Geometry Problem and Isosceles Triangle $ABC$ is an isosceles triangle with $AB=AC$, $\angle BAC=96^\circ$. $D$ is a point such that $\angle DCA=48^\circ$, $AD=BC$ and angle $DAC$ is obtuse. What is the measure (in degrees) of $\angle DAC$?

First, draw. Then calculate the angles you might already know and name the unknown ones. So, using that $ABC$ is isosceles, we have $\angle BAC=42^\circ$. Since $42+48=90$, we have $\cos 48^\circ=\sin 42^\circ$. Let $\delta:=\angle ADC$, which is $<90^\circ$ since $\angle DAC$ is obtuse. After finding $\delta$, we will have
$$\angle DAC=180^\circ-48^\circ-\delta\ .$$
Use the theorem of sines, for the triangles $ABC$ and $ACD$, and use the identity for $\sin(2x)$ with $x=48^\circ$:
$$\frac{\sin\delta}{\sin 48^\circ}=\frac{AC}{AD}=\frac{AC}{BC}= 
\frac{\sin 42^\circ}{\sin 96^\circ}$$
From this, as $\sin 96^\circ=2\sin 48^\circ\cos 48^\circ=2\sin 48^\circ\sin 42^\circ $, we get that $\sin\delta=1/2$, that is $\delta=30^\circ$.
