Suppose the angle $s$ is 60 degrees, then it is easy to calculate the length of the diagonal and from it all the angles in the diagram. You might like to check your answer by drawing the quadrilateral accurately, using ruler and compasses only, and then measuring the angles.

To calculate the angles of the cyclic quadrilateral formed by keeping the lengths of the sides the same and changing the angles so that opposite angles add up to 180 degrees you simply need to use the fact that, in this case, $\cos s=-\cos q$.

Congratulations to Katharina Jurges from the International School in Strasbourg and to Chris Hardstaff from Heckmondwike Grammar School for their solutions to Bendy Quad. Here is Katharina's solution:

To solve this problem I only used one formula which is called the Al Kashi theorem in the country I go to school (France). It is actually an extension of Pythagoras' theorem and is true for every triangle, rectangular or not:

${\displaystyle a^2=b^2+c^2-2\mathrm{bc}\times \cos (k)}$

where a is the side opposite to the vertex A and k the angle at A; b and c are the two other sides of the triangle. By transforming it one obtains:

${\displaystyle \cos (k)=(b^2+c^2-a^2)/(2\mathrm{bc})}$

First part (convex quadrilateral):
Note $p$, $q$, $r$ are the other three angles in the quad
$p$ is formed by sides 4 and 5 (i.e. the sides with the lengths 4 and 5),
$q$ is formed by the sides 5 and 6, and r by the sides 6 and 3).

Note $d$ the diagonal opposite to the angle of 60. Using the first equation, one finds $d$ equals the square root of 13. Now it is easy to calculate the other angles, although one has to calculate $p$ and $r$ by first calculating the two parts of each angle because they are cut by the diagonal $d$. I just give the approximate values for the angles:

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{p}& =& 132.{9222836}^o\\ \multicolumn{1}{c}{q}& =& 36.{86989765}^o\\ \multicolumn{1}{c}{r}& =& 130.{2078187}^o\end{array}}$

Second Part (cyclic quadrilateral):
I use the same notations as before and note s to be the angle which measured $60\backslash circle$ in the first part.
$s$ and $q$ are opposite angles (in a cyclic quadrilateral), therefore $\cos (s)=-\cos (q)$.

According to the Al Kashi theorem:

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{d^2}& =& 3^2+6^2-2\times 3\times 6\times \cos (s)\\ \multicolumn{1}{c}{d^2}& =& 4^2+5^2-2\times 4\times 5\times \cos (q)\\ \multicolumn{1}{c}{}& =& 4^2+5^2+2\times 4\times 5\times \cos (s)\\ \multicolumn{1}{c}{3^2+6^2-2\times 3\times 6\times \cos (s)}& =& 4^2+5^2+2\times 4\times 5\times \cos (s)\end{array}}$

Thus one can calculate $\cos (s)$, then $\cos (q)$, $d$, and the other two angles which must be calculated "by parts" once again (as in the first part). I give the approximate values for the four angles and d:

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{d}& =& 6.565459859\\ \multicolumn{1}{c}{p}& =& 115.{3769335}^o\\ \multicolumn{1}{c}{q}& =& 93.{01696131}^o\\ \multicolumn{1}{c}{r}& =& 64.{62306648}^o\\ \multicolumn{1}{c}{s}& =& 86.{98303869}^o\end{array}}$