If `Delta ABC`, if `tan A + tan B + tan C = 3sqrt(3)`, then the triangle is :

Options

A. Isosceles
B. Right angled
C. Equilateral
D. None of these

Answer

To solve the problem, we need to determine the type of triangle given that \( \tan A + \tan B + \tan C = 3\sqrt{3} \).

### Step-by-Step Solution:

1. **Understanding the Angles of a Triangle**:
We know that in any triangle, the sum of the angles \( A + B + C = 180^\circ \) or in radians, \( A + B + C = \pi \).

2. **Using the Tangent Addition Formula**:
We can express \( \tan(A + B) \) using the formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Since \( C = 180^\circ - (A + B) \), we have:
\[ \tan C = \tan(\pi - (A + B)) = -\tan(A + B) \]
Therefore, we can write:
\[ \tan A + \tan B + \tan C = \tan A + \tan B - \tan(A + B) \]

3. **Substituting the Given Condition**:
We are given that:
\[ \tan A + \tan B + \tan C = 3\sqrt{3} \]
This implies:
\[ \tan A + \tan B - \tan(A + B) = 3\sqrt{3} \]

4. **Using the Identity for \( \tan(A + B) \)**:
From the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
We can express:
\[ \tan A + \tan B = \tan(A + B)(1 - \tan A \tan B) \]

5. **Assuming Equal Angles**:
Let's assume \( A = B = C = 60^\circ \). Then:
\[ \tan 60^\circ = \sqrt{3} \]
Therefore:
\[ \tan A + \tan B + \tan C = \sqrt{3} + \sqrt{3} + \sqrt{3} = 3\sqrt{3} \]

6. **Conclusion**:
Since \( \tan A + \tan B + \tan C = 3\sqrt{3} \) holds true when \( A = B = C = 60^\circ \), we conclude that the triangle is an **equilateral triangle**.

### Final Answer:
The triangle is an **equilateral triangle**.

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