Clearly theta will be less than Pi/2.

Input := 


BPC[t_] = (Pi - t)/(2 Pi) Pi R^2 + 1/2 R Cos[t] R Sin[t]

Output =


 2             2

R  (Pi - t)   R  Cos[t] Sin[t]

----------- + ----------------

     2               2

Input := 


eq4 = Expand[BPC[t]/R^2] == 1/3 Pi R^2/R^2

Output =


Pi   t   Cos[t] Sin[t]    Pi

-- - - + ------------- == --

2    2         2          3

Input := 


sol = FindRoot[eq4,{t,.7}]

Output =


{t -> 1.30266}

Input := 


tsol = t/.sol[[1]]

Output =


1.30266