If tan A, tan B, tan C are the solutions of the equation `x^(3)-k^(2)x^(2)-px+2k+1=0`, then `Delta ABC` is

Options

A. an isosceles triangle
B. an equilateral triangle
C. a right angled triangle
D. none of these

Answer

To solve the problem, we need to analyze the given cubic equation and the relationships between the angles A, B, and C based on the roots of the equation.

### Step-by-Step Solution:

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

2. **Use Vieta's Formulas**:
According to Vieta's formulas for a cubic equation \(x^3 + ax^2 + bx + c = 0\):
- The sum of the roots (tan A + tan B + tan C) is equal to \(-\) (coefficient of \(x^2\)) which is \(k^2\).
- The product of the roots (tan A \(\cdot\) tan B \(\cdot\) tan C) is equal to \(-\) (constant term) which is \(- (2k + 1)\).

Therefore, we have:
\[ \tan A + \tan B + \tan C = k^2 \]
\[ \tan A \cdot \tan B \cdot \tan C = - (2k + 1) \]

3. **Relate the Angles**:
Since \(\tan A + \tan B + \tan C\) can be expressed in terms of the angles of a triangle, we can use the identity:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \cdot \tan(A + B + C) \]
Given that \(A + B + C = \pi\) (for a triangle), we can substitute:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \cdot 0 = 0 \]
However, this is not directly useful, so we will focus on the relationships we derived from Vieta's formulas.

4. **Set Up the Equation**:
From the first equation, we have:
\[ k^2 = \tan A + \tan B + \tan C \]
From the second equation:
\[ - (2k + 1) = \tan A \cdot \tan B \cdot \tan C \]

5. **Solve for k**:
To find the value of \(k\), we can equate:
\[ k^2 = - (2k + 1) \]
Rearranging gives:
\[ k^2 + 2k + 1 = 0 \]
This factors to:
\[ (k + 1)^2 = 0 \]
Thus, \(k = -1\).

6. **Determine the Nature of the Triangle**:
Now, substituting \(k = -1\) back into the equations:
\[ \tan A + \tan B + \tan C = 1 \]
\[ \tan A \cdot \tan B \cdot \tan C = -1 \]
We can analyze the angles:
- If \(A = B\) and \(C\) is different, it suggests an isosceles triangle.
- If \(A\) and \(B\) are both acute and \(C\) is obtuse, it is also possible.

7. **Conclusion**:
Given the relationships and the nature of the angles, we conclude that the triangle formed by angles A, B, and C is an isosceles triangle.

### Final Answer:
The area of triangle ABC, denoted as \(\Delta ABC\), can be concluded to be that of an isosceles triangle based on the derived relationships.