In the $\triangle ABC$ shown, $D$ is some interior point, and $x, y, z, w$ are the measures of angles
in degrees. Solve for $x$ in terms of $y, z$ and $w$.

$\textbf{(A)}\ w-y-z \qquad \textbf{(B)}\ w-2y-2z \qquad \textbf{(C)}\ 180-w-y-z \qquad \\ \textbf{(D)}\ 2w-y-z\qquad \textbf{(E)}\ 180-w+y+z$
By angles in a quadrilateral, $x = 360^{\circ} - \text{reflex } \angle ADB - y - z$, and by angles at a point, $\text{reflex } \angle ADB = 360^{\circ} - w$, so our expression becomes $360^{\circ} - (360^{\circ} - w) - y - z = w - y - z$, which is $\boxed{\text{A}}$.