The magnitude to two vectors are `sqrt(61) and sqrt(78)` .If their scalar prodcut is ` - 59` , what is the angle between the two vector ?

Explanation

To find the angle between two vectors given their magnitudes and scalar product, we can use the formula for the scalar (dot) product of two vectors. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Identify the given information:**
- Magnitude of vector A, \( |A| = \sqrt{61} \)
- Magnitude of vector B, \( |B| = \sqrt{78} \)
- Scalar product (dot product), \( A \cdot B = -59 \)

2. **Use the formula for the scalar product:**
The scalar product of two vectors can be expressed as:
\[ A \cdot B = |A| |B| \cos(\theta) \]
where \( \theta \) is the angle between the two vectors.

3. **Substitute the known values into the formula:**
\[ -59 = (\sqrt{61})(\sqrt{78}) \cos(\theta) \]

4. **Calculate the product of the magnitudes:**
First, calculate \( |A| |B| \):
\[ |A| |B| = \sqrt{61} \times \sqrt{78} = \sqrt{61 \times 78} \]
Now calculate \( 61 \times 78 \):
\[ 61 \times 78 = 4758 \]
Thus,
\[ |A| |B| = \sqrt{4758} \]

5. **Calculate \( \cos(\theta) \):**
Rearranging the equation gives:
\[ \cos(\theta) = \frac{-59}{\sqrt{4758}} \]
Now calculate \( \sqrt{4758} \):
\[ \sqrt{4758} \approx 68.97 \]
Therefore,
\[ \cos(\theta) = \frac{-59}{68.97} \approx -0.855 \]

6. **Find the angle \( \theta \):**
To find \( \theta \), take the inverse cosine:
\[ \theta = \cos^{-1}(-0.855) \]
Using a calculator,
\[ \theta \approx 148^\circ \]

### Final Answer:
The angle between the two vectors is approximately \( 148^\circ \).