$\triangle ABC$ has $AC=BC$, and $\angle ACB=96^\circ$. $D$ is a point in $\triangle ABC$, such that $\angle DAB=18^\circ, \angle DBA=30^\circ$. What is $\angle ACD$?
My attempt:
$$\angle ABC=\angle BAC=\frac{(180^\circ-96^\circ)}{2}=42^\circ.$$
$$\angle ADB=180^\circ-18^\circ-30^\circ=132^\circ.$$
From here onwards, I have no idea how to carry on. Can anyone help?