In `Delta ABC,` if `(cos A)/(a)=(cos B)/(b)=(cos C)/(c ) and a=2`, then area of the triangle is

Options

A. 1 sq unit
B. 2 sq units
C. `(sqrt(3))/(2)` sq units
D. `sqrt(3)` sq units

Answer

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 \(a = 2\).

### Step-by-step Solution:

1. **Understanding the Given Condition**:
We start with the condition:
\[ \frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c} = k \quad \text{(for some constant } k\text{)} \]
This implies:
\[ \cos A = ka, \quad \cos B = kb, \quad \cos C = kc \]

2. **Using the Law of Sines**:
From the Law of Sines, we know:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Let’s denote this common ratio as \(R\):
\[ a = R \sin A, \quad b = R \sin B, \quad c = R \sin C \]

3. **Relating Cosine and Sine**:
Substituting \(a\) in the cosine equations:
\[ \cos A = kR \sin A, \quad \cos B = kR \sin B, \quad \cos C = kR \sin C \]

4. **Setting Up the Equations**:
Dividing the cosine equations by the sine equations gives:
\[ \frac{\cos A}{\sin A} = kR, \quad \frac{\cos B}{\sin B} = kR, \quad \frac{\cos C}{\sin C} = kR \]
This means:
\[ \cot A = \cot B = \cot C \]

5. **Conclusion on Angles**:
The only way for \(\cot A = \cot B = \cot C\) is if:
\[ A = B = C = 60^\circ \]
Thus, triangle ABC is equilateral.

6. **Finding the Area of the Triangle**:
The area \(A\) of an equilateral triangle with side length \(s\) is given by:
\[ \text{Area} = \frac{\sqrt{3}}{4} s^2 \]
Here, we know \(a = 2\) (which is the side length of the triangle):
\[ \text{Area} = \frac{\sqrt{3}}{4} \times 2^2 = \frac{\sqrt{3}}{4} \times 4 = \sqrt{3} \]

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

---