Let the angle between the two arms be 2.  When u draw d free body diagram u 'll find that the vertically downward vector mg (weight of the boy) bisects this angle (if the two arms are equal).  So, angle between each of the two arms with mg is . Let the tension in each arm be T.

             So, for vertical equilibrium,

                      Tcos + Tcos = mg

                So,                 T = mg/2cos

                So, when cos is least, i.e when  = 90 degree,  then tension T is the maximum.

                     So, the required angle is 2 = 2X90 degree = 180 degree 

Let us suppose that the angle between the arms is  2.  

 

upon drawing the FBD, it's apparent that the vertically downward vector mg is bisecting this angle assuming that the two arms are of equal length.  So, angle between each of the two arms with mg is . Let the tension in each arm be T.

             So, balancing the forces in vertical direction,

                      Tcos + Tcos = mg

            thus        T = mg/2cos

            thus ::      when cos is least, i.e when  = 90 degree,  then tension T is the maximum.

                     So, the required angle is 2 = 2X90 degree = 180 degree