A projectile is thrown with a speed ‘u’ at angle $\text{'}\theta \text{'}$ to an inclined plane of inclination $\beta$. The angle $\text{'}\theta \text{'}$  at which the projectile is thrown such that it strikes the inclined plane horizontally is

 
 

 
 
  $$\begin{gathered} Taking horizontal as X-axis aand vertical as Y-axis, If particle strikes the plane horizontally then \\ its vertical velocity w.r.t ground is zero\Rightarrow v_x=0 \\ {v}_x=u_x+at \\ 0=\text{u sin}(\theta +\beta )-\text{gt} \\ time=t=\frac{u\sin \left(\theta +\beta \right)}{g} \\ time of flight on incine plane, T=\frac{2u\sin \theta }{g\cos \beta } \\ equating with time of flight of projection on inclined plane \\ \frac{2u\sin \theta }{g\cos \beta }=\frac{u\sin \left(\theta +\beta \right)}{g} \\ \text{2sin}\theta =\text{ sin}(\theta +\beta )\text{cos}\beta \\ let (\theta +\beta )=k \\ \begin{array}{l}\theta =k-\beta \\ 2\sin (k-\beta )=\sin k\cdot \cos \beta \\ 2\sin k\cdot \cos \beta -2\cos k\cdot \sin \beta =\sin k\cdot \cos \beta \\ \tan k=2\tan \beta \\ substitute k value\\ \tan (\theta +\beta )=2\tan \beta \\ \tan (\theta +\beta )=2\tan \beta )\\ \theta +\beta ={\tan }^{-1}(2\tan \beta )\\ \theta ={\tan }^{-1}(2\tan \beta )-\beta \end{array} \end{gathered}$$