Tricky chase

I initially used trig and asked, is there an easier way? Michael replied Yes, so I figured it out. The (probably) intended method is below the trig one.

$c = a\cdot\frac{\sin(100)}{\sin(40)} = a\cdot\frac{\sin(80)}{\sin(40)}$

$\overline{BO} = c\cdot\frac{\sin(30)}{\sin(130)} = c\cdot\frac{1}{2\cos(40)}$

$= a\cdot\frac{\sin(80)}{2\sin(40)\cos(40)} = a\cdot\frac{\sin(80)}{\sin(80)} = a$

So $BOC$ is isosceles and $\angle BCO = (180-20)/2 = 80$ , and hence the needed angle is $\boxed{20}$ degrees.

Here the values filled in are easy to find. The trick that can be hard to see if you don't have the diagram drawn to scale is the symmetry that entails that the two angles labeled "x" are equal.