How I find the angle 'a'. 
Please let me know if you can prove the angle a is 150°.

Let $\triangle ABC$ satisfy $\angle BAC=96^\circ$, $\angle ABC = \angle ACB=42^\circ$. Let $CK$ be a line with K on $AB$ such that $\angle BCK=18^\circ$ and $A,K$ are on the same side of $BC$. Let $F$ be the point on $BC$ such that $\angle AFC=60^\circ$. Let $AF$ meet $KC$ at $D$. Our goal is to prove that $\angle DBC=30^\circ$ and hence $A,B,C,D$ are the four points depicted in your graph.
First, let $E$ be the point such that $BE//CD$ and meets $AF$ at $E$. notice that $$\angle BAE=\angle AFC-\angle ABC=60^\circ-42^\circ=18^\circ=\angle DCF$$
$$\angle ABE = \angle ABC+\angle CBE=\angle ABC + \angle KCF=60^\circ=\angle DFC$$
Hence, we have $\triangle DCF \sim \triangle EAB$. 
Notice that $\angle CAD = \angle BAC-\angle BAE= 96^\circ-18^\circ=78^\circ = \angle DFC + \angle DCF = \angle ADC$, so $\triangle ACD$ is an isosceles triangle and $AC=CD$.
With $\triangle DCF \sim \triangle EAB$, $AC=CD$, and $BE//CD$ we have
$$\frac{BF}{BE}=\frac{CF}{CD}=\frac{CF}{AC}=\frac{CF}{AB}=\frac{DF}{BE}$$
Hence, $BF=DF$ and $\angle DBF = \angle BDF = \frac{1}{2}\cdot \angle DFC = 30^\circ$.
Therefore, we now have proved that our constructed $\triangle ABC$ is in fact the same as the given graph with the two lines of angles $30^\circ$ and $18^\circ$! 
Hence, $a=\angle BDA = 180^\circ-\angle BDF = 150^\circ$ as desired.
