已知A,B,C是△ABC的三个内角,向量m=(1,-√3),n=(cosA,sinA),且m·n=-1, (1)求角A(2)若(sinB+cosB)/(sinB-cosB)=3,求tanC的值请教步骤
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![已知A,B,C是△ABC的三个内角,向量m=(1,-√3),n=(cosA,sinA),且m·n=-1, (1)求角A(2)若(sinB+cosB)/(sinB-cosB)=3,求tanC的值请教步骤](/uploads/image/z/3684837-21-7.jpg?t=%E5%B7%B2%E7%9F%A5A%2CB%2CC%E6%98%AF%E2%96%B3ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92%2C%E5%90%91%E9%87%8Fm%3D%EF%BC%881%2C-%E2%88%9A3%EF%BC%89%2Cn%3D%EF%BC%88cosA%2CsinA%EF%BC%89%2C%E4%B8%94m%C2%B7n%3D-1%2C+%281%29%E6%B1%82%E8%A7%92A%EF%BC%882%29%E8%8B%A5%28sinB%2BcosB%29%2F%28sinB-cosB%29%3D3%2C%E6%B1%82tanC%E7%9A%84%E5%80%BC%E8%AF%B7%E6%95%99%E6%AD%A5%E9%AA%A4)
已知A,B,C是△ABC的三个内角,向量m=(1,-√3),n=(cosA,sinA),且m·n=-1, (1)求角A(2)若(sinB+cosB)/(sinB-cosB)=3,求tanC的值请教步骤
已知A,B,C是△ABC的三个内角,向量m=(1,-√3),n=(cosA,sinA),且m·n=-1, (1)求角A
(2)若(sinB+cosB)/(sinB-cosB)=3,求tanC的值
请教步骤
已知A,B,C是△ABC的三个内角,向量m=(1,-√3),n=(cosA,sinA),且m·n=-1, (1)求角A(2)若(sinB+cosB)/(sinB-cosB)=3,求tanC的值请教步骤
(1)
m*n=-1
1*cosA-√3sinA=-1
1/2cosA-√3/2sinA=-1/2
cosAcosπ/3-sinAsinπ/3=-1/2
cos(A+π/3)=-1/2
所以A+π/3=2π/3
A=π/3
(2)
(sinB+cosB)/(sinB-cosB)=3
sinB+cosB=3sinB-3cosB
2sinB=4cosB
sinB=2cosB
tanB=2
tanA=tanπ/3=√3
tanC=tan(180-(A+B))=
=-tan(A+B)
=-(tanA+tanB)/(1-tanAtanB)
=-(√3+2)/(1-2√3)
=(√3+2)/(2√3-1)
=(√3+2)(2√3+1)/(2√3-1)(2√3+1)
=(6+√3+4√3+2)/(12-1)
=(8+5√3)/11
m*n=cosA-根号3sinA=2(sin30cosA-cos30sinA)=2sin(30-A)=-1
即sin(30-A)=-1/2, sin(A-30)=1/2
所以,A-30=30
即:角A=60度。
(sinB+cosB)/(sinB-cosB)=3
上下同除以cosB得:
(tanB+1)/(tanB-1)=3
tanB+1=...
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m*n=cosA-根号3sinA=2(sin30cosA-cos30sinA)=2sin(30-A)=-1
即sin(30-A)=-1/2, sin(A-30)=1/2
所以,A-30=30
即:角A=60度。
(sinB+cosB)/(sinB-cosB)=3
上下同除以cosB得:
(tanB+1)/(tanB-1)=3
tanB+1=3tanB-3
tanB=2
tanA=根号3
所以,tanC=tan[180-(A+B)]=-tan(A+B)
=-(tanA+tanB)/(1-tanAtanB)
=-(根号3+2)/(1-2根号3)
=-(2+根号3)(1+2根号3)/(1-12)
=(2+4根号3+根号3+6)/11
=(8+5根号3)/11
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