The vertices of the triangle ABC are A(1, 2), B (0, 0) and C (2, 3), then the greatest angle of the triangle is

Options

A. `75^(@)`
B. `105^(@)`
C. `120^(@)`
D. none of these

Answer

To find the greatest angle of triangle ABC with vertices A(1, 2), B(0, 0), and C(2, 3), 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 the sides \( a \), \( b \), and \( c \) of triangle ABC.

- Length of side \( a \) (opposite angle A):
\[ a = BC = \sqrt{(2 - 0)^2 + (3 - 0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

- Length of side \( b \) (opposite angle B):
\[ b = AC = \sqrt{(2 - 1)^2 + (3 - 2)^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

- Length of side \( c \) (opposite angle C):
\[ c = AB = \sqrt{(1 - 0)^2 + (2 - 0)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 2: Use the Cosine Rule to find the angles
We will use the cosine rule to find the angles \( A \), \( B \), and \( C \).

- For angle \( A \):
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Substituting the values:
\[ \cos A = \frac{(\sqrt{2})^2 + (\sqrt{5})^2 - (\sqrt{13})^2}{2 \cdot \sqrt{2} \cdot \sqrt{5}} = \frac{2 + 5 - 13}{2 \cdot \sqrt{10}} = \frac{-6}{2\sqrt{10}} = \frac{-3}{\sqrt{10}} \]

- For angle \( B \):
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Substituting the values:
\[ \cos B = \frac{(\sqrt{13})^2 + (\sqrt{5})^2 - (\sqrt{2})^2}{2 \cdot \sqrt{13} \cdot \sqrt{5}} = \frac{13 + 5 - 2}{2 \cdot \sqrt{65}} = \frac{16}{2\sqrt{65}} = \frac{8}{\sqrt{65}} \]

- For angle \( C \):
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Substituting the values:
\[ \cos C = \frac{(\sqrt{13})^2 + (\sqrt{2})^2 - (\sqrt{5})^2}{2 \cdot \sqrt{13} \cdot \sqrt{2}} = \frac{13 + 2 - 5}{2 \cdot \sqrt{26}} = \frac{10}{2\sqrt{26}} = \frac{5}{\sqrt{26}} \]

### Step 3: Determine the angles
Now we will find the angles using the inverse cosine function.

- Angle \( A \):
\[ A = \cos^{-1}\left(-\frac{3}{\sqrt{10}}\right) \approx 161.44^\circ \]

- Angle \( B \):
\[ B = \cos^{-1}\left(\frac{8}{\sqrt{65}}\right) \approx 7.16^\circ \]

- Angle \( C \):
\[ C = \cos^{-1}\left(\frac{5}{\sqrt{26}}\right) \approx 11.33^\circ \]

### Step 4: Identify the greatest angle
Comparing the angles:
- \( A \approx 161.44^\circ \)
- \( B \approx 7.16^\circ \)
- \( C \approx 11.33^\circ \)

The greatest angle is \( A \).

### Conclusion
The greatest angle of triangle ABC is approximately \( 161.44^\circ \).

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