Two forces 13kg, weight and `3 sqrt""3kg` weight act on a particle at an angle `theta` and are equal to a resultant force of 14kg weight then the angle between the forces is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to find the angle \( \theta \) between the two forces acting on a particle. We are given the magnitudes of the forces and the resultant force. The forces are:

- \( P = 13 \, \text{kg} \) (weight)
- \( Q = 3\sqrt{3} \, \text{kg} \) (weight)
- Resultant \( R = 14 \, \text{kg} \) (weight)

### Step-by-Step Solution

1. **Write the formula for the resultant of two forces**:
The resultant \( R \) of two forces \( P \) and \( Q \) acting at an angle \( \theta \) is given by the formula:
\[ R^2 = P^2 + Q^2 + 2PQ \cos \theta \]

2. **Substitute the known values**:
We know \( R = 14 \, \text{kg} \), \( P = 13 \, \text{kg} \), and \( Q = 3\sqrt{3} \, \text{kg} \). Plugging these values into the formula gives:
\[ 14^2 = 13^2 + (3\sqrt{3})^2 + 2 \cdot 13 \cdot 3\sqrt{3} \cos \theta \]

3. **Calculate the squares**:
- \( 14^2 = 196 \)
- \( 13^2 = 169 \)
- \( (3\sqrt{3})^2 = 27 \)

Now substituting these values:
\[ 196 = 169 + 27 + 2 \cdot 13 \cdot 3\sqrt{3} \cos \theta \]

4. **Simplify the equation**:
Combine the constants on the right side:
\[ 196 = 196 + 78\sqrt{3} \cos \theta \]

5. **Rearranging the equation**:
Subtract 196 from both sides:
\[ 0 = 78\sqrt{3} \cos \theta \]

6. **Solve for \( \cos \theta \)**:
Since \( 78\sqrt{3} \) is not zero, we can conclude:
\[ \cos \theta = 0 \]

7. **Find the angle \( \theta \)**:
The cosine of an angle is zero at:
\[ \theta = 90^\circ \]

### Final Answer
The angle between the two forces is \( \theta = 90^\circ \).