When a fluid container accelerates upward with acceleration $a$ along an incline of angle $\alpha$, the effective gravity becomes $g+a$ in the downward direction. The free surface will be perpendicular to the net effective gravity vector. The component of this effective gravity parallel to the incline is $(g+a)\sin \alpha$ and perpendicular is $(g+a)\cos \alpha$. The angle of inclination $\theta$ of the free surface is given by the ratio of the horizontal to vertical components of acceleration relative to the container. The correct expression is derived from the effective acceleration components along and perpendicular to the incline.

Final Answer: ${\tan }^{-1}\left[\frac{a+g\sin \alpha }{g\cos \alpha }\right]$