In a triangle ABC a= 4 , b= 3 , `angle A =60 ^@` ,Then c is root of 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 for \( c \) 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:

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

### Step 1: Substitute the known values into the formula.
Given:
- \( a = 4 \)
- \( b = 3 \)
- \( \angle A = 60^\circ \)

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

Substituting these values into the Law of Cosines:

\[ c^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \cos 60^\circ \]

### Step 2: Calculate \( a^2 \) and \( b^2 \).
Calculating \( a^2 \) and \( b^2 \):

\[ c^2 = 16 + 9 - 2 \cdot 4 \cdot 3 \cdot \frac{1}{2} \]

### Step 3: Simplify the equation.
Now, simplify the expression:

\[ c^2 = 16 + 9 - 2 \cdot 4 \cdot 3 \cdot \frac{1}{2} \]
\[ = 16 + 9 - 12 \]
\[ = 13 \]

### Step 4: Rearranging the equation.
Now we can rewrite the equation:

\[ c^2 - 13 = 0 \]

### Step 5: Rearranging to standard form.
To express this in the form of a quadratic equation, we can rearrange it:

\[ c^2 - 0c - 13 = 0 \]

### Final Equation:
The final equation we derived is:

\[ c^2 - 0c - 13 = 0 \]

### Conclusion:
Thus, the equation whose roots will give us the value of \( c \) is:

\[ c^2 - 0c - 13 = 0 \]