b2+c2-bc=a2 b/c=1/2+根号3 求tanB

2个回答

  • 在三角形ABC中,

    2bc*cosA = b^2 + c^2 - a^2 = bc,

    cosA = 1/2.

    A = 60度.

    B + C = 180度 - 60度 = 120度

    C = 120度 - B.

    sinC = sin(120度 - B) = sin(120度)cosB - cos(120度)sinB

    = [3^(1/2)cosB + sinB]/2,

    sinC/sinB = [3^(1/2)/tanB + 1]/2,

    1/2 + 3^(1/2) = b/c = sinB/sinC = 2/[3^(1/2)/tanB + 1],

    2 = [3^(1/2)/tanB + 1][1/2 + 3^(1/2)]

    = 3^(1/2)/(2tanB) + 1/2 + 3/tanB + 3^(1/2)

    3/2 - 3^(1/2) = [3^(1/2) + 6]/(2tanB)

    tanB = [3^(1/2) + 6]/[3 - 2*3^(1/2)]

    = [3^(1/2) + 6][3 + 2*3^(1/2)]/[9 - 12]

    = [3*3^(1/2) + 18 + 6 + 12*3^(1/2)]/(-3)

    = -[5*3^(1/2) + 8]