The vertices of triangle ABC are A (4, 4), B (6, 3), C (2, -1), then angle ABC is equal to

Answer

To find the angle \( \angle ABC \) for triangle ABC with vertices A(4, 4), B(6, 3), and C(2, -1), we will use the slope formula and the tangent of the angle between two lines.

### Step-by-Step Solution:

1. **Identify the Points**:
- A = (4, 4)
- B = (6, 3)
- C = (2, -1)

2. **Calculate the Slopes**:
- The slope \( m_1 \) of line AB:
\[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 4}{6 - 4} = \frac{-1}{2} \]

- The slope \( m_2 \) of line BC:
\[ m_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{2 - 6} = \frac{-4}{-4} = 1 \]

3. **Use the Tangent Formula**:
- The formula for the tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
- Substitute \( m_1 \) and \( m_2 \):
\[ \tan \theta = \left| \frac{-\frac{1}{2} - 1}{1 + (-\frac{1}{2}) \cdot 1} \right| \]

4. **Simplify the Expression**:
- Calculate the numerator:
\[ -\frac{1}{2} - 1 = -\frac{1}{2} - \frac{2}{2} = -\frac{3}{2} \]
- Calculate the denominator:
\[ 1 + (-\frac{1}{2}) \cdot 1 = 1 - \frac{1}{2} = \frac{1}{2} \]
- Now substitute back:
\[ \tan \theta = \left| \frac{-\frac{3}{2}}{\frac{1}{2}} \right| = \left| -3 \right| = 3 \]

5. **Find the Angle**:
- To find \( \theta \), take the arctangent:
\[ \theta = \tan^{-1}(3) \]

### Final Result:
The angle \( \angle ABC \) is \( \tan^{-1}(3) \).