Pinwheel Pattern

Note: In the last line, it should read $\sin E = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}} \Rightarrow E = 45 ^ \circ$ .

Based on symmetry, the quadrilateral has to be a parallelogram, so the diagonals will intersect in the center $E$ , with $DE = EB$ . We construct a point $F$ on $DC$ so that $EF = DE$ . Angle $DFE$ will therefore be also $30^\circ$ , giving us angle $DEF$ as $120^\circ$ , therefore $\angle FEB = 60^\circ$ . Since $EF = EB$ , triangle $EBF$ is equilateral. Angle $EFB$ will therefore be $60^\circ$ giving us angle $ECB$ as $30^\circ$ , as required. The distance $FC$ , as a radius of the circle centered at $F$ , will be equal to $EF$ . With angle $BFC$ as $90^\circ$ , the obtuse angle in the isosceles triangle $EFC$ is $150^\circ$ , and angle $FEC$ is $15^\circ$ . Angle FEB (60 degrees) minus angle CEB (15 degrees) = 45 degrees.

This is only one possible solution, however, since the circle intersects the horizontal line at the bottom of the image in two possible locations of point C, the other being much closer to point D, as shown below.

Angle DFE is still 30 degrees, and EFB is 60, triangle EBF is equilateral, and CF = EF. In the isosceles triangle CEF angle CEF is $\frac{180^\circ-30^\circ}2 = 75^\circ$ , so $x=180^\circ - 60^\circ - 75 = \boxed{45^\circ}$ .