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A particle is projected with an angle of projection $θ$ 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 [AIIMS 2015]

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$\tan^{- 1} [\frac{P^{2} + Q^{2}}{2 P Q}]$

$\tan^{- 1} [\frac{P^{2} + Q^{2} - P Q}{P Q}]$

$\tan^{- 1} [\frac{P^{2} + P Q + Q^{2}}{P Q}]$

$\sin^{- 1} [\frac{P^{2} + Q^{2} + P Q}{2 P Q}]$

answer is A.

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Detailed Solution

The equation of trajectory,

$y = x \tan α [1 - \frac{x}{R}]$

$Q = P \tan θ [1 - \frac{P}{R}]$

$and P = Q \tan θ [1 - \frac{Q}{R}]$

On dividing Eq. (i) by Eq. (ii), we get

$\frac{Q^{2}}{P^{2}} = \frac{[1 - P / R]}{[1 - Q / R]}$

$\frac{1}{R} [P^{3} - Q^{3}] = P^{2} - Q^{2}$

$\Rightarrow R = \frac{P^{3} - Q^{3}}{P^{2} - Q^{2}} = \frac{P^{2} + P Q + Q^{2}}{P + Q}$

Now, $\frac{Q}{P} = \tan θ [1 - \frac{P (P + Q)}{P^{2} + P Q + Q^{2}}]$

$= \tan θ [\frac{P^{2} + P Q + Q^{2} - P^{2} - P Q}{P^{2} + P Q + Q^{2}}]$

$\Rightarrow \tan θ = \frac{P^{2} + Q^{2} + P Q}{P Q}$

$\Rightarrow θ = \tan^{- 1} [\frac{P^{2} + P Q + Q^{2}}{P Q}]$

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