tan⁻¹[P²+PQ+Q²/PQ]
tan⁻¹[P²+Q²-PQ/PQ]
sin⁻¹[P²+Q²+PQ/2PQ]
tan⁻¹[P²+Q²/2PQ]
Correct option is A. tan⁻¹[P²+PQ+Q²/PQ]
Given that, A particle is projected with an angle of projection to the horizontal line passing through the points (P,Q) and (Q,P).
The general equation of path for projectile motion is
y = x tanθ + gx²/2u²(cos θ)²
Now, since the above equation passes through (P,Q) and (Q,P) so, we will get two equations as
P = Qtanθ - gQ²/2u²(cos θ)² (I)
Q = Ptanθ - gP²/2u²(cos θ)² (II)
From equation (I) and (II) after solving
θ = tan⁻¹[P²+PQ+Q²/PQ]
So, The angle of projection is θ = tan⁻¹[P²+PQ+Q²/PQ]
Hence, A is correct option.