Two forces P and Q have a resultant perpendicular to P. The angle between the forces is

Options

A. `tan^(-1)((-P)/Q)`
B. `tan^(-1)((P)/Q)`
C. `sin^(-1)((P)/Q)`
D. `cos^(-1)(-(P)/Q)`

Explanation

To solve the problem of finding the angle between two forces \( P \) and \( Q \) when their resultant is perpendicular to \( P \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have two forces \( P \) and \( Q \) with a resultant \( R \) that is perpendicular to \( P \). We need to find the angle \( \theta \) between the two forces.

2. **Setting Up the Diagram**:
- Let \( P \) be represented as a vector along the x-axis.
- Let \( Q \) be represented as a vector making an angle \( \theta \) with \( P \).
- The resultant \( R \) of the two forces \( P \) and \( Q \) is perpendicular to \( P \).

3. **Using Vector Addition**:
- The resultant \( R \) can be expressed using the law of cosines:
\[ R^2 = P^2 + Q^2 + 2PQ \cos(\theta) \]
- Since \( R \) is perpendicular to \( P \), we know that:
\[ R \cdot P = 0 \]

4. **Finding the Angle**:
- The angle \( \alpha \) that \( R \) makes with \( P \) is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.
- We can use the tangent of the angle \( \alpha \):
\[ \tan(\alpha) = \frac{Q \sin(\theta)}{P + Q \cos(\theta)} \]
- Since \( \alpha = 90^\circ \), we have:
\[ \tan(90^\circ) = \infty \]
- This implies that the denominator must approach zero:
\[ P + Q \cos(\theta) = 0 \]

5. **Solving for \( \cos(\theta) \)**:
- Rearranging the equation gives:
\[ Q \cos(\theta) = -P \]
- Thus,
\[ \cos(\theta) = -\frac{P}{Q} \]

6. **Finding \( \theta \)**:
- Finally, we find the angle \( \theta \):
\[ \theta = \cos^{-1}\left(-\frac{P}{Q}\right) \]

### Final Answer:
The angle \( \theta \) between the two forces \( P \) and \( Q \) is given by:
\[ \theta = \cos^{-1}\left(-\frac{P}{Q}\right) \]