解析:∵S=a^2-(b-c)^2=2bc-(b^2+c^2-a^2)
=2bc-2bc*cosA
=2bc(1-cosA)
=1/2*bc*sinA
∴4(1-cosA)=sinA
即cosA=1-sinA/4
∴(sinA)^2+(cosA)^2
=(sinA)^2+(1-sinA/4)^2=1
解得sinA=8/17,sinA=0(舍去)
∵sinA/a=sinB/b=sinC/c=1/2R
∴a=2RsinA=16√17/17
cosA=1-sinA/4=15/17
(sinB+sinC)/(b+c)=1/2R
即b+c=2R(sinB+sinC)=2√17*8/√17=16
又cosA=(b^2+c^2-a^2)/2bc
=[(b+c)^2-16*16/17-2bc]/2bc
=[16*16-16*16/17-2bc]/2bc
=15/17
得,bc=64
∴S=1/2*bc*sinA=1/2*64*8/17=256/17