If you draw a vertical line from P through the circle centre,
mark where it touches the circle circumference, label it X.
You can calculate the diameter as clearly the triangle QPR has 90 and 45 degree angles,
so $\displaystyle Sin45^o=\frac{R}{40}\ \Rightarrow\ Diameter=2R=80Sin45^o$
Next, triangle PXT is a right-angled triangle since PX is a diameter.
You can calculate the angle XPS from
$\displaystyle Cos\theta=\frac{R}{25}\ \Rightarrow\ \theta=Cos^{-1}\left(\frac{R}{25}\right)$
The rest should be ok