<s>[INST] Measure of an Interior Angle Triangle $ABC$ has $AC = BC$ , $\angle ACB = 96^\circ$ . $D$  is a point in $ABC$ such that $\angle DAB = 18^\circ$  and $\angle DBA = 30^\circ$ . What is the measure (in degrees) of $\angle ACD$ ?
 [/INST] If we put $AB=1$ and apply the low of sines, we get 
$$AC=\frac{\sin 42^\circ}{\sin 96^\circ}=\frac{\sin 42^\circ}{\cos 6^\circ},
\\ AD = \frac{\sin 30^\circ}{\sin 132^\circ}=\frac{1}{2\cos 42^\circ}.$$
But then it follows that $AC=AD$ from $2 \sin 42^\circ \cos 42^\circ=\sin 84^\circ = \cos 6^\circ$.
So $\triangle ADC$ is isosceles with the angle $24^\circ$ at $A$, which means the two equal angles are each $78^\circ$, and one of these is the desired $\angle ACD$.
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