The vertices of a triangle ABC are (0,0),(2,-1) and (9,2) respectively, then `cos B=`………..

Answer

To find \( \cos B \) for triangle ABC with vertices at \( A(0,0) \), \( B(2,-1) \), and \( C(9,2) \), 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 \( AB \), \( AC \), and \( BC \).

1. **Length of side \( AB \)**:
\[ AB = \sqrt{(2 - 0)^2 + (-1 - 0)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]

2. **Length of side \( AC \)**:
\[ AC = \sqrt{(9 - 0)^2 + (2 - 0)^2} = \sqrt{9^2 + 2^2} = \sqrt{81 + 4} = \sqrt{85} \]

3. **Length of side \( BC \)**:
\[ BC = \sqrt{(9 - 2)^2 + (2 - (-1))^2} = \sqrt{(7)^2 + (3)^2} = \sqrt{49 + 9} = \sqrt{58} \]

### Step 2: Use the cosine rule to find \( \cos B \)
The cosine rule states:
\[ \cos B = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} \]

Now substituting the lengths we calculated:

1. **Calculate \( AB^2 \)**:
\[ AB^2 = (\sqrt{5})^2 = 5 \]

2. **Calculate \( BC^2 \)**:
\[ BC^2 = (\sqrt{58})^2 = 58 \]

3. **Calculate \( AC^2 \)**:
\[ AC^2 = (\sqrt{85})^2 = 85 \]

Now substitute these values into the cosine rule formula:
\[ \cos B = \frac{5 + 58 - 85}{2 \cdot \sqrt{5} \cdot \sqrt{58}} \]
\[ = \frac{-22}{2 \cdot \sqrt{5} \cdot \sqrt{58}} \]
\[ = \frac{-22}{2 \sqrt{290}} = \frac{-11}{\sqrt{290}} \]

### Final Answer:
\[ \cos B = \frac{-11}{\sqrt{290}} \]

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