在三角形ABC中,已知A,B,C成等差数列,且b=根号3 ,(1)若sinA+cosA=根号2,求a;(2)求三角形AB

1个回答

  • (1)

    A,B,C成等差数列,A + C = 2B

    A+ B + C = B + 2B = 180˚

    B = 60²

    sinA+cosA = √2

    (sinA + cosA)² = 2

    sin²A + cos²A + 2sinAcosA = 2

    1 + sin(2A) = 2

    sin(2A) = 1

    2A = 90˚,A = 45˚

    a/sinA = b/sinB

    a = bsinA/sinB = √3sin45˚/sin60˚ = √3(√2/2)/(√3/2) = √2

    (2)

    a/sinA = b/sinB = csinC

    a = bsinA/sinB

    c = bsinC/sinB

    S = (1/2)ac*sinB = (1/2)(sinB)(bsinA/sinB)(bsinC/sinB)

    = (1/2)b²sinAsinC/sinB

    = [(1/2)*3/(√3/2)]sinAsinC

    = √3sinAsinC

    = (√3/2)[cos(A - C) - cos(A + C)]

    = (√3/2)[cos(A - C) -cos120˚]

    = (√3/2)[cos(A - C) + 1/2]

    A - C = 0,S最大

    S = (√3/2)(cos0˚ + 1/2)

    = (√3/2)(1 + 1/2)

    = 3√3/4