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 $\angle ACD$?

$\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$
As  $\angle DAB=18^\circ, \angle DAC=(42-18)^\circ=24^\circ, $
Similarly, $\angle CBD=18^\circ$
Let $\angle ACD=x,$ so, $\angle DCB=96^\circ-x$
So, in $\triangle ADC, \angle ADC=180^\circ-(24^\circ+x)=156^\circ-x$
Similarly, from $\triangle BCD, \angle BDC=72^\circ-x$
Applying sine law in  $\triangle BCD,$ $$\frac {CD}{\sin 12^\circ}=\frac {BC}{\sin(72^\circ+x)}$$
Similarly, from $\triangle ADC,$ $$\frac {CD}{\sin 24^\circ}=\frac {AC}{\sin(156^\circ-x)}$$
On division, $$\frac{\sin 24^\circ}{\sin 12^\circ}=\frac{\sin(156^\circ-x)}{\sin(72^\circ+x)}$$ as $AC=BC$
But $\sin(156^\circ-x)=\sin(180^\circ-(24^\circ+x))=\sin(24^\circ+x)$ and $\sin 24^\circ=2\sin 12^\circ\cos 12^\circ$
So, $$2\cos 12^\circ=\frac{\sin(24^\circ+x)}{\sin(72^\circ+x)}$$
Applying $2\sin A\cos B=\sin(A+B)+\sin(A-B),$
$$\sin(84^\circ+x)+\sin(60^\circ+x)=\sin(24^\circ+x)$$
or applying $\sin(A+B)$ formula and separating sine and cosine,  $$\sin x (\cos24^\circ-\cos84^\circ-\cos60^\circ)=\cos x(\sin84^\circ-\sin24^\circ+\sin60^\circ)$$
$$\tan x=\frac{\sin84^\circ-\sin24^\circ+\sin60^\circ}{\cos24^\circ-\cos84^\circ-\cos60^\circ}=\frac{2\sin30^\circ\cos54^\circ+\sin60^\circ}{2\sin30^\circ\sin54^\circ-\cos60^\circ}$$ (applying $\sin C-\sin D,\cos C-\cos D$ formula)
$$=\frac{\cos54^\circ+\cos30^\circ}{\sin54^\circ-\sin30^\circ}=\frac{2\cos42^\circ\cos12^\circ}{2\cos42^\circ\sin12^\circ}$$
(applying $\sin C-\sin D,\cos C+\cos D$ formula)
$$=\cot 12 ^\circ=\tan(90-12)^\circ$$
$\implies x=78^\circ$ as $0<x<180^\circ$
