A particle is projected with an angle of projection `theta` to the horizontal line passing through the points (P,Q) and (Q,P) referred to horizontal and vertical axes (can be treated as X-axis and Y-axis respectively). The angle of projection can be given by

Options

A. `tan^(2)[(P^(2)+PQ+Q^(2))/(PQ)]`
B. `tan^(-1) [(P^(2)+Q^(2)-PQ)/(PQ)]`
C. `tan^(-1) [(P^(2)+Q^(2))/(2PQ)]`
D. `sin^(-1) [(P^(2)+Q^(2)+PQ)/(2PQ)]`

Answer

To solve the problem, we need to find the angle of projection \( \theta \) of a particle projected through the points \( (P, Q) \) and \( (Q, P) \). We will use the equations of projectile motion to derive the angle of projection.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We have two points: \( (P, Q) \) and \( (Q, P) \). We can treat these as coordinates in a Cartesian plane, where \( P \) is along the x-axis and \( Q \) is along the y-axis.

2. **Equation of Trajectory:**
The equation of the trajectory of a projectile is given by:
\[ y = x \tan \theta - \frac{g}{2v^2 \cos^2 \theta} x^2 \]
where \( g \) is the acceleration due to gravity, \( v \) is the initial velocity, and \( \theta \) is the angle of projection.

3. **Using the Points:**
For the point \( (P, Q) \):
\[ Q = P \tan \theta - \frac{g}{2v^2 \cos^2 \theta} P^2 \quad \text{(Equation 1)} \]
For the point \( (Q, P) \):
\[ P = Q \tan \theta - \frac{g}{2v^2 \cos^2 \theta} Q^2 \quad \text{(Equation 2)} \]

4. **Dividing the Equations:**
Divide Equation 1 by Equation 2:
\[ \frac{Q}{P} = \frac{P \tan \theta - \frac{g}{2v^2 \cos^2 \theta} P^2}{Q \tan \theta - \frac{g}{2v^2 \cos^2 \theta} Q^2} \]

5. **Cross Multiplying:**
Cross-multiply to simplify:
\[ Q(Q \tan \theta - \frac{g}{2v^2 \cos^2 \theta} Q^2) = P(P \tan \theta - \frac{g}{2v^2 \cos^2 \theta} P^2) \]

6. **Rearranging:**
Rearranging gives us:
\[ Q^2 \tan \theta - \frac{g}{2v^2 \cos^2 \theta} Q^3 = P^2 \tan \theta - \frac{g}{2v^2 \cos^2 \theta} P^3 \]

7. **Isolating \( \tan \theta \):**
Rearranging terms to isolate \( \tan \theta \):
\[ (Q^2 - P^2) \tan \theta = \frac{g}{2v^2 \cos^2 \theta} (P^3 - Q^3) \]

8. **Using the Identity:**
Using the identity \( P^3 - Q^3 = (P - Q)(P^2 + PQ + Q^2) \):
\[ (Q^2 - P^2) \tan \theta = \frac{g}{2v^2 \cos^2 \theta} (P - Q)(P^2 + PQ + Q^2) \]

9. **Final Expression for \( \tan \theta \):**
Solving for \( \tan \theta \):
\[ \tan \theta = \frac{g (P - Q)(P^2 + PQ + Q^2)}{2v^2 \cos^2 \theta (Q^2 - P^2)} \]

10. **Conclusion:**
The angle of projection \( \theta \) can be expressed as:
\[ \theta = \tan^{-1}\left(\frac{g (P - Q)(P^2 + PQ + Q^2)}{2v^2 \cos^2 \theta (Q^2 - P^2)}\right) \]