Solve the angle $\angle{DCB}$ in triangle $\triangle{ABC}$ with $\angle{A}=84^{\circ}$

Draw isosceles triangle $ACE$ such that $CA=CE$.

Let the circumcentre of $\triangle DEC$ be $F$ and draw the equilateral triangle $DEF$.

$\angle CEF=\angle ECF=36^\circ$.

Now construct another equilateral triangle with side $CE$ as in the figure.

$\angle DEG=36^\circ$

$\therefore\triangle DEG\cong\triangle FEC\text{ (S-A-S)}\\ \implies DE=DG.$

Also $\angle BDG=72^\circ.$

Then mark point $H$ on $CE$ such that $CF=CH$.

$\therefore \angle CHF=\angle CFH=72^\circ$.

Also $\triangle EHF$ is isosceles.

Now we can see $\triangle EDG\sim\triangle EHF$. From there we can prove that $\triangle BGE\sim \triangle CFE$.

From similar isosceles triangles above, $\frac{ED}{EG}=\frac{EH}{EF}.$
Adding $1$ to both sides, $\frac{EB}{EG}=\frac{EC}{EF}\implies\small \triangle BGE\sim \triangle CFE.$

It shows that $BG=GE$ and then we can easily show $\angle EBC=\angle ABC=30^\circ.$