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A large, heavy box is sliding down a smooth plane of inclination \( \theta \). From a point \( P \) on the bottom of the box,
\( \mathrm{P} \)
W a particle is projected inside the box. The initial speed of the particle with respect to the box is \( u \), and the direction of projection makes an angle \( \alpha \) with the bottom as shown in figure. The distance along the bottom of the box between the point of projection \( P \) and the point \( Q \) where the particle lands is \( \frac{u^{2} \sin x \alpha}{g \cos \theta} \). Then the value of \( x \) is (Assume that the particle does not hit any other surface of the box. Neglect air resistance.)
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