三角形ABC为等腰三角形,顶角A=80度,O为三角形ABC内一点,角OBC=10度,角OCB=20度,求角OAC的度数

1个回答

  • 角OAC的度数 x,角OAB=80-x,

    △AOB,OA/sin40=OB/sin(80-x) (正弦定理)

    △AOC,OA/sin30=OC/sinx

    sin40/sin30=(OB/OC)*sinx/sin(80-x) (正弦定理)

    (OB/OC)*sinx/sin(80-x)=sin40/(1/2)

    (OB/OC)*sinx/sin(80-x)=2sin40.(1)

    △OBC,OB/sin20=OC/sin10,(正弦定理)

    OB/OC=sin20/sin10=2sin10cos10/sin10=2cos10代入(1)

    2cos10sinx/sin(80-x)=2sin40

    sin80sinx=sin40sin(80-x)

    2sin40cos40=sin40sin(80-x)

    2cos40sinx=sin(80-x)=sin80cosx-cos80sinx

    (2cos40+cos80)sinx=sin80cosx

    tanx=sin80/(2cos40+cos80)

    x=arctan[sin80/(2cos40+cos80)]