Given acute triangle $\triangle ABC$ in plane $P$, a point $Q$ in space is defined such that $\angle AQB = \angle BQC = \angle CQA = 90^\circ.$  Point $X$ is the point in plane $P$ such that $QX$ is perpendicular to plane $P$. Given $\angle ABC = 40^\circ$ and $\angle ACB = 75^\circ,$ find $\angle AXC.$