For reference: Calculate the measure of the smallest angle that determine the diagonals of a quadrilateral, inscribed in a circle, if the sides and are equivalent to the sides of the equilateral triangle and regular pentagon, inscribed in said circle. (Answer:$6^o$)
My progress:
I made the drawing and put the information that I could deduce
$\theta=?\\L_3=r\sqrt3\\ cos72=sin18= \frac{\sqrt5-1}{4}\\ cos36 = sin54 = \frac{\sqrt5+1}{4}\\ L_5=2rcos54 = \frac{\sqrt10-2\sqrt5}{4}$
I don't know if the way is this way
Comment: as you see in figure if triangle is equilateral then minimum of $\theta$ is ($\theta=24^o$)(triangle DEF and quadrilateral FIJE). ($\theta=6^o$) is possible if triangle is isosceles(quadrilateral BCIJ) in this case one side of quadrilateral is the base of isosceles triangle ABC.