The resultant of two forces P and Q makes an angle of `30^(@)` with P. What is the angle between P and Q, if the magnitude of the resultant is equal to `Q sqrt(3)` ?

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Explanation

To solve the problem, we need to analyze the forces \( P \) and \( Q \) and their resultant. Let's denote the angle between the forces \( P \) and \( Q \) as \( \theta \).

### Step 1: Understand the given information
We know:
- The resultant \( R \) of the forces \( P \) and \( Q \) makes an angle of \( 30^\circ \) with \( P \).
- The magnitude of the resultant \( R \) is equal to \( Q \sqrt{3} \).

### Step 2: Use the formula for the magnitude of the resultant
The magnitude of the resultant \( R \) when two forces \( P \) and \( Q \) are acting at an angle \( \theta \) is given by:
\[ R = \sqrt{P^2 + Q^2 + 2PQ \cos(\theta)} \]
We can also express \( R \) in terms of the angle \( \phi \) (the angle between \( R \) and \( P \)):
\[ R = Q \sqrt{3} \]

### Step 3: Set up the equations
Since \( R \) makes an angle of \( 30^\circ \) with \( P \), we can use the sine and cosine rules to relate the angles:
\[ R \cos(30^\circ) = P + Q \cos(\theta) \]
\[ R \sin(30^\circ) = Q \sin(\theta) \]

### Step 4: Substitute the value of \( R \)
Substituting \( R = Q \sqrt{3} \) into the equations:
1. For the cosine component:
\[ Q \sqrt{3} \cos(30^\circ) = P + Q \cos(\theta) \]
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ Q \sqrt{3} \cdot \frac{\sqrt{3}}{2} = P + Q \cos(\theta) \]
\[ \frac{3Q}{2} = P + Q \cos(\theta) \quad \text{(Equation 1)} \]

2. For the sine component:
\[ Q \sqrt{3} \sin(30^\circ) = Q \sin(\theta) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ Q \sqrt{3} \cdot \frac{1}{2} = Q \sin(\theta) \]
\[ \frac{Q \sqrt{3}}{2} = Q \sin(\theta) \]
Dividing both sides by \( Q \) (assuming \( Q \neq 0 \)):
\[ \sin(\theta) = \frac{\sqrt{3}}{2} \]
This implies:
\[ \theta = 60^\circ \quad \text{(Equation 2)} \]

### Step 5: Use Equation 1 to find \( P \)
Substituting \( \theta = 60^\circ \) into Equation 1:
\[ \frac{3Q}{2} = P + Q \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ \frac{3Q}{2} = P + Q \cdot \frac{1}{2} \]
\[ \frac{3Q}{2} = P + \frac{Q}{2} \]
Subtracting \( \frac{Q}{2} \) from both sides:
\[ \frac{3Q}{2} - \frac{Q}{2} = P \]
\[ \frac{2Q}{2} = P \]
\[ P = Q \]

### Conclusion
The angle between the forces \( P \) and \( Q \) is \( 60^\circ \).