Two forces of magnitudes P and Q are inclined at an angle `(theta)`. The magnitude of their resultant is 3Q. When the inclination is changed to `(180^(@)-theta)`, the magnitude of the resultant force becomes Q. Find the ratio of the forces.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the equation for the resultant of two forces at angle θ
The magnitude of the resultant \( R \) of two forces \( P \) and \( Q \) inclined at an angle \( \theta \) is given by:
\[ R = \sqrt{P^2 + Q^2 + 2PQ \cos \theta} \]

### Step 2: Set up the equation for the first case
According to the problem, when the angle is \( \theta \), the resultant is \( 3Q \):
\[ 3Q = \sqrt{P^2 + Q^2 + 2PQ \cos \theta} \]
Squaring both sides, we get:
\[ (3Q)^2 = P^2 + Q^2 + 2PQ \cos \theta \]
This simplifies to:
\[ 9Q^2 = P^2 + Q^2 + 2PQ \cos \theta \]
Rearranging gives:
\[ P^2 + 2PQ \cos \theta = 9Q^2 - Q^2 \]
\[ P^2 + 2PQ \cos \theta = 8Q^2 \quad \text{(Equation 1)} \]

### Step 3: Set up the equation for the second case
When the angle changes to \( 180^\circ - \theta \), the resultant becomes \( Q \):
\[ Q = \sqrt{P^2 + Q^2 - 2PQ \cos \theta} \]
Squaring both sides, we have:
\[ Q^2 = P^2 + Q^2 - 2PQ \cos \theta \]
Rearranging gives:
\[ P^2 - 2PQ \cos \theta = Q^2 - Q^2 \]
\[ P^2 - 2PQ \cos \theta = 0 \]
This can be rewritten as:
\[ P^2 = 2PQ \cos \theta \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 2 into Equation 1
Now we substitute \( P^2 \) from Equation 2 into Equation 1:
\[ 2PQ \cos \theta + 2PQ \cos \theta = 8Q^2 \]
This simplifies to:
\[ 4PQ \cos \theta = 8Q^2 \]
Dividing both sides by \( 4Q \) (assuming \( Q \neq 0 \)):
\[ P \cos \theta = 2Q \]
Thus, we can express \( P \) in terms of \( Q \):
\[ P = \frac{2Q}{\cos \theta} \]

### Step 5: Find the ratio of the forces
To find the ratio \( \frac{P}{Q} \):
\[ \frac{P}{Q} = \frac{2}{\cos \theta} \]
However, we need to find a specific ratio. From the previous equations, we have \( P^2 = 2PQ \cos \theta \). Substituting \( P = 2Q \) gives us:
\[ (2Q)^2 = 2(2Q)Q \cos \theta \]
\[ 4Q^2 = 4Q^2 \cos \theta \]
This implies \( \cos \theta = 1 \), which means \( \theta = 0^\circ \) or \( P = 2Q \).

Thus, the ratio of the forces \( P \) and \( Q \) is:
\[ \frac{P}{Q} = 2:1 \]

### Final Answer
The ratio of the forces \( P \) and \( Q \) is \( 2:1 \).

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