How to find missing angles in a quadrilateral I have a quadrilateral ABCD, with diagonals AC and BD.
Given are four angles: ∠DAC = 20°, ∠CAB = 60°, ∠ABD = 50°, and ∠DBC = 30°.

Those are the red angles in the above image.
I need to fill in all the other angles.  Most are trivial – the angles in blue – but how do I find ∠BDC and ∠ACD?  Their sum is 110, obviously, but I can't figure out how to find the individual angles.
Edit:  Note that the red angles are examples; I'm looking for a general solution given any values for these angles that form a convex quadrilateral.  (They do if ∠DAC + ∠CAB + ∠ABD < 180° and ∠CAB + ∠ABD + ∠DBC < 180; in that case you can draw triangles ABD and ABC and then quadrilateral ABCD.)
 A: I have four different and one general solution to this problem, I hope it will help you.





A: A numeric computation using the Laws
of Sines
and Cosines yields
$\angle ACD=30^{\circ}$ and $\angle BDC=80^{\circ}$.
Let the intersection of $AC$ and $BD$ be $E$ and, wlog, let $AB=1$.
\begin{align}
AD &= \frac{\sin 50^{\circ}}{\sin 50^{\circ}} = 1 \\
DE &= AD\frac{\sin 20^{\circ}}{\sin 110^{\circ}} = 0.36397023426620234 \\
BC &= \frac{\sin 60^{\circ}}{\sin 40^{\circ}} = 1.3472963553338608 \\
CE &= BC\frac{\sin 30^{\circ}}{\sin 110^{\circ}} = 0.71688141714205145 \\
CD &= \sqrt{CE^2+DE^2-2\times CE\times DE\times\cos 70^{\circ}} = 0.68404028665133743 \\
\angle ACD &= \arcsin\big(\frac{DE}{CD}\sin 70^{\circ}\big) = 29.999999999999996 \\
\angle BDC &= \arcsin\big(\frac{CE}{CD}\sin 70^{\circ}\big) = 79.999999999999986
\end{align}
Note that $\angle DAB = \angle ABC$. I am not sure how to use this to reach an elegant solution, but I am sure that a non-computation solution will use both this observation and $AB=AD$ in a non-trivial way.
Note also that, post-factum, $CD$ is tangent to the circumcircle of $\triangle ABD$ (because we computed that $\angle ACD=\angle BAD$).
A: Use sine theorem for $\Delta ABC$ and $\Delta ACD:$
$$\frac{AB}{\sin{40}}=\frac{AC}{\sin{80}}$$
$$\frac{AD}{\sin{ACD}}=\frac{AC}{\sin{(20+ACD)}}$$
Noting $AB=AD$, we can find $mACD=30, mBDC=80$.
A: Let $AC\cap DB=\{F\}$ and $L\in FC$ such that $\measuredangle ABL=60^{\circ}$.
Also, let $BL\cap AD=\{G\}$.
Since $\measuredangle DAB=\measuredangle CBA$, we see that $GC||AB$ and from here $GC=GL=LC$.
Now, since $\measuredangle DBA=\measuredangle ADB=50^{\circ}$, we obtain: $AL=AB=AD$, which says that $\measuredangle ADL=\frac{180^{\circ}-20^{\circ}}{2}=80^{\circ}$ and since $\measuredangle AGB=180^{\circ}-80^{\circ}-60^{\circ}=40^{\circ}$,
we obtain: $\measuredangle GLD=40^{\circ}$, which says that $DG=DL$.
But also $GC=CL$, which gives that $DC$ is a bisector of $\angle GCL$.
Thus, $\measuredangle DCA=\frac{1}{2}\cdot60^{\circ}=30^{\circ}$ and the rest is smooth.
A: If you extend AD and BC to meet at E, this becomes the famous "Langley's adventitious angles" problem.  I have seen at least four proofs (other than non-proofs using numerical applications of the law of cosines) deriving angle ACD.  A good one is this:
Draw a line meeting AB at A, and meeting BC at F such that angle BAF is $20^\circ$. Thus AB = AF, and angle CAF is $40^\circ$.
Angle ABD and angle ADB are both $50^\circ$ so AD = AB, thus AF = AD.  In triangle AFD, sides AF and AD are equal and meet at an angle of $60^\circ$ so triangle AFD is equilateral and AF = FD. 
Angle FBC = $80^\circ - 20^\circ - 20^\circ = 40^\circ$.  In triangle ACB, 
angle ACB is $180^\circ - 80^\circ - 60^\circ = 40^\circ$. So triangle AFC is isosceles and CF = AF.  Then CF = FD. 
That shows that triangle CFD is isosceles and angles FCD and FDC are equal.  Angle CFD is $180^\circ$ minus angle AFD minus angle AFB which is  $180^\circ - 80^\circ - 60^\circ = 40^\circ$.  So angle FDC = angle FCD = $70^\circ$.
Finally since angle ACD plus angle ACF add to $70^\circ$,  angle is  $30^\circ$
(and angle ADC is $80^\circ$).
A: You don't need trig to solve this, it's completely doable by basic geometry, although you do have to draw a lot of stuff. Calculating those angles is actually a problem already online known as the "world's hardest easy problem" (problem 2)
https://www.duckware.com/tech/worldshardesteasygeometryproblem.html
There's a solution on the site, but I've shortly explained it alone.
(the quadrilateral from your problem is ABED on the picture, those sequences of 3 letters are supposed to be angles as I couldnt write them normally in paint)
http://imgur.com/a/mq6Nb
