Question

在三角形ABC中,角A,B,C所对的边分别为a,b,c,且1+tanA/tanb=2c/b,求∠A

Answer 1

由1+tanA/tanB=2c/b得,
tanB+tanA=2tanB*c/b,由正弦定理得c/b=sinC/sinB,故得
tanB+tanA=2tanB*sinC/sinB=2sinC/cosB
即tanB+tanA=2sinC/cosB
sinB*cosA+sinA*cosB=2sinC*cosA
sin(A+B)=2sinC*cosA,
sinC=sin(180°-A-B)=sin(A+B)=2sinC*cosA,sinC=2sinC*cosA,
由sinC不等于零,故得cosA=1/2,A=30°,

Answer 2

最后一步错了吧，cosA=1/2,A应该等于60度.
其他的都是对的

Answer 3

|m+n|取最小值√2.
由1+tanA/tanB=2c/b得,
tanB+tanA=2tanB*c/b,由正弦定理得c/b=sinC/sinB,故得
tanB+tanA=2tanB*sinC/sinB=2sinC/cosB
即tanB+tanA=2sinC/cosB
sinB*cosA+sinA*cosB=2sinC*cosA
sin(A+B)=2sinC*cosA,
sinC=sin(180-A-B)=sin(A+B)=2sinC*cosA,sinC=2sinC*cosA,
由sinC不等于零,故得cosA=1/2,A=30,
m+n=(cosB,-1+2(cosC/2)^2)=(cosB,-cosC),B+C=150,C=150-B,
|m+n|^2=(cosB)^2+(cosC)^2=(cosB)^2+(cos(150-B))^2
=2+cos2B+cos(300-2B)=2+cos2B+cos300cos2B+sin300sin2B
=2+cos2B+(1/2)cos2B-(√3/2)sin2B=2+(3cos2B-√3sin2B)/2
=2+√3sin(D-2B),其中tanD=3/(2√3),
当D-2B=0时,|m+n|^2取得最小值2,即|m+n|取得最小值√2,当D-2B=90时,|m+n|^2取得最大值2+√3,即|m+n|取得最大值√(2+√3).