Square of the resultant of two forces of equal magnitude is equal to three times the product of their magnitude. The angle between them is :

Options

A. `0^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to find the angle between two forces of equal magnitude given that the square of their resultant is equal to three times the product of their magnitudes.

### Step-by-Step Solution:

1. **Let the Magnitude of Forces**:
Let the magnitude of each force be \( F \). Since both forces are equal, we have:
\[ F_1 = F \quad \text{and} \quad F_2 = F \]

2. **Resultant of Two Forces**:
The formula for the resultant \( R \) of two forces \( F_1 \) and \( F_2 \) at an angle \( \theta \) is given by:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
Substituting \( F_1 = F \) and \( F_2 = F \):
\[ R = \sqrt{F^2 + F^2 + 2F \cdot F \cos \theta} = \sqrt{2F^2 + 2F^2 \cos \theta} \]
This simplifies to:
\[ R = \sqrt{2F^2(1 + \cos \theta)} = F\sqrt{2(1 + \cos \theta)} \]

3. **Square of the Resultant**:
Now, squaring the resultant:
\[ R^2 = 2F^2(1 + \cos \theta) \]

4. **Given Condition**:
According to the problem, the square of the resultant is equal to three times the product of their magnitudes:
\[ R^2 = 3F^2 \]

5. **Setting the Equations Equal**:
We can set the two expressions for \( R^2 \) equal to each other:
\[ 2F^2(1 + \cos \theta) = 3F^2 \]

6. **Dividing by \( F^2 \)**:
Assuming \( F \neq 0 \), we can divide both sides by \( F^2 \):
\[ 2(1 + \cos \theta) = 3 \]

7. **Solving for \( \cos \theta \)**:
Rearranging the equation gives:
\[ 1 + \cos \theta = \frac{3}{2} \]
\[ \cos \theta = \frac{3}{2} - 1 = \frac{1}{2} \]

8. **Finding the Angle \( \theta \)**:
The angle \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

### Final Answer:
The angle between the two forces is \( 60^\circ \).