已知ABC为△ABC的三个内角,向量m=(1,-根号3) ,n=(cosA,sinA)m*n=-1.求∠A.若1+sin2B/sin^2 B-cos^2 B
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![已知ABC为△ABC的三个内角,向量m=(1,-根号3) ,n=(cosA,sinA)m*n=-1.求∠A.若1+sin2B/sin^2 B-cos^2 B](/uploads/image/z/981654-6-4.jpg?t=%E5%B7%B2%E7%9F%A5ABC%E4%B8%BA%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%281%2C-%E6%A0%B9%E5%8F%B73%29+%2Cn%3D%EF%BC%88cosA%2CsinA%EF%BC%89m%2An%3D-1.%E6%B1%82%E2%88%A0A.%E8%8B%A51%2Bsin2B%2Fsin%5E2+B-cos%5E2+B)
已知ABC为△ABC的三个内角,向量m=(1,-根号3) ,n=(cosA,sinA)m*n=-1.求∠A.若1+sin2B/sin^2 B-cos^2 B
已知ABC为△ABC的三个内角,向量m=(1,-根号3) ,n=(cosA,sinA)m*n=-1.求∠A.若1+sin2B/sin^2 B-cos^2 B
已知ABC为△ABC的三个内角,向量m=(1,-根号3) ,n=(cosA,sinA)m*n=-1.求∠A.若1+sin2B/sin^2 B-cos^2 B
m*n=-cosA+根号3sinA=2sin(A-π/6)=1
sin(A-π/6)=1/2
A=π/3,或者 π
故A=π/3
tanA=根号3
(1+sin2B)/(cos^2B-sin^2B)=-3
1+sin2B=3sin^2B-3cos^2B
sinBcosB=sin^2B-2cos^2B
(2cosB-sinB)(cosB+sinB)=0
2cosB-sinB=0
tanB=2
tanC=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=-(根号3+2)/(1-2根号3)
=(2+根号3)/(2根号3-1)
依题意有
1*cosA+ -√3sinA=-1
cosA-√3sinA+1=0
2(1/2cosA-√3/2sinA)+1=0
2cos(A+π/3)+1=0
cos(A+π/3)=-1/2
三角形内角不大于180°不小于0°所以A+π/3=120°
A=120°-60°=60°
第二问的问题是什么?
根据题意,m*n=cosA-根号3*sinA=-1
-1/2cosA-(根号3)/2*sinA=-1/2
cos(A+π/3)=-1/2
则A+π/3=2π/3
∠A=π/3