 Solution

Use coordinates $s$ down the inclined plane and $n$ along its upward normal, with the launch point as origin. Gravity has components $g\sin \varphi$ and $-g\cos \varphi$. For launch angle $\theta$ above the downhill direction,The nonzero root of $n(t)=0$ is $t_f=2v_0\sin \theta /(g\cos \varphi )$. Substituting into $s(t)$ and combining the two trigonometric terms gives the distance along the slope:This assumes $0\le \varphi <\pi /2$, $\sin \theta >0$, and a downhill landing with $\cos (\theta -\varphi )>0$, as well as the neglected air resistance. It reduces to the horizontal projectile range $v_0^2\sin (2\theta )/g$ at $\varphi =0$. The angle relative to the horizontal is $\theta -\varphi$, which explains the shifted cosine. The original diagram uses stipulated illustrative angles; no examination figure is being reconstructed.