How I find the angle 'a'. 
Please let me know if you can prove the angle a is 150°.
 A: Hint: The large triangle is isosceles. What does that tell you about the base angles?
A: 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.
A: After very straightforward angle-chasing one can calculate all the angles from the picture below, so I am going to directly refer to them. 

Let $M$ be the midpoint of edge $AB$. Since $ABC$ is isosceles ($AC = BC$), line $CM$ is the orthogonal bisector of $AB$. Take point $E$ on line $CM$, so that it is on the same side of $AB$ as point $C$, such that triangle $ABE$ is equilateral, i.e. $AB = BE = EA$. Produce segment $AD$ until it intersects $CM$ at point $F$. As $CM$ is orthogonal bisector of $AB$, it is immediate that $AF = BF$, so $$\angle \, ABF = \angle \, BAF = \angle \, BAD = 30^{\circ}$$ Thus $\angle \, DBF = \angle \, CBF = 12^{\circ}$ i.e. $BF$ is the angle bisector of $\angle \, CBD$. 
Observe that triangles $ABD$ and $BEC$ are congruent, because $AB = BE$ as well as  $\angle \, ABD = \angle \, EBC = 18^{\circ}$ and $\angle \, BAD = \angle \, BEC = 30^{\circ}$. Therefore $BC = BD$ and thus triangle $BCD$ is isosceles. Consequently, the angle bisector $BF$ of $\angle \, CBD$ is also the altitude, the median and the orthogonal bisector of edge $CD$ of triangle $BCD$. Therefore, $CF = DF$ which means that triangle $CDF$ is isosceles and $\angle \, CDF = \angle \, DCF$. But since $\angle \, DFM = 60^{\circ}$ one can immediately conclude that  $$\angle \, CDF = \angle \, DCF = 30^{\circ}$$ Consequently, $$a = \angle \, ADC = 180^{\circ} - \angle \, CDF = 180^{\circ} - 30^{\circ} = 150^{\circ}$$      
Alternatively, one can also prove that $BC = BD$ without the construction of the equilateral triangle $ABE$ ( however, the construction of the points $M$ and $F$ is important ) by simply noticing that $\angle \, BDF = 48^{\circ} = \angle \, BCF$. Combined with the angle bisector property of $BF$, which implies that $\angle \, CBF = \angle \, DBF = 18^{\circ}$, one concludes that triangles $BCF$ and $BDF$ are congruent and thus $BC=BD$. The rest is the same as before.  
A: Let AM be the perpendicular bisector of BC; where M is the midpoint of BC.
BX extended cuts AM and AC at P and Q respectively. CX cuts AM at R. QR cuts CP at S.

By angle chasing, all the marked angles have their values as indicated.
Therefore, AXRQ is cyclic.
By ASA, $\triangle RPC \cong \triangle QPC$. This means CR = CQ. This further means $\angle CSR = 90^0$.
Then, $\beta = \alpha = 30^0$. Result follows.
