Find the measure of the angle A of the `DeltaABC` with vertices `A(2,3)`, `B(-2,2)` and `C(0,2)`

Answer

To find the measure of angle A in triangle ABC with vertices A(2,3), B(-2,2), and C(0,2), we will follow these steps:

### Step 1: Find the slopes of lines AB and AC
The slope of a line through two points (x1, y1) and (x2, y2) is given by the formula:
\[ m = \frac{y2 - y1}{x2 - x1} \]

**For line AB:**
- Points A(2,3) and B(-2,2)
\[ m_{AB} = \frac{2 - 3}{-2 - 2} = \frac{-1}{-4} = \frac{1}{4} \]

**For line AC:**
- Points A(2,3) and C(0,2)
\[ m_{AC} = \frac{2 - 3}{0 - 2} = \frac{-1}{-2} = \frac{1}{2} \]

### Step 2: Use the formula for the angle between two lines
The formula for the angle θ between two lines with slopes m1 and m2 is:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Here, \( m_1 = \frac{1}{4} \) and \( m_2 = \frac{1}{2} \).

### Step 3: Substitute the slopes into the formula
\[ \tan(A) = \left| \frac{\frac{1}{4} - \frac{1}{2}}{1 + \frac{1}{4} \cdot \frac{1}{2}} \right| \]

Calculating the numerator:
\[ \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = \frac{-1}{4} \]

Calculating the denominator:
\[ 1 + \frac{1}{4} \cdot \frac{1}{2} = 1 + \frac{1}{8} = \frac{8}{8} + \frac{1}{8} = \frac{9}{8} \]

### Step 4: Combine the results
\[ \tan(A) = \left| \frac{-\frac{1}{4}}{\frac{9}{8}} \right| = \left| -\frac{1}{4} \cdot \frac{8}{9} \right| = \frac{2}{9} \]

### Step 5: Find angle A
To find angle A, we take the arctangent:
\[ A = \tan^{-1}\left(\frac{2}{9}\right) \]

### Final Answer
The measure of angle A is:
\[ A = \tan^{-1}\left(\frac{2}{9}\right) \]

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