In a triangle ABC, tanA and tanB satisfy the inequality `sqrt(3)x^(2)-4x+sqrt(3) lt 0`, then which of the following must be correct ?
(where symbols used have usual meanings)

Options

A. `a^(2)+b^(2)-ab lt c^(2)`
B. `a^(2)+b^(2) gt c^(2)`
C. `a^(2)+b^(2)+ab gt c^(2)`
D. `a^(2)+b^(2) lt c^(2)`

Answer

To solve the problem, we need to analyze the inequality given for the triangle ABC, where \( \tan A \) and \( \tan B \) satisfy the inequality:

\[ \sqrt{3}x^2 - 4x + \sqrt{3} < 0 \]

### Step 1: Identify the quadratic equation
The inequality can be treated as a quadratic equation:

\[ \sqrt{3}x^2 - 4x + \sqrt{3} = 0 \]

### Step 2: Find the roots of the quadratic equation
We will use the quadratic formula to find the roots:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = \sqrt{3} \), \( b = -4 \), and \( c = \sqrt{3} \).

Calculating the discriminant:

\[ b^2 - 4ac = (-4)^2 - 4(\sqrt{3})(\sqrt{3}) = 16 - 12 = 4 \]

Now substituting into the quadratic formula:

\[ x = \frac{4 \pm \sqrt{4}}{2\sqrt{3}} = \frac{4 \pm 2}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} \text{ and } \frac{2}{2\sqrt{3}} \]

This simplifies to:

\[ x = \frac{3}{\sqrt{3}} = \sqrt{3} \quad \text{and} \quad x = \frac{1}{\sqrt{3}} \]

### Step 3: Analyze the inequality
The roots of the quadratic are \( x = \sqrt{3} \) and \( x = \frac{1}{\sqrt{3}} \). The quadratic opens upwards (since the coefficient of \( x^2 \) is positive). Therefore, the inequality \( \sqrt{3}x^2 - 4x + \sqrt{3} < 0 \) holds between the roots:

\[ \frac{1}{\sqrt{3}} < x < \sqrt{3} \]

### Step 4: Determine the range for \( \tan A \) and \( \tan B \)
Since \( \tan A \) and \( \tan B \) must satisfy this inequality, we have:

\[ \frac{1}{\sqrt{3}} < \tan A, \tan B < \sqrt{3} \]

### Step 5: Convert the tangent values to angle measures
Using the known values of tangent:

- \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)
- \( \tan 60^\circ = \sqrt{3} \)

Thus, the angles \( A \) and \( B \) must satisfy:

\[ 30^\circ < A, B < 60^\circ \]

### Step 6: Determine the range for angle \( C \)
Since the angles in a triangle sum to \( 180^\circ \):

\[ C = 180^\circ - (A + B) \]

The minimum value of \( A + B \) occurs when both are \( 30^\circ \):

\[ A + B > 60^\circ \implies C < 120^\circ \]

The maximum value of \( A + B \) occurs when both are \( 60^\circ \):

\[ A + B < 120^\circ \implies C > 60^\circ \]

Thus, we have:

\[ 60^\circ < C < 120^\circ \]

### Step 7: Final inequalities for \( C \)
From the above analysis, we conclude:

- \( C \) lies between \( 60^\circ \) and \( 120^\circ \).

### Conclusion
The correct options regarding the angles \( A \), \( B \), and \( C \) based on the given inequality are:

- \( A \) and \( B \) are between \( 30^\circ \) and \( 60^\circ \).
- \( C \) is between \( 60^\circ \) and \( 120^\circ \).