find the angle by geometry

First, we have $\angle ABD=150$ and $\angle BCD=96$

Next by the law of sines on $\triangle ABD$ and $\triangle BCD$

$\frac{AB}{BD}=\frac{\sin24}{\sin150}=2\sin24$ and $\frac{CD}{BD}=\frac{\sin54}{\sin96}=\frac{\sin54}{\sin84}$

Which means $\frac{AD}{DC}=\frac{2\sin24\sin84}{\sin54}$

A calculator shows this equals $1$ . I was unable to prove this using identities. I was able to by using the exact values. The details are rather messy.

So $AD=AC$ and we have an isosceles triangle. $\angle ACD=(180-36)/2 = 72$ and $x=96-74=\boxed{24}$

Incidentally, $\triangle ABC$ is also isosceles as is the littlest triangle.

PS Whoever inked the picture did an amazing job.

Let $O$ denote the circumcenter of $\triangle BCD$ . Note that $O \in AD, CBO$ is an equilateral $\triangle, \angle ABO= 18^{\circ}$ and $\angle AOB = 12^{\circ}.$ Let $P$ be a point : $\angle PBA= \angle ABO =18^{\circ}, \angle POA = \angle AOB = 12^{\circ}.$ [Just doubling the angles..] Note that $\angle BPO= 120^{\circ} \Rightarrow CBPO$ is cyclic. Also, $A$ is the incenter of $\Delta PBO$ and $AC$ bisects $\angle BPO \Rightarrow P \in CA \Rightarrow \angle PCB = \angle POB = 24^{\circ}.$ $\boxed{K.I.P.K.I.G}$