In a triangle ABC, angle A is greater than angle B. If the measures of angles A and B satisfy the equation ` 2tanx -k(1+tan^2x)=0`, where k E (0, 1), then the measure of the angle Cis

Options

A. `pi/3`
B. `(2pi)/(3)`
C. `(pi)/(2)`
D. `(3pi)/(4)`

Answer

To solve the problem, we need to find the measure of angle C in triangle ABC given that angle A is greater than angle B and the angles satisfy the equation \( 2\tan x - k(1 + \tan^2 x) = 0 \), where \( k \in (0, 1) \).

### Step-by-step Solution:

1. **Rewrite the Given Equation**:
The equation given is:
\[ 2\tan x - k(1 + \tan^2 x) = 0 \]
Rearranging this gives:
\[ 2\tan x = k(1 + \tan^2 x) \]

2. **Express \( k \)**:
From the rearranged equation, we can express \( k \):
\[ k = \frac{2\tan x}{1 + \tan^2 x} \]
We recognize that the right-hand side can be rewritten using the double angle identity for sine:
\[ k = \sin(2x) \]

3. **Set Up Relationships**:
Since \( k = \sin(2A) \) and \( k = \sin(2B) \) (because \( x \) represents both angles A and B), we have:
\[ \sin(2A) = \sin(2B) \]

4. **Analyze the Sine Function**:
The equation \( \sin(2A) = \sin(2B) \) can yield two cases:
- Case 1: \( 2A = 2B \) which simplifies to \( A = B \)
- Case 2: \( 2A = \pi - 2B \)

Since we know from the problem statement that \( A > B \), the first case \( A = B \) is not possible. Therefore, we consider the second case:
\[ 2A + 2B = \pi \]

5. **Find the Sum of Angles**:
From the equation \( 2A + 2B = \pi \), we can divide by 2:
\[ A + B = \frac{\pi}{2} \]

6. **Calculate Angle C**:
In a triangle, the sum of the angles is \( \pi \):
\[ A + B + C = \pi \]
Substituting \( A + B = \frac{\pi}{2} \) into this equation gives:
\[ \frac{\pi}{2} + C = \pi \]
Therefore, solving for \( C \):
\[ C = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

### Final Answer:
The measure of angle C is:
\[ C = \frac{\pi}{2} \text{ or } 90^\circ \]