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  • 向量m=(2cosC/2,-sinC) n=(cosC/2,2sinC) 向量m⊥n 角C=60° 若a²=2b²+c²求tanA的值

    向量m=(2cosC/2,-sinC) n=(cosC/2,2sinC) 向量m⊥n 角C=60° 若a²=2b²+c²求tanA的值
    數(shù)學(xué)人氣:889 ℃時(shí)間:2019-10-19 19:28:29
    優(yōu)質(zhì)解答
    m⊥n
    =>m.n=0
    (2cos(C/2),-sinC).(cos(C/2),2sinC)=0
    2(cosC/2)^2-2(sinC)^2=0
    (2(cosC/2)^2-1)-2(sinC)^2+1=0
    cosC-2(sinC)^2+1=0
    cosC - (2-2(cosC)^2)+1=0
    2(cosC)^2+cosC-1=0
    cosC = (-1+3)/4 = -1/2
    C=2π/3
    sinC = √3/2
    a^2 =2b^2+c^2
    c^2 = a^2+b^2 - 3b^2
    by cosine rule
    -3b^2 =-2abcosC
    cosC = (3b/(2a)) = -1/2
    b= -a/3
    a^2 =2b^2+c^2
    = b^2 + c^2 +b^2
    by cosine -rule
    -2bccosA = b^2
    cosA = -b/(2c)
    = -b/ (2asinC/sinA)
    tanA = -2asinC/b
    = -2a(√3/2))/( -a/3)
    = 3√3
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