jmc

We can see that $\angle ACB=40^{\circ }$ must be half of the central angle formed by the arc $AB,$ or $80^{\circ }.$ Likewise, $\angle CAD=30^{\circ }$ must be half of the central angle formed by the arc $CD,$ or $60^{\circ }.$ Then, we can see that the angles formed by arcs $BC$ and $DA$ must sum to $360^{\circ }-(80^{\circ }+60^{\circ })=220^{\circ }.$ That means the sum $\angle CAB+\angle ACD$ must be half of that, or $\boxed{110^{\circ }}.$