The lengths of the sides of a convex quadrilateral are $$\sqrt{a}$$, $$\sqrt{a + 3}$$, $$\sqrt{a + 2}$$ and $$\sqrt{2a + 5}$$, in this order. The length of each diagonal is $$\sqrt{2a + 5}$$. If $$\theta^\circ$$ is the difference between the largest angle and the second largest angle of the quadrilateral, determine the value of $$\theta$$.

Label the consecutive vertices $$A, B, C, D$$, so $$BC^2 + CD^2 = (a + 3) + (a + 2) = BD^2$$ and the converse of Pythagoras gives $$\angle BCD = 90^\circ$$. Drawing $$AH \perp CD$$ and $$AK \perp CB$$ and applying Pythagoras leads to $$\frac{(a + 4)^2}{a + 3} + \frac{a + 2}{4} = 2a + 5$$, which simplifies to $$(3a + 10)(a - 1) = 0$$, so $$a = 1$$. Then $$BK = \tfrac{1}{2}$$ and $$AK = \frac{\sqrt{3}}{2}$$ give $$\angle ABK = 60^\circ$$ and hence $$\angle ABC = 120^\circ$$, while the angles at $$A$$ and $$D$$ are acute base angles of isosceles triangles. The two largest angles are $$120^\circ$$ and $$90^\circ$$, so $$\theta = 30$$.