How to decide dimensions of cyclic quadrilateral?

Circle with center O		PQ = a	QR=b	RS=c	SP=d
Radius of circle	R	5			Sin	R Sin theta	Cos	R Cos theta	Height of isosceles triangle	Angle between Radius & side
Angle bet P & Q		60	30	0.523599	0.5	2.5	0.866025	4.330127	4.330127019	60	Angle OPQ
Angle bet Q & R		120	60	1.047198	0.866025404	4.330127	0.5	2.5	2.5	30	Angle OQR
Angle bet R & S		90	45	0.785398	0.707106781	3.5355339	0.707107	3.535534	3.535533906	45	Angle ORS
Angle bet S & P		90	45	0.785398	0.707106781	3.5355339	0.707107	3.535534	3.535533906	45	Angle OSP
PQ		5
QR		8.660254
RS		7.071068
SP		7.071068
Area of Triangle OPQ		10.82532	10.825
Area of Triangle OQR		10.82532
Area of Triangle OPQ		12.5
Area of Triangle OPQ		12.5
Are of Quad PQRS		46.65064
Angle bet PQ & QR		90
Angle bet QR & RS		75
Angle bet RS & SP		90
Angle bet SP & PQ		105
Angle Q + Angle S		180	Because it is a cyclic quadrilateral
Angle P + Angle R		180	Because it is a cyclic quadrilateral
Sum ofangles of quad.		360				rcumfewrence
Circumference		31.41593
Perimeter of quad		27.80239	It is always less than circumference of the circle in which it is inscribed
Diagonal PR					Sin	R Sin Theta
Angle bet P and R	Angle OPQ + Angle OQR	180	90	1.570796	1	5
	(1/2PR)/R= Sin (1/2*Angle OPR)
Diagonal PR =	10
Diagonal QS					Sin	R Sin
Angle bet Q and S	Angle OQR + Angle ORS	210	105	1.832596	0.965925826	4.8296291
	(1/2QS)/R= Sin (1/2*Angle OQS)
Diagonal QS =	9.659258263
Let E bepoint of intersection of the diagonals
PE/ER= da /cb	0.577350269			By Similarity of triagles
PR/ER =	1.577350269
ER =	6.339745962
PE = PR-ER	3.660254038
SE/EQ= cd/ba	1.154700538
SQ/EQ=	2.154700538
EQ=	4.482877361
SE=SQ-EQ	5.176380902
PE*ER=	23.20508076
SE*EQ=	23.20508076
Internal intersecting chord property proved.