
\[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \]
Substituting \(a = 2\):
\[ \text{Area} = \frac{\sqrt{3}}{4} \times (2^2) = \frac{\sqrt{3}}{4} \times 4 = \sqrt{3} \]

### Final Answer
The area of triangle ABC is \(\sqrt{3}\).

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