Geometry snacks
Let the polygon be ABCD, with \angle A = 90^\circ, \angle B = 150^\circ, and \angle D = x.
Let M be the intersection point of the perpendicular from C to AB with AB.
Since \angle CBM = 30^\circ and \angle CMB = 90^\circ, we have
CM = \frac{1}{2}BC = \frac{1}{2}AD.
Let T be the intersection point of AB with CD.
Because CM \parallel AD and CM = \frac{1}{2}AD, points C and M are the midpoints of DT and AT respectively.
Thus AC = CD = CT ( median in right-angled triangle ).
Therefore x = \angle CDA = \angle CAD = 90^\circ - \angle CAB = 75^\circ
