在三角形abc中,内角A,B,C所对的边分别是a,b,c已知A=派/4,b平方-a平方=1/2c平方(1)求tanC的值

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  • (1)b平方-a平方=1/2c平方,

    由正弦定理,(sinB)^2-(sinA)^2=(1/2)(sinC)^2,

    A=π/4,B=3π/4-C,

    ∴[(√2/2)cosC+(√2/2)sinC]^2-1/2=(1/2)(sinC)^2,

    ∴(cosC+sinC)^2-1=(sinC)^2,

    ∴2sinCcosC-(sinC)^2=0,sinC>0,

    ∴2cosC=sinC,cosC≠0,

    ∴tanC=2.

    (2)sinC=2/√5,cosC=1/√5,

    sinB=(cosC+sinC)/√2=3/√10,

    由正弦定理,a=bsinA/sinB,c=bsinC/sinB,

    ∴S△ABC=(1/2)acsinB

    =(1/2)b^2*sinAsinC/sinB

    =(b^2/2)*1/√2*2/√5*√10/3

    =b^2/3=3,

    ∴b^2=9,b=3.