In a `Delta ABC`, `2 sinA cos B + 2 sin B cos C + 2 sin C cos A` = `sin A + sin B + sin C,` then `Delta ABC` is (a) Isoceles (b) Right Angled (c) Acute (d) None of these

Options

A. isosceles
B. right angled
C. acute angled
D. none of these

Explanation

To solve the problem, we start with the equation given in the triangle ABC:

\[ 2 \sin A \cos B + 2 \sin B \cos C + 2 \sin C \cos A = \sin A + \sin B + \sin C \]

We will manipulate this equation step by step.

### Step 1: Use the identity for sine and cosine
We know the identity:
\[ \sin A \cos B = \frac{1}{2} (\sin(A + B) + \sin(A - B)) \]
Using this identity, we can rewrite each term on the left side of the equation.

### Step 2: Rewrite the equation
Applying the identity, we have:
\[ \begin{align*} 2 \sin A \cos B & = \sin(A + B) + \sin(A - B) \\ 2 \sin B \cos C & = \sin(B + C) + \sin(B - C) \\ 2 \sin C \cos A & = \sin(C + A) + \sin(C - A) \end{align*} \]
Thus, the left side becomes:
\[ \sin(A + B) + \sin(A - B) + \sin(B + C) + \sin(B - C) + \sin(C + A) + \sin(C - A) \]

### Step 3: Substitute angles in terms of π
In triangle ABC, we know:
\[ A + B + C = \pi \]
Thus, we can express:
\[ \begin{align*} A + B & = \pi - C \\ B + C & = \pi - A \\ C + A & = \pi - B \end{align*} \]

### Step 4: Substitute back into the equation
Substituting these into our equation gives:
\[ \sin(\pi - C) + \sin(A - B) + \sin(\pi - A) + \sin(B - C) + \sin(\pi - B) + \sin(C - A) \]
Using the property \(\sin(\pi - x) = \sin x\), we simplify:
\[ \sin C + \sin(A - B) + \sin A + \sin(B - C) + \sin B + \sin(C - A) \]

### Step 5: Combine terms
Now we have:
\[ \sin A + \sin B + \sin C + \sin(A - B) + \sin(B - C) + \sin(C - A) = \sin A + \sin B + \sin C \]
This implies:
\[ \sin(A - B) + \sin(B - C) + \sin(C - A) = 0 \]

### Step 6: Analyze the sine terms
The equation \(\sin(A - B) + \sin(B - C) + \sin(C - A) = 0\) suggests that at least one of the sine terms must be zero. This leads us to conclude that:
1. \(A = B\)
2. \(B = C\)
3. \(C = A\)

### Conclusion
Since any two angles are equal, triangle ABC is isosceles.

Thus, the answer is:
**(a) Isosceles**