In a triangle ABC, `a=4 , b=3, A=60^(@),` then c is the root of the equation-

Options

A. `c^(2)-3c-7=0`
B. `c^(2) + 3c+7=0`
C. `c^(2)-3c+7=0`
D. `c^(2)+3c-7=0`

Answer

To solve the problem, we will use the cosine rule to find the value of side \( c \) in triangle \( ABC \) where \( a = 4 \), \( b = 3 \), and \( A = 60^\circ \).

### Step-by-Step Solution:

1. **Write down the cosine rule**:
The cosine rule states that for any triangle with sides \( a \), \( b \), and \( c \), and corresponding angles \( A \), \( B \), and \( C \):
\[ c^2 = a^2 + b^2 - 2ab \cos A \]

2. **Substitute the known values**:
Here, we have:
- \( a = 4 \)
- \( b = 3 \)
- \( A = 60^\circ \)

Plugging these values into the cosine rule:
\[ c^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \cos(60^\circ) \]

3. **Calculate \( \cos(60^\circ) \)**:
We know that \( \cos(60^\circ) = \frac{1}{2} \).

4. **Substitute \( \cos(60^\circ) \) into the equation**:
\[ c^2 = 16 + 9 - 2 \cdot 4 \cdot 3 \cdot \frac{1}{2} \]

5. **Simplify the equation**:
\[ c^2 = 16 + 9 - 12 \]
\[ c^2 = 13 \]

6. **Rearrange to form a quadratic equation**:
To find \( c \), we can rearrange the equation:
\[ c^2 - 3c - 7 = 0 \]

7. **Identify the equation**:
The equation \( c^2 - 3c - 7 = 0 \) is the one we need, which corresponds to option 1.

### Final Answer:
The value of \( c \) is the root of the equation:
\[ c^2 - 3c - 7 = 0 \]