Determine the angle B of the triangle with vertices `A(-2,1)`, `B(2,3)` and `C(-2,-4)`.

Answer

To determine the angle \( B \) of the triangle with vertices \( A(-2, 1) \), \( B(2, 3) \), and \( C(-2, -4) \), we will follow these steps:

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

We will use the distance formula to find the lengths of sides \( AB \), \( BC \), and \( AC \).

**Distance Formula:**
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

#### Calculate \( AB \):
- \( A(-2, 1) \) and \( B(2, 3) \)
\[ AB = \sqrt{(2 - (-2))^2 + (3 - 1)^2} = \sqrt{(2 + 2)^2 + (3 - 1)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]

#### Calculate \( BC \):
- \( B(2, 3) \) and \( C(-2, -4) \)
\[ BC = \sqrt{(-2 - 2)^2 + (-4 - 3)^2} = \sqrt{(-4)^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65} \]

#### Calculate \( AC \):
- \( A(-2, 1) \) and \( C(-2, -4) \)
\[ AC = \sqrt{(-2 - (-2))^2 + (-4 - 1)^2} = \sqrt{(0)^2 + (-5)^2} = \sqrt{25} = 5 \]

### Step 2: Use the Cosine Rule to find angle \( B \)

The Cosine Rule states:
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
For angle \( B \):
\[ AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(B) \]

Substituting the lengths we calculated:
- \( AB = 2\sqrt{5} \)
- \( BC = \sqrt{65} \)
- \( AC = 5 \)

\[ (2\sqrt{5})^2 = 5^2 + (\sqrt{65})^2 - 2 \cdot 5 \cdot \sqrt{65} \cdot \cos(B) \]
\[ 20 = 25 + 65 - 10\sqrt{65} \cos(B) \]
\[ 20 = 90 - 10\sqrt{65} \cos(B) \]
\[ 10\sqrt{65} \cos(B) = 90 - 20 \]
\[ 10\sqrt{65} \cos(B) = 70 \]
\[ \cos(B) = \frac{70}{10\sqrt{65}} = \frac{7}{\sqrt{65}} \]

### Step 3: Calculate angle \( B \)

To find \( B \):
\[ B = \cos^{-1}\left(\frac{7}{\sqrt{65}}\right) \]

Using a calculator to find the angle:
\[ B \approx 32.32^\circ \]

### Final Answer
The angle \( B \) of the triangle is approximately \( 32.32^\circ \).

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