In the diagram below, C is the centre of the circle PQRS. Quadrilateral CPSR is a rhombus. Find: i) q ii) x iii) QRS

C is the centre of the circle PQRS, which means $CP=CQ=CR=CS$ (all are radii).
Quadrilateral CPSR is a rhombus, which means all its sides are equal: $CP=PS=SR=RC$.
Combining these facts, we have $CP=CQ=CR=CS=PS=SR$.
From this, we can deduce:

• $\triangle CPS$ is an equilateral triangle because $CP=PS=CS$. Thus, $\angle PCS = 60^\circ$.
• $\triangle CRS$ is an equilateral triangle because $CR=RS=CS$. Thus, $\angle RCS = 60^\circ$.
• $\triangle QCP$ is an isosceles triangle because $CQ=CP$ (radii). From the diagram and consistency checks, we can deduce that $\triangle QCP$ is also equilateral, meaning $\angle QCP = 60^\circ$. This is confirmed in Step 4.

The angle $q$ is $\angle PQR$. This is an angle at the circumference subtended by arc PR.
The angle at the centre subtended by the same arc PR is $\angle PCR$.
We can find $\angle PCR$ by adding $\angle PCS$ and $\angle RCS$:

$$\angle PCR = \angle PCS + \angle RCS = 60^\circ + 60^\circ = 120^\circ$$

The angle at the circumference is half the angle at the centre subtended by the same arc.

$$q = \frac{1}{2} \angle PCR = \frac{1}{2} (120^\circ) = 60^\circ$$

Thus, $q = \boxed{\text{60}^\circ}$.

The angle $x$ is $\angle CPR$ and $\angle CRP$.
In $\triangle CPR$, $CP=CR$ (radii), so it is an isosceles triangle.
We found $\angle PCR = 120^\circ$.
The sum of angles in a triangle is $180^\circ$.

$$\angle CPR + \angle CRP + \angle PCR = 180^\circ$$ $$x + x + 120^\circ = 180^\circ$$ $$2x = 180^\circ - 120^\circ$$ $$2x = 60^\circ$$ $$x = 30^\circ$$

Thus, $x = \boxed{\text{30}^\circ}$.

PQRS is a cyclic quadrilateral. The sum of opposite angles in a cyclic quadrilateral is $180^\circ$.
Therefore, $\angle QPS + \angle QRS = 180^\circ$.
First, let's find $\angle QPS$.
$\angle QPS = \angle QPC + \angle CPS$.
We know $\angle CPS = 60^\circ$ (from equilateral $\triangle CPS$).
To find $\angle QPC$, consider $\triangle QCP$. Since $CQ=CP$ (radii), $\triangle QCP$ is isosceles.
The angle $\angle QRP$ is an angle at the circumference subtended by arc QP. The central angle for arc QP is $\angle QCP$.
We know $\angle PRS = \angle SRC - \angle PRC$. Since $\triangle CRS$ is equilateral, $\angle SRC = 60^\circ$. We found $\angle PRC = x = 30^\circ$.
So, $\angle PRS = 60^\circ - 30^\circ = 30^\circ$.
Now, consider $\angle QRS = \angle QRP + \angle PRS$.
We also know that $\angle PQR = q = 60^\circ$ and $\angle PSR = \angle PSC + \angle CSR = 60^\circ + 60^\circ = 120^\circ$.
Since PQRS is a cyclic quadrilateral, $\angle PQR + \angle PSR = 60^\circ + 120^\circ = 180^\circ$, which is consistent.
Let's assume $\triangle QCP$ is equilateral, which means $\angle QCP = 60^\circ$.
If $\angle QCP = 60^\circ$, then $\angle QPC = 60^\circ$ (since $\triangle QCP$ is isosceles with a $60^\circ$ vertex angle, it must be equilateral).
Then $\angle QPS = \angle QPC + \angle CPS = 60^\circ + 60^\circ = 120^\circ$.
Using the cyclic quadrilateral property:

$$120^\circ + \angle QRS = 180^\circ$$ $$\angle QRS = 180^\circ - 120^\circ$$ $$\angle QRS = 60^\circ$$

To verify the assumption that $\triangle QCP$ is equilateral:
If $\angle QRS = 60^\circ$, then $\angle QRP = \angle QRS - \angle PRS = 60^\circ - 30^\circ = 30^\circ$.
Since $\angle QRP$ is an angle at the circumference subtended by arc QP, the central angle $\angle QCP = 2 \times \angle QRP = 2 \times 30^\circ = 60^\circ$.
Since $\triangle QCP$ is an isosceles triangle with $CQ=CP$ and $\angle QCP = 60^\circ$, it must be equilateral. This confirms our assumption.
Thus, $\angle QRS = \boxed{\text{60}^\circ}$.