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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
人气:361 ℃ 时间:2020-01-26 12:01:45
解答
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)
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