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,sin

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  • 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