已知等边三角形ABC中,0
问题描述：
已知等边三角形ABC中,0
答
先吐槽...不可能是等边三角形吧- -
sinA=√2/10
cosA=7√2/10
tanA=1/7
tan(A-B)=(tanA-tanB)/(1+tanAtanB)=-2/11
(1/7-tanB)/[1+(1/7)*tanB]=-2/11
tanB=1/3
tanB=sinB/cosB=1/3
3sinB=cosB
√(1-sin^2B)=3sinB
1-sin^2B=9sin^2B
sin^2B=1/10
sinB=√10/10
cosB=3√10/10
cosC=cos[180°-(A+B)]=-cos(A+B)
=-(cosAcosB-sinAsinB)
=-(7√2/10*3√10/10-√2/10*√10/10)
=-2√5/5
tan2B=2tanB/(1-tan^2B)=3/4
tan(A+2B)=(tanA+tan2B)/(1-tanAtan2B)
=(1/7+3/4)/(1-1/7*3/4)=1
∵cosC＜0,sinA=√2/10＜1/2,A-B＜0
∴A+2B=45°