Two vectors inclined at an angle `theta` have magnitude `3 N` and `5 N` and their resultant is of magnitude `4 N`. The angle `theta` is

Options

A. `90^(@)`
B. `cos^(-1)(4/5)`
C. `cos^(-1)(3/5)`
D. `cos^(-1)(-3/5)`

Explanation

To solve the problem, we will use the law of cosines, which relates the sides and angles of a triangle formed by two vectors and their resultant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnitude of the first vector, \( A = 3 \, \text{N} \)
- Magnitude of the second vector, \( B = 5 \, \text{N} \)
- Magnitude of the resultant vector, \( R = 4 \, \text{N} \)
- Angle between the vectors, \( \theta \)

2. **Apply the Law of Cosines:**
The law of cosines states that for any triangle with sides \( a \), \( b \), and \( c \), and the angle \( \theta \) opposite side \( c \):
\[ c^2 = a^2 + b^2 - 2ab \cos(\theta) \]
In our case:
\[ R^2 = A^2 + B^2 - 2AB \cos(\theta) \]

3. **Substitute the Known Values:**
\[ 4^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos(\theta) \]
This simplifies to:
\[ 16 = 9 + 25 - 30 \cos(\theta) \]

4. **Simplify the Equation:**
Combine the terms on the right side:
\[ 16 = 34 - 30 \cos(\theta) \]
Rearranging gives:
\[ 30 \cos(\theta) = 34 - 16 \]
\[ 30 \cos(\theta) = 18 \]

5. **Solve for \( \cos(\theta) \):**
Divide both sides by 30:
\[ \cos(\theta) = \frac{18}{30} = \frac{3}{5} \]

6. **Find the Angle \( \theta \):**
Use the inverse cosine function:
\[ \theta = \cos^{-1}\left(\frac{3}{5}\right) \]

### Final Answer:
The angle \( \theta \) is \( \cos^{-1}\left(\frac{3}{5}\right) \).

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