∵在△ABC中,acosB=bcosA,
∴
a
b
=
cosA
cosB
,又由正弦定理可得
a
b
=
sinA
sinB
,
∴
cosA
cosB
=
sinA
sinB
,sinAcosB-cosAsinB=0,sin(A-B)=0.
由-π<A-B<π 得,A-B=0,
则△ABC为等腰三角形,
http://www.***.com/html/qDetail/02/g0/201401/avtgg002557253.html
∵在△ABC中,acosB=bcosA,
∴
a
b
=
cosA
cosB
,又由正弦定理可得
a
b
=
sinA
sinB
,
∴
cosA
cosB
=
sinA
sinB
,sinAcosB-cosAsinB=0,sin(A-B)=0.
由-π<A-B<π 得,A-B=0,
则△ABC为等腰三角形,
http://www.***.com/html/qDetail/02/g0/201401/avtgg002557253.html