Triangle Problem

Since I got a conflicting answer, I suppose it's time I show my work:

 Spoiler

- Prove that triangle BEC is isoceles, (angle BEC = 50*) -> BC = CE
- Law of Sines -> sin(theta)/CE = sin(DCA)/AD (with DCA = 20*) -> sin(theta) = sin20*BC/AD (1)
- Prove Triangle ADC is isoceles at D -> AD = DC -> need DC
- Law of Sines, part 2 -> sin80/DC = sin(BDC)/BC -> DC (= CE) = BC * sin80/sin40 (2)

(1) & (2) => sin(theta) = sin20*(BC/(BC * sin80/sin40)) = sin40*sin20/sin80 = 0.223
=> theta = arcsin(0.223) = ~13*

I'll go dig a hole now. Nevertheless, I think it's possible to find AD and AE with the above direction (in terms of BC's length).. then proceed to find angle ADE and AED (2 equations with 2 unknowns) then theta.

Thing is, I got 30 from degrees from actually drawing the triangle to scale and measuring it. Using just geometry and algebra isn't enough to solve it, and I'm too lazy to use trig. I'll leave that part up to someone else.