Triangle Problem

Trig method, not sure why it isn't getting that 30 degree answer? Did this independently of Kalovale.
DC/sin(80) = BC/sin(40)
EC = BC (since EBC is isosceles)
DCE = 20 degrees
DE^2 = DC^2 + EC^2 - 2DC*EC*cos(20)
DE/sin(20) = EC/sin(theta)
Now, working back:
sin(theta) = EC/DE*sin(20)
sub in for EC and DE
sin(theta) = BC*sin(20)/sqrt(DC^2 + BC^2 - 2DC*BC*cos(20))
keep going to get all BCs
sin(theta) = BC*sin(20)/sqrt((BC*sin(80)/sin(40))^2 + BC^2 - 2*(BC*sin(80)/sin(40))*BC*cos(20))
then we have BC/sqrt(BC^2) which factors out
sin(theta) = sin(20)/sqrt(sin^2(80)/sin^2(40) + 1 - 2*sin(80)*cos(20)/sin(40))
multiply a sin^2(40) into the bottom
sin(theta) = sin(20)*sin(40)/sqrt(sin^2(80) + sin^2(40) - 2*sin(80)*cos(20)*sin(40))

This gives theta = 30 degrees...