If vectors `vec(A),vec(B)` and `vec(C)` have magnitudes 8,15 and 17 units and `vec(A) +vec(B) = vec(C)`, find the angle between `vec(A)` and `vec(B)`.

Answer

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 as follows:

- Magnitude of \(\vec{A} = 8\) units
- Magnitude of \(\vec{B} = 15\) units
- Magnitude of \(\vec{C} = 17\) units

### Step-by-step Solution:

1. **Understand the relationship between the vectors**:
Since \(\vec{A} + \vec{B} = \vec{C}\), we can use the law of cosines which states:
\[ C^2 = A^2 + B^2 + 2AB \cos(\theta) \]
where \(C\) is the magnitude of \(\vec{C}\), \(A\) is the magnitude of \(\vec{A}\), \(B\) is the magnitude of \(\vec{B}\), and \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

2. **Substitute the known values**:
Substitute \(C = 17\), \(A = 8\), and \(B = 15\) into the equation:
\[ 17^2 = 8^2 + 15^2 + 2 \cdot 8 \cdot 15 \cdot \cos(\theta) \]

3. **Calculate the squares**:
\[ 17^2 = 289, \quad 8^2 = 64, \quad 15^2 = 225 \]
Now substitute these values into the equation:
\[ 289 = 64 + 225 + 2 \cdot 8 \cdot 15 \cdot \cos(\theta) \]

4. **Combine the terms**:
\[ 289 = 289 + 240 \cos(\theta) \]

5. **Rearrange the equation**:
Subtract 289 from both sides:
\[ 0 = 240 \cos(\theta) \]

6. **Solve for \(\cos(\theta)\)**:
Since \(240 \cos(\theta) = 0\), we can conclude:
\[ \cos(\theta) = 0 \]

7. **Find the angle \(\theta\)**:
The angle whose cosine is 0 is:
\[ \theta = 90^\circ \]

### Final Answer:
The angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).