To determine the values of \( \beta \) for which angle \( A \) of triangle \( ABC \) is obtuse, we will use the cosine rule and the properties of triangles. Let's follow the steps systematically.
### Step 1: Identify the coordinates of the vertices
The vertices of the triangle are given as:
- \( A(0, \beta) \)
- \( B(-2, 0) \)
- \( C(1, 1) \)
### Step 2: Use the cosine rule
The cosine rule states that for any triangle with sides \( a \), \( b \), and \( c \) opposite to angles \( A \), \( B \), and \( C \) respectively:
\[
\cos A = \frac{b^2 + c^2 - a^2}{2bc}
\]
For angle \( A \) to be obtuse, we need \( \cos A < 0 \).
### Step 3: Calculate the lengths of the sides
We will calculate the lengths of the sides \( AB \), \( AC \), and \( BC \) using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
#### Length \( BC \):
\[
BC = \sqrt{(1 - (-2))^2 + (1 - 0)^2} = \sqrt{(1 + 2)^2 + (1 - 0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}
\]
Thus, \( BC^2 = 10 \).
#### Length \( AB \):
\[
AB = \sqrt{(0 - (-2))^2 + (\beta - 0)^2} = \sqrt{(0 + 2)^2 + \beta^2} = \sqrt{4 + \beta^2}
\]
Thus, \( AB^2 = 4 + \beta^2 \).
#### Length \( AC \):
\[
AC = \sqrt{(1 - 0)^2 + (1 - \beta)^2} = \sqrt{1^2 + (1 - \beta)^2} = \sqrt{1 + (1 - 2\beta + \beta^2)} = \sqrt{2 - 2\beta + \beta^2}
\]
Thus, \( AC^2 = 2 - 2\beta + \beta^2 \).
### Step 4: Apply the cosine rule condition
We need to find when:
\[
\cos A < 0 \implies AB^2 + AC^2 < BC^2
\]
Substituting the values we found:
\[
(4 + \beta^2) + (2 - 2\beta + \beta^2) < 10
\]
This simplifies to:
\[
6 + 2\beta^2 - 2\beta < 10
\]
\[
2\beta^2 - 2\beta - 4 < 0
\]
Dividing through by 2:
\[
\beta^2 - \beta - 2 < 0
\]
### Step 5: Factor the quadratic
Factoring the quadratic:
\[
(\beta - 2)(\beta + 1) < 0
\]
### Step 6: Determine the intervals
To find the intervals where the product is negative, we can test the intervals determined by the roots \( \beta = -1 \) and \( \beta = 2 \):
- For \( \beta < -1 \): both factors are negative, product is positive.
- For \( -1 < \beta < 2 \): one factor is negative and the other is positive, product is negative.
- For \( \beta > 2 \): both factors are positive, product is positive.
Thus, the solution to the inequality is:
\[
-1 < \beta < 2
\]
### Step 7: Exclude collinear points
We need to check if \( \beta = \frac{2}{3} \) makes the points collinear:
Using the determinant method for collinearity:
\[
\begin{vmatrix}
0 & \beta & 1 \\
-2 & 0 & 1 \\
1 & 1 & 1
\end{vmatrix} = 0
\]
Calculating this determinant gives:
\[
0(\text{det}) + \beta(-2 - 1) + 1(0 + 2) = -3\beta + 2 = 0 \implies \beta = \frac{2}{3}
\]
Since \( \beta = \frac{2}{3} \) makes the points collinear, we exclude it from the interval.
### Final Interval
Thus, the values of \( \beta \) for which angle \( A \) is obtuse are:
\[
\beta \in (-1, \frac{2}{3}) \cup (\frac{2}{3}, 2)
\]
