The maqunitudes of vecotr `overset(rarrA), overset(rarr)B` and `overset(rarr)C` are respectively 12,5 and 13 units and `overset(rarr)A+overset(rarr)B=overset(rarr)C` then the angle between `overset(rarr)A` and `overset(rarr)B` is :

Options

A. 0
B. `pi`
C. `pi//2`
D. `pi//4`

Explanation

To solve the problem, we need to find the angle between vectors \(\overset{\rarr}{A}\) and \(\overset{\rarr}{B}\) given that their magnitudes are 12 units and 5 units respectively, and the resultant vector \(\overset{\rarr}{C}\) has a magnitude of 13 units. The relationship between these vectors is given by the equation:

\[ \overset{\rarr}{A} + \overset{\rarr}{B} = \overset{\rarr}{C} \]

### Step-by-Step Solution:

1. **Identify the Magnitudes**:
- Let the magnitude of \(\overset{\rarr}{A} = 12\) units.
- Let the magnitude of \(\overset{\rarr}{B} = 5\) units.
- Let the magnitude of \(\overset{\rarr}{C} = 13\) units.

2. **Apply the Law of Cosines**:
According to the law of cosines, we can express the relationship between the magnitudes of the vectors as follows:
\[ |\overset{\rarr}{C}|^2 = |\overset{\rarr}{A}|^2 + |\overset{\rarr}{B}|^2 + 2 |\overset{\rarr}{A}| |\overset{\rarr}{B}| \cos(\theta) \]
where \(\theta\) is the angle between \(\overset{\rarr}{A}\) and \(\overset{\rarr}{B}\).

3. **Substitute the Values**:
Substitute the known magnitudes into the equation:
\[ 13^2 = 12^2 + 5^2 + 2 \cdot 12 \cdot 5 \cdot \cos(\theta) \]

4. **Calculate the Squares**:
Calculate the squares:
\[ 169 = 144 + 25 + 120 \cos(\theta) \]

5. **Simplify the Equation**:
Combine the terms:
\[ 169 = 169 + 120 \cos(\theta) \]
Subtract 169 from both sides:
\[ 0 = 120 \cos(\theta) \]

6. **Solve for \(\cos(\theta)\)**:
Divide both sides by 120:
\[ \cos(\theta) = 0 \]

7. **Determine the Angle**:
The angle \(\theta\) for which \(\cos(\theta) = 0\) is:
\[ \theta = 90^\circ \]

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

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