To find Angles of a quadrilateral Assume the following angles are known:
$ABD$,$DBC$,$BAC$,$ACD$.
Is it possible to compute $CDA$?

 A: No since no matter what equations you create you will always get
$$\measuredangle BDC+\measuredangle DCA$$
that cannot be isolated and depend on the length of $[AB]$
However, if the quadrilateral was a parallelogram then definitely since in that case
$$\measuredangle BCA=\measuredangle DAC$$
A: Generally speaking, it is possible to use the sine rule in $\triangle ABD$, $\triangle DBC$ and $\triangle BAC$ to develop a system of trigonometric equations which also develops some sort of relation between the sides of the quadrilateral, and then solving the system by isolating the sine functions with the missing angle (say $\angle x$) to obtain the answer. I believe such trigonometric generalizations already exist, I'd suggest looking up "Langley's adventitious angles" on Wikipedia, there, you'll find several links and more information regarding these "adventitious quadrangles" as well.
On the other hand, developing generalized geometric (purely euclidean) solutions is very hard if not near-impossible without further constraints, and even with that, the problem (and by extension its solution) can change significantly just by altering one or two angles. This applies to both quadrilaterals and triangles. Overall, there are several techniques and methods you can use to attack a problem using euclidean geometry, but they're very specific as I said and won't work all the time. What techniques? Well, let's take a general example. Suppose, in this scenario, we're given that $AB=AD$ and that $\angle ACD=30$, in this case, my very first move would be to locate a point (say, point $P$) outside $\triangle ADC$ and let it be the circumcenter for said triangle. Doing this would give us several connections such as $AB=AD=AP=DP=CP$ and the $\triangle APD$ would be equilateral. This would be a very helpful connection and help me proceed further with the solution in such a specific scenario. Alternatively, you can also reflect the $\triangle ADC$ about the line segment $DC$ and reflect point $A$ onto point $A'$ which would give you $\angle ACA'=60$ and thus $\triangle ACA'$ would be equilateral. This is yet another technique that you could use in such a specific scenario. There are definitely more, some even relying on previous results of problems I've solved (which I do at least mention whenever I use them and I don't think doing otherwise is fair), but these two would definitely things I'd try almost immediately in a scenario like this before, if this fails, proceeding to either some different techniques depending on other constraints or trying to do something completely different and being "creative" which involves staring the problem silently for 15 minutes.
