As title suggests, the problem is as follows:
We're given a triangle $\triangle ABC$ where $D$ lies inside the triangle. We know that $\angle DBA=10^\circ, \angle DBC=20^\circ, \angle DAB=30^\circ$ and $\angle DAC=50^\circ$. The goal is to find the measure of $\angle DCA=x$
I will share my own solution below as an answer, please let me know if there is anything wrong in it. Furthermore, I'd like to see any other different ways to approach this (such as via trigonometry, analytical geometry, etc) so please share your own approaches as well.
Best Answer
Here is a (completely changed) solution using trigonometry.
Let $I$ be the intersection point of $AC$ and $BD$.
The main observation is that $AIB$ is a right triangle because the sum of angles in $A$ and $B$ is $90°$ ; $BI \perp AD$; as a consequence, $BIC$ is also a right triangle.
Let $BD=a, DI=b, AI=c, IC=d$.
The sine law gives :
$$\begin{cases}\text{in triangle DIA : } \frac{b}{c}&=&\tan(50°)\\ \text{in triangle DIC : }\frac{b}{d}&=&\tan(x)\end{cases} \implies \frac{c}{d}=\frac{\tan(x)}{\tan(50°)}\tag{1}$$
$$\begin{cases}\text{in triangle AIB : } \frac{c}{a+b}&=&\tan(10°)\\ \text{in triangle CIB : }\frac{d}{a+b}&=&\tan(20°)\end{cases} \implies \frac{c}{d}=\frac{\tan(10°)}{\tan(20°)}\tag{2}$$
Equating values in relationships (1) and (2) :
$$\tan(x)=\frac{\tan(10°)\tan(50°)}{\tan(20°)}=\frac{\tan(10°)}{\tan(20°)\tan(40°)}$$ $$\implies x=30°,\tag{3}$$
the last implication being due to a relation which is a close parent of the so-called "Morrie's law" (mentioned at the bottom of this article) saying that :
$$\underbrace{\tan(20°)\tan(40°)\tan(80°)}_{\frac{\tan(20°)\tan(40°)}{\tan(10°)}}=\tan(60°)$$