Two force in the ratio `1:2` act simultaneously on a particle. The resultant of these forces is three times the first force. The angle between them is

Options

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

Answer

To solve the problem, we need to determine the angle between two forces acting on a particle, given that their magnitudes are in the ratio 1:2 and the resultant force is three times the magnitude of the first force.

### Step-by-Step Solution:

1. **Define the Forces:**
Let the magnitude of the first force \( F_1 \) be \( F \). Since the forces are in the ratio \( 1:2 \), the magnitude of the second force \( F_2 \) will be \( 2F \).

2. **Define the Resultant Force:**
According to the problem, the resultant force \( R \) is three times the first force. Therefore, we can write:
\[ R = 3F \]

3. **Use 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:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
Substituting the values of \( F_1 \) and \( F_2 \):
\[ R = \sqrt{F^2 + (2F)^2 + 2 \cdot F \cdot 2F \cdot \cos \theta} \]
Simplifying this, we get:
\[ R = \sqrt{F^2 + 4F^2 + 4F^2 \cos \theta} \]
\[ R = \sqrt{5F^2 + 4F^2 \cos \theta} \]

4. **Set the Resultant Equal to Three Times the First Force:**
Since \( R = 3F \), we can set up the equation:
\[ 3F = \sqrt{5F^2 + 4F^2 \cos \theta} \]

5. **Square Both Sides:**
Squaring both sides to eliminate the square root gives:
\[ (3F)^2 = 5F^2 + 4F^2 \cos \theta \]
\[ 9F^2 = 5F^2 + 4F^2 \cos \theta \]

6. **Rearrange the Equation:**
Rearranging the equation, we find:
\[ 9F^2 - 5F^2 = 4F^2 \cos \theta \]
\[ 4F^2 = 4F^2 \cos \theta \]

7. **Divide by \( 4F^2 \):**
Dividing both sides by \( 4F^2 \) (assuming \( F \neq 0 \)):
\[ 1 = \cos \theta \]

8. **Find the Angle:**
The cosine of an angle is equal to 1 when the angle \( \theta \) is:
\[ \theta = 0^\circ \]

### Conclusion:
The angle between the two forces is \( 0^\circ \).