For the points #A(3,3), B(1,2)# and #C(8,9)#
#AB=sqrt((1-3)^2+(2-3)^2)=sqrt(4+1)=sqrt(5)~=2.2#
#BC=sqrt((8-1)^2+(9-2)^2)=sqrt(7^2+7^2)=sqrt(98)=7sqrt(2)~=9.9#
#AC=sqrt((3-8)^2+(3-9)^2)=sqrt(25+36)=sqrt(61)~=7.8#
The case is schematically represented by the figure below
As the sides of the triangle are 5, 8 and 9:
#x+y=7sqrt(2)#
#x+z=sqrt(61)#
#y+z=sqrt(5)# => #z=sqrt(5)-y#
#-> x+sqrt(5)-y=sqrt(61)# => #x-y=sqrt(61)-sqrt(5)#
Adding the first and last equations
#2x=7sqrt(2)+sqrt(61)-sqrt(5)# => #x=(7sqrt(2))/2+sqrt(61)/2-sqrt(5)~=7.737#
Using the Law of Cosines:
#(sqrt(5))^2=(7sqrt(2))^2+(sqrt(61))^2-2*7sqrt(2)*sqrt(61)*cos alpha#
#cos alpha=(98+61-5)/(14*sqrt(122))=154/(14*sqrt(122))=11/sqrt(122)#
#alpha=5.194^@#
In the right triangle with #x# as cathetus, we can see that
#tan (alpha/2)=r/x#
#r=x*tan (alpha/2)=7.737*tan (5.194^@/2)# => #r=.351#
