Beauty Square

Let $O$ the point of interseccion between $AC$ and $DB$ .

We have that $\angle AOR=\angle ROC=90° \therefore$ $\triangle ARC$ is isosceles with $AR=RC$ , but $AC=RC$ because both are radius of the circumference. So $\triangle ARC$ is equilateral $\Rightarrow \angle PRQ=30°$ and $\angle RPQ=90° \therefore \boxed{\angle PQR=60°}$ .

Just by analyzing the picture,

If we draw line the diagonal AC of the square, we see is equal to the radius R and is parallel to PQ since the other diagonal is contain in PR.

Then segments AC=CQ=QP =PA all are equal to radius R forming a rhombus which diagonal equal to CP= R so our rhombus is formed by to equilateral triangles then angle PQR must equal to 60 degrees.

sin PQR = cosPRQ =PR / QR Either PQR = 30 and PRQ = 60 or vice versa or both = 45° we assume that PRQ = 60° as BD bisects angle ADC the angle BDC =45° and angle RDC = 135° So in the triangle CRD sum of angle RDC and angle DRC is equal to 135 + 60 = 195 but it is impossible for the same reason PRQ can't be equal to 45° so PRQ has to be equal to 30 and PQR=90-30=60 °

If we join AC, O is the point of intersection of the diagonals AC and BD, then, OC= 1/2 AC (OR) 1/2 (radius). So in triangle ORC, sin r = 1/2 or 30 degrees. So, the answer is 60 degrees.

Applying the sin rule to the triangle RBC gives sin R = (BC sin B)/RC = (BC/AC) sin 45 = 1/2. So R is 30. We know P is 90 as RQ is a diameter, so Q is 60.

If you rotate the figure so that $AC$ is vertical, you might find it easier to visualise.

It can readily be seen that $AC$ is a line of symmetry of the figure, ignoring $Q$ and the line segments connected to it for the moment. As such, $AC$ is a perpendicular bisector of $RP$ .

Let $X$ be the centre of the square. Obviously $X$ lies on both $AC$ and $RP$ . Let $Y$ be the point opposite $A$ on the circle.

By the intersecting chord formula, $AX.XY = RX.XP$ . If we take the radius of the circle to be 1, then this gives $\frac12 . \frac32 = RX.XP$ . But $X$ bisects $RP$ , therefore $\frac34 = RX^2 \Rightarrow RX = \frac{\sqrt3}2 \Rightarrow RP = \sqrt3$ .

$\angle RPQ = 90°$ since it's an angle in a semicircle. Hence $AC \parallel PQ$ , and we can see that $PQR$ is half of the circle's inscribed rectangle, leading to $PQ = 2XC = AC = 1$ . You might recognise $PQ = 1$ and $RP = \sqrt3$ as the sides of a 90°/60°/30° triangle, hence $\angle PQR = \boxed{60°}$ .

Actually that square is useless. As angle PCQ=angle PQC and= 2PRQ by (angle at centre twice angle at circumference) then you can solve angle PQR=60degrees