In addition to the figure -
let I be the enter of the circle
and P the point of tangency of RT
with the circle.

If t = tan((angle PIC)/2), then it 
is easy to show that the area of
triangle RST is

   sqrt(3)*(3 - t^2)
  -------------------
       1 - 3*t^2

Note: Angle PIC can be measured CW 
      or CCW. Thus, we can have one
      or two positions for point P.

1a)   sqrt(3)*(3 - t^2)
     ------------------- = 3*sqrt(3). 
          1 - 3*t^2

     or t = 0.

     Thus, P = C.

1b)   sqrt(3)*(3 - t^2)
     ------------------- = 8. 
          1 - 3*t^2

                   8 - 3*sqrt(3)
     or t = sqrt[ --------------- ].
                    24 - sqrt(3)

     Thus, angle PIC =

                      8 - 3*sqrt(3)
     2*arctan{ sqrt[ --------------- ] }
                       24 - sqrt(3)

     ~= 39.073817 degrees.

1c)   sqrt(3)*(3 - t^2)
     ------------------- = 6*sqrt(3) 
          1 - 3*t^2

     or t = sqrt[ 3/17 ].

     Thus, angle PIC = 

     2*arctan{ sqrt[ 3/17 ] }

     ~= 45.572996 degrees.   

2)   For angle RTS to be a right angle,
     angle PIC must equal 30 degrees.

     Thus, area [RST] 

        sqrt(3)*(3 - tan(15)^2)
     = -------------------------
            1 - 3*tan(15)^2

     = 2*sqrt(3) + 3

     ~= 6.464102