△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinc),向量n=(cos c/2,2sinc),且...△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinC),向量n=(cos c/2,2sinc),且向量m⊥向量n,(1)求角C的大小
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![△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinc),向量n=(cos c/2,2sinc),且...△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinC),向量n=(cos c/2,2sinc),且向量m⊥向量n,(1)求角C的大小](/uploads/image/z/5166900-36-0.jpg?t=%E2%96%B3ABC%E4%B8%AD%2C%E8%A7%92A%2CB%2CC%2C%E6%89%80%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2C%E5%90%91%E9%87%8Fm%3D%EF%BC%882cos+c%2F2%2C-sinc%EF%BC%89%2C%E5%90%91%E9%87%8Fn%3D%EF%BC%88cos+c%2F2%2C2sinc%EF%BC%89%2C%E4%B8%94...%E2%96%B3ABC%E4%B8%AD%2C%E8%A7%92A%2CB%2CC%2C%E6%89%80%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2C%E5%90%91%E9%87%8Fm%3D%EF%BC%882cos+c%2F2%2C-sinC%EF%BC%89%2C%E5%90%91%E9%87%8Fn%3D%EF%BC%88cos+c%2F2%2C2sinc%EF%BC%89%2C%E4%B8%94%E5%90%91%E9%87%8Fm%E2%8A%A5%E5%90%91%E9%87%8Fn%2C%EF%BC%881%EF%BC%89%E6%B1%82%E8%A7%92C%E7%9A%84%E5%A4%A7%E5%B0%8F)
△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinc),向量n=(cos c/2,2sinc),且...△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinC),向量n=(cos c/2,2sinc),且向量m⊥向量n,(1)求角C的大小
△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinc),向量n=(cos c/2,2sinc),且...
△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinC),向量n=(cos c/2,2sinc),且向量m⊥向量n,
(1)求角C的大小,(2)若a平方=2b平方+c平方,求tanA的值
△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinc),向量n=(cos c/2,2sinc),且...△ABC中,角A,B,C,所对边分别为a,b,c,向量m=(2cos c/2,-sinC),向量n=(cos c/2,2sinc),且向量m⊥向量n,(1)求角C的大小
因为向量m⊥向量n所以,2cos c/2*cos c/2-sinc*2sinc=0 可得,2(cos c/2)^2-2(sinc)^2=0
1+cos c-2*(1-cosc^2)=0 cos c=1/2 c=π/3
(2)a^2=(2b)^2+c^2 . 1
加余弦定理和正弦定理
a^2=b^2-2bcxosA+c^2.2
b/sinB=c/sinC=2R .3
1-2得
0=3b+2ccosA
带入3
tanA=(-5根号3)/3