Question

在△ABC中,角A,B,C所对的边分别为a,b,c,若3bcosA=ccosA+acosC,则tanA的值是

在△ABC中,角A,B,C所对的边分别为a,b,c,若3bcosA=ccosA+acosC,则tanA的值是

Answer 1

3bcosA=ccosA+acosC
∵a/sinA=b/sinB=c/sinC
∴原式可化为：3sinBcosA=sinCcosA+sinAcosC = sin(A+C) = sin(π-B) = sinB
∵sinB≠0
∴两边同除以sinB得：3cosA=1
cosA=1/3＞0，A为锐角
tanA = √(1-cos²A)/cosA = √(1-1/9)/(1/3) = 2√2

Answer 2

解：首先ccosA+acosC=b

3bcosA=b
cosA=1/3
sinA=2√2/3
tanA=sinA/cosA=2√2/3/(1/3)=2√2

推广acosB+bcosA=c
bcosC+ccosB=a