Two forces, each of magnitude F have a resultant of the same magnitude F. The angle between the two forces is

Options

A. `30^(@)`
B. `60^(@)`
C. `120^(@)`
D. `150^(@)`

Answer

To solve the problem, we need to find the angle between two forces of equal magnitude \( F \) that result in a resultant force also of magnitude \( F \).

### Step-by-Step Solution:

1. **Understanding the Problem**: We have two forces, both with magnitude \( F \), and we need to find the angle \( \theta \) between them such that their resultant is also \( F \).

2. **Using the Formula for Resultant of Two Forces**: The formula for the resultant \( R \) of two forces \( F_1 \) and \( F_2 \) acting at an angle \( \theta \) is given by:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
In our case, since both forces are equal:
\[ R = \sqrt{F^2 + F^2 + 2F \cdot F \cos \theta} \]

3. **Substituting Values**: Since both forces are \( F \), we can substitute \( F_1 = F \) and \( F_2 = F \):
\[ R = \sqrt{F^2 + F^2 + 2F^2 \cos \theta} \]
This simplifies to:
\[ R = \sqrt{2F^2 + 2F^2 \cos \theta} \]

4. **Setting the Resultant Equal to \( F \)**: We know the resultant \( R \) is also \( F \):
\[ F = \sqrt{2F^2 + 2F^2 \cos \theta} \]

5. **Squaring Both Sides**: To eliminate the square root, we square both sides:
\[ F^2 = 2F^2 + 2F^2 \cos \theta \]

6. **Rearranging the Equation**: Rearranging gives:
\[ F^2 - 2F^2 = 2F^2 \cos \theta \]
This simplifies to:
\[ -F^2 = 2F^2 \cos \theta \]

7. **Dividing by \( F^2 \)**: Dividing both sides by \( F^2 \) (assuming \( F \neq 0 \)):
\[ -1 = 2 \cos \theta \]

8. **Solving for \( \cos \theta \)**: This gives:
\[ \cos \theta = -\frac{1}{2} \]

9. **Finding the Angle \( \theta \)**: The angle \( \theta \) for which \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \quad \text{(in the second quadrant)} \]

10. **Conclusion**: Therefore, the angle between the two forces is \( 120^\circ \).

### Final Answer:
The angle between the two forces is \( 120^\circ \).