Dont know how to set up the Cosine Equation to solve this...
Check page 7 of this. If it is confusing, I will try to work you through the process.
http://missharvey.weebly.com/uploads/4/4/8/2/4482376/t3.pdf
After 2 hours:
RE^2 =18^2+24^2 (24 side length- 6 BE= 18)
RE=30
PR^2=24^2+24^2
PR=33.94112.....
Cos(PER)=30^2+30^2-33.9411...^2 / 2 (30) (30)
cos(PER) = 0.36
acos (0.36) = Angle PER
PER = 68.899803975907 degrees
HOPE THIS IS IT....... Please confirm
Conside the line from E to RC that is parallel to BC. Let the point of intersection be X
EX=24
XR=24-6=18
find ER
\(ER=\sqrt{24^2+18^2}\\ ER=\sqrt{900}\\ ER=30 \)
Now PR is just the diagonal of one side of the square
\(PR=\sqrt{24^2+24^2}\\ PR=\sqrt{1152}\\ \)
Now I have all 3 sides of the triangle so I can use cosine rule to find the angle between them
the sides are \(30, \quad30, \quad \sqrt{1152}\)
\(1152=900+900-2*900*Cos(PER)\\ 1152=1800-1800*Cos(PER)\\ -648=-1800*Cos(PER)\\ 0.36=Cos(PER)\\ \angle{PER}=cos^{-1}0.36=68.8998^\circ=68^\circ54' \)
That is assuming I've not made any stupid mistakes.