Find the interior angle of the triangle whose vertics are A(4, 3), B(-2, 2) and C(2, -8)

Answer

To find the interior angles of the triangle with vertices A(4, 3), B(-2, 2), and C(2, -8), we will follow these steps:

### Step 1: Calculate the slopes of the sides of the triangle

1. **Calculate the slope of side AB:**
- The formula for the slope (m) between two points (x1, y1) and (x2, y2) is:
\[ m = \frac{y2 - y1}{x2 - x1} \]
- For points A(4, 3) and B(-2, 2):
\[ m_{AB} = \frac{2 - 3}{-2 - 4} = \frac{-1}{-6} = \frac{1}{6} \]

2. **Calculate the slope of side BC:**
- For points B(-2, 2) and C(2, -8):
\[ m_{BC} = \frac{-8 - 2}{2 - (-2)} = \frac{-10}{4} = -\frac{5}{2} \]

3. **Calculate the slope of side CA:**
- For points C(2, -8) and A(4, 3):
\[ m_{CA} = \frac{3 - (-8)}{4 - 2} = \frac{11}{2} \]

### Step 2: Calculate the angles using the slopes

To find the angles of the triangle, we will use the formula for the tangent of the angle between two lines:
\[ \tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2} \]

1. **Calculate angle at A (∠CAB):**
- Using slopes \(m_{AB}\) and \(m_{CA}\):
\[ \tan \theta_A = \frac{\left|\frac{1}{6} - \frac{11}{2}\right|}{1 + \left(\frac{1}{6} \cdot \frac{11}{2}\right)} \]
- Simplifying:
\[ = \frac{\left|\frac{1}{6} - \frac{33}{6}\right|}{1 + \frac{11}{12}} = \frac{\left|-\frac{32}{6}\right|}{\frac{23}{12}} = \frac{\frac{32}{6}}{\frac{23}{12}} = \frac{32 \cdot 12}{6 \cdot 23} = \frac{64}{23} \]
- Therefore, angle A:
\[ \theta_A = \tan^{-1}\left(\frac{64}{23}\right) \]

2. **Calculate angle at B (∠ABC):**
- Using slopes \(m_{AB}\) and \(m_{BC}\):
\[ \tan \theta_B = \frac{\left|\frac{1}{6} - \left(-\frac{5}{2}\right)\right|}{1 + \left(\frac{1}{6} \cdot -\frac{5}{2}\right)} \]
- Simplifying:
\[ = \frac{\left|\frac{1}{6} + \frac{15}{6}\right|}{1 - \frac{5}{12}} = \frac{\left|\frac{16}{6}\right|}{\frac{7}{12}} = \frac{\frac{16}{6}}{\frac{7}{12}} = \frac{16 \cdot 12}{6 \cdot 7} = \frac{32}{7} \]
- Therefore, angle B:
\[ \theta_B = \tan^{-1}\left(\frac{32}{7}\right) \]

3. **Calculate angle at C (∠BCA):**
- Using slopes \(m_{BC}\) and \(m_{CA}\):
\[ \tan \theta_C = \frac{\left|-\frac{5}{2} - \frac{11}{2}\right|}{1 + \left(-\frac{5}{2} \cdot \frac{11}{2}\right)} \]
- Simplifying:
\[ = \frac{\left|-\frac{16}{2}\right|}{1 - \frac{55}{4}} = \frac{8}{\frac{-51}{4}} = \frac{32}{-51} \]
- Therefore, angle C:
\[ \theta_C = \tan^{-1}\left(\frac{32}{51}\right) \]

### Final Angles
- The angles of the triangle are:
- ∠A = \(\tan^{-1}\left(\frac{64}{23}\right)\)
- ∠B = \(\tan^{-1}\left(\frac{32}{7}\right)\)
- ∠C = \(\tan^{-1}\left(\frac{32}{51}\right)\)