Triangle Problem

Answer is 30, but when I tried to solve it numerically, I would get zero, and when I solved it using matrices, I would get 55, so I don't know what's up with that.

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.