已知三角形ABC的面积为S,外接圆半径R等于根号17,a.b.c分别是角ABC对边,设S=a^2-(b-c)^2,sinB+ sinC=8/√17,求

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=...