Find an angle in a given triangle $\triangle ABC$ has sides $AC = BC$ and $\angle ACB = 96^\circ$. $D$ is a point in $\triangle ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$. What is the measure (in degrees) of $\angle ACD$?
In $\triangle ADB,\angle ADB=(180-18-30)^\circ=132^\circ$

Applying sine law in $\triangle ADB,$ $$\frac{AB}{\sin 132^\circ}=\frac{AD}{\sin30^\circ}\implies AD=\frac{AB}{2\sin48^\circ}$$ as $\sin132^\circ=\sin(180-132)^\circ=\sin48^\circ$

$\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$

Applying sine law in $\triangle ABC,$ $$\frac{AC}{\sin 42^\circ}=\frac{AB}{\sin 96^\circ}\implies \frac{AC}{AB}=\frac{\sin 42^\circ}{\sin 96^\circ}=\frac{\cos 48^\circ}{2\sin 48^\circ \cos 48^\circ}$$ (applying $\sin 2A=2\sin A\cos A$)

So, $$AC=\frac{AB}{2\sin48^\circ} \implies AC=AD$$

So, $\angle ACD=\angle ADC=\frac{(180-24)^\circ}2=78^\circ$
Hint: 

Have you drawn your picture? 

I drew the picture out and I got an isosceles triangle, with base angles being ∠A and ∠B. Then I placed point D inside the triangle and connected that point to the 3 vertices of the triangle. 

The first statement tells us that we are in an isosceles triangle ( meaning base angles are the congruent ). Therefore we can find the measure of the base angles since we know that ∠ACB=96∘. Each base angle turns out to be 42∘ each. 

Consider triangle ABD. Can you find all angles of that triangle? And can you then find the angles of triangle BDC? and ADC?

I hope this helped! If you are still stuck let us know!