If magnitudes of vectors `vec A, vec B and vec C` are 5, 4 and 3 units respectively and `vec A = vec B + vec C`, then angle between `vec A and vec B` is

Options

A. `cos^(-1) ((3)/(5))`
B. `cos^(-1) ((4)/(5))`
C. `sin^(-1)((4)/(5))`
D. `tan^(-1)((3)/(5))`

Explanation

To solve the problem, we need to find the angle between vectors \(\vec{A}\) and \(\vec{B}\) given that \(\vec{A} = \vec{B} + \vec{C}\) and the magnitudes of the vectors are \(|\vec{A}| = 5\), \(|\vec{B}| = 4\), and \(|\vec{C}| = 3\).

### Step-by-step Solution:

1. **Write the equation for vector \(\vec{C}\)**:
Since \(\vec{A} = \vec{B} + \vec{C}\), we can rearrange this to find \(\vec{C}\):
\[ \vec{C} = \vec{A} - \vec{B} \]

2. **Use the dot product**:
Taking the dot product of both sides with themselves gives:
\[ \vec{C} \cdot \vec{C} = (\vec{A} - \vec{B}) \cdot (\vec{A} - \vec{B}) \]
This expands to:
\[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2 \vec{A} \cdot \vec{B} \]

3. **Substitute the magnitudes**:
Substitute the known magnitudes into the equation:
\[ 3^2 = 5^2 + 4^2 - 2 |\vec{A}| |\vec{B}| \cos \theta \]
This simplifies to:
\[ 9 = 25 + 16 - 2 \cdot 5 \cdot 4 \cos \theta \]

4. **Combine and simplify**:
Combine the terms on the right side:
\[ 9 = 41 - 40 \cos \theta \]
Rearranging gives:
\[ 40 \cos \theta = 41 - 9 \]
\[ 40 \cos \theta = 32 \]

5. **Solve for \(\cos \theta\)**:
Divide both sides by 40:
\[ \cos \theta = \frac{32}{40} = \frac{4}{5} \]

6. **Find the angle \(\theta\)**:
To find the angle \(\theta\), take the inverse cosine:
\[ \theta = \cos^{-1} \left(\frac{4}{5}\right) \]

### Conclusion:
The angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(\theta = \cos^{-1} \left(\frac{4}{5}\right)\).