The magnitude of vectors A, B and C are 3, 4 and 5 units respectively. If A+ B =C , the angle between A and B is

Options

A. `(pi)/(2)`
B. `cos^(-1)(0.6)`
C. `tan^(-1)((7)/(5))`
D. `(pi)/(4)`

Answer

To solve the problem, we need to find the angle between vectors A and B given that their magnitudes are 3 units and 4 units respectively, and that A + B = C, where the magnitude of C is 5 units.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Magnitude of vector A (|A|) = 3 units
- Magnitude of vector B (|B|) = 4 units
- Magnitude of vector C (|C|) = 5 units
- The relationship given is A + B = C.

2. **Use the Law of Cosines:**
The law of cosines relates the magnitudes of the vectors and the angle between them. It states:
\[ |C|^2 = |A|^2 + |B|^2 + 2|A||B|\cos(\theta) \]
where \(\theta\) is the angle between vectors A and B.

3. **Substitute the Known Values:**
Substitute the magnitudes of the vectors into the equation:
\[ 5^2 = 3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot \cos(\theta) \]

4. **Calculate the Squares:**
Calculate the squares of the magnitudes:
\[ 25 = 9 + 16 + 24\cos(\theta) \]

5. **Simplify the Equation:**
Combine the terms on the right side:
\[ 25 = 25 + 24\cos(\theta) \]

6. **Isolate the Cosine Term:**
Rearrange the equation to isolate the cosine term:
\[ 25 - 25 = 24\cos(\theta) \]
\[ 0 = 24\cos(\theta) \]

7. **Solve for Cosine:**
Since \(24\cos(\theta) = 0\), we find:
\[ \cos(\theta) = 0 \]

8. **Determine the Angle:**
The cosine of an angle is zero at \(\theta = 90^\circ\) (or \(\frac{\pi}{2}\) radians). Therefore, the angle between vectors A and B is:
\[ \theta = 90^\circ \]

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

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