To solve the problem of finding the values of \( \beta \) for which angle \( A \) of triangle \( ABC \) is the largest, we will follow these steps:
### Step 1: Understand the Geometry
We have the vertices of the triangle as follows:
- \( A(0, \beta) \)
- \( B(-2, 0) \)
- \( C(1, 1) \)
We want to find the conditions under which angle \( A \) is the largest angle in triangle \( ABC \).
### Step 2: Use the Law of Sines
For angle \( A \) to be the largest angle, the side opposite to it, which is \( BC \), must be the longest side. Therefore, we need to establish the relationships between the lengths of the sides \( BC \), \( AC \), and \( AB \).
### Step 3: Calculate the Lengths of the Sides
Using the distance formula, we calculate the lengths of the sides:
1. **Length of \( BC \)**:
\[
BC = \sqrt{(1 - (-2))^2 + (1 - 0)^2} = \sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10}
\]
2. **Length of \( AC \)**:
\[
AC = \sqrt{(1 - 0)^2 + (1 - \beta)^2} = \sqrt{(1)^2 + (1 - \beta)^2} = \sqrt{1 + (1 - \beta)^2}
\]
3. **Length of \( AB \)**:
\[
AB = \sqrt{(-2 - 0)^2 + (0 - \beta)^2} = \sqrt{(-2)^2 + (-\beta)^2} = \sqrt{4 + \beta^2}
\]
### Step 4: Set Up the Inequalities
For angle \( A \) to be the largest, we need:
\[
BC > AC \quad \text{and} \quad BC > AB
\]
This gives us two inequalities:
1. \( \sqrt{10} > \sqrt{1 + (1 - \beta)^2} \)
2. \( \sqrt{10} > \sqrt{4 + \beta^2} \)
### Step 5: Solve the First Inequality
Squaring both sides of the first inequality:
\[
10 > 1 + (1 - \beta)^2
\]
\[
10 > 1 + (1 - 2\beta + \beta^2)
\]
\[
10 > 1 + 1 - 2\beta + \beta^2
\]
\[
10 > 2 - 2\beta + \beta^2
\]
\[
\beta^2 - 2\beta - 8 < 0
\]
Now, we can factor this quadratic inequality:
\[
(\beta - 4)(\beta + 2) < 0
\]
The critical points are \( \beta = -2 \) and \( \beta = 4 \). The solution to this inequality is:
\[
-2 < \beta < 4
\]
### Step 6: Solve the Second Inequality
Now, squaring both sides of the second inequality:
\[
10 > 4 + \beta^2
\]
\[
10 - 4 > \beta^2
\]
\[
6 > \beta^2
\]
This gives us:
\[
-\sqrt{6} < \beta < \sqrt{6}
\]
### Step 7: Combine the Results
Now we combine the intervals:
1. From the first inequality: \( -2 < \beta < 4 \)
2. From the second inequality: \( -\sqrt{6} < \beta < \sqrt{6} \)
The combined intervals are:
\[
\beta \in (-2, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, 4)
\]
### Final Result
Thus, the values of \( \beta \) for which angle \( A \) of triangle \( ABC \) is the largest lie in the interval:
\[
\beta \in (-2, \frac{2}{3}) \cup (\frac{2}{3}, \sqrt{6})
\]
