Find the side AB of the trapezoid ABCD if the angles ABC and BCD are 60 ° and 135 ° respectively, and CD = 36

Find the segments as shown in the figure.
DE = AF, because these are the heights of the trapezoid.
∠DCE = 180 ° -∠BCD = 180 ° -135 ° = 45 ° (since these are adjacent angles).
sin (∠DCE) = ED / CD (by definition)
sin45 ° = ED / CD (sin45 ° = √2 / 2 according to the table)
√2 / 2 = ED / 36
ED = 36√2 / 2 = 18√2
sin (∠ABF) = AF / AB (by definition)
sin60 ° = ED / AB
AB = ED / sin60 ° (sin60 ° = √3 / 2 according to the table)
AB = (18√2) / (√3 / 2) = 18 * 2 * √2 / √3 = 36√2 / √3 = √362 * 2 / √3 = √1296 * 2/3 = √864 = 4√54 = 4 * 3√6 = 12√6
Answer: AB = 12√6