In a `DeltaABC`, if a = 4, b = 3 and `angleA=60^(@)`, then c is a 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`

Explanation

To find the value of \( c \) in triangle \( ABC \) where \( a = 4 \), \( b = 3 \), and \( \angle A = 60^\circ \), we can use the Law of Cosines. The Law of Cosines states that:

\[ c^2 = a^2 + b^2 - 2ab \cos A \]

### Step-by-Step Solution:

1. **Identify the given values:**
- \( a = 4 \)
- \( b = 3 \)
- \( \angle A = 60^\circ \)

2. **Calculate \( \cos A \):**
- Since \( \angle A = 60^\circ \), we know that:
\[ \cos 60^\circ = \frac{1}{2} \]

3. **Substitute the values into the Law of Cosines formula:**
\[ c^2 = a^2 + b^2 - 2ab \cos A \]
Substituting the known values:
\[ c^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \frac{1}{2} \]

4. **Calculate each term:**
- \( 4^2 = 16 \)
- \( 3^2 = 9 \)
- \( 2 \cdot 4 \cdot 3 \cdot \frac{1}{2} = 12 \)

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

6. **Solve for \( c \):**
\[ c = \sqrt{13} \]

7. **Formulate the equation:**
- Rearranging gives us:
\[ c^2 - 13 = 0 \]
This is the equation where \( c \) is a root.

### Final Answer:
The equation is \( c^2 - 13 = 0 \).

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