Statement-1: If `triangleABC, 3bc=(a-b+c)(a+b-c)` then `A=120^(@)`.
and Statement-2: `cos 120^(@)=-1/2`

Options

A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement-1
B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct expanation for statement-1
C. Statement-1 is True, Statement-2 is false
D. Statement-1 is False, Statement-2 is true

Answer

To solve the problem, we need to analyze the given statements and derive the necessary conclusions step by step.

### Step 1: Understand the Given Condition
We are given the condition for triangle ABC:
\[ 3BC = (A - B + C)(A + B - C) \]

### Step 2: Rewrite the Equation
We can rewrite the right-hand side:
\[ (A - B + C)(A + B - C) = A^2 - (B - C)^2 \]
This is based on the identity \( (x - y)(x + y) = x^2 - y^2 \).

### Step 3: Substitute into the Condition
Substituting this back into our equation gives:
\[ 3BC = A^2 - (B - C)^2 \]

### Step 4: Expand and Rearrange
Now, we can expand \( (B - C)^2 \):
\[ (B - C)^2 = B^2 - 2BC + C^2 \]
Thus, we have:
\[ 3BC = A^2 - (B^2 - 2BC + C^2) \]
This simplifies to:
\[ 3BC = A^2 - B^2 + 2BC - C^2 \]

### Step 5: Rearranging Further
Rearranging gives:
\[ A^2 - B^2 - C^2 = BC \]

### Step 6: Use the Cosine Rule
Using the cosine rule:
\[ A^2 = B^2 + C^2 - 2BC \cos A \]
Substituting this into our equation:
\[ B^2 + C^2 - 2BC \cos A - B^2 - C^2 = BC \]
This simplifies to:
\[ -2BC \cos A = BC \]

### Step 7: Divide by BC
Assuming \( BC \neq 0 \), we can divide both sides by \( BC \):
\[ -2 \cos A = 1 \]
Thus:
\[ \cos A = -\frac{1}{2} \]

### Step 8: Determine the Angle A
The angle \( A \) for which \( \cos A = -\frac{1}{2} \) is:
\[ A = 120^\circ \]

### Conclusion
Thus, we have shown that if \( 3BC = (A - B + C)(A + B - C) \), then \( A = 120^\circ \).

### Verification of Statements
- **Statement 1**: True, since we derived \( A = 120^\circ \).
- **Statement 2**: True, as \( \cos 120^\circ = -\frac{1}{2} \).

Both statements are correct.