tan A, tan B , tan C are the roots of the cubic equation `x^(3) - 7x^(2) + 11x - 7 =0` then `A+B+C=`

Options

A. `(pi)/(2)`
B. `pi`
C. 0
D. none

Answer

To solve the problem, we need to find the sum of angles \( A + B + C \) given that \( \tan A, \tan B, \tan C \) are the roots of the cubic equation \( x^3 - 7x^2 + 11x - 7 = 0 \).

### Step-by-Step Solution:

1. **Identify the roots of the cubic equation**:
The roots of the equation \( x^3 - 7x^2 + 11x - 7 = 0 \) are given as \( \tan A, \tan B, \tan C \).

2. **Use Vieta's Formulas**:
According to Vieta's formulas, for a cubic equation \( x^3 + px^2 + qx + r = 0 \):
- The sum of the roots \( r_1 + r_2 + r_3 = -p \).
- The sum of the product of the roots taken two at a time \( r_1r_2 + r_2r_3 + r_3r_1 = q \).
- The product of the roots \( r_1r_2r_3 = -r \).

For our equation:
- Sum of roots: \( \tan A + \tan B + \tan C = 7 \) (since the coefficient of \( x^2 \) is -7).
- Product of roots: \( \tan A \tan B \tan C = 7 \) (since the constant term is -7).

3. **Relate the sum of tangents to angles**:
We know that for any triangle with angles \( A, B, C \):
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \]
This means:
\[ \tan A + \tan B + \tan C = 7 \]
and
\[ \tan A \tan B \tan C = 7 \]

4. **Use the identity for angles in a triangle**:
In a triangle, the sum of the angles is always \( 180^\circ \) or \( \pi \) radians:
\[ A + B + C = 180^\circ = \pi \text{ radians} \]

5. **Conclusion**:
Therefore, the sum of the angles \( A + B + C \) is:
\[ A + B + C = \pi \]

### Final Answer:
\[ A + B + C = \pi \]