In trapezium $$ABCD$$, it is given that $$AB$$ is parallel to $$CD$$. Assume that $$AB = 3CD$$, $$CD = DA$$, and $$\angle CDA = 120^\circ$$. If the largest angle of $$ABCD$$ is $$x^\circ$$ and the smallest angle is $$y^\circ$$, what is the value of $$x/y$$?

Put $$CD = DA = s$$, so $$\angle DAB = 180^\circ - 120^\circ = 60^\circ$$, and divide $$AB = 3s$$ into three equal parts at $$E$$ and $$F$$. Triangle $$ADE$$ is equilateral, and $$DCFE$$ is a parallelogram, so $$CF = s = FB$$ and triangle $$CFB$$ is isosceles with apex angle $$120^\circ$$. Hence $$\angle B = 30^\circ$$ and $$\angle C = 150^\circ$$, so the four angles are $$60^\circ, 30^\circ, 150^\circ, 120^\circ$$ and $$x/y = 150/30 = 5$$.