>3

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>3
>3

>3
1、由正弦定理及b=ccosA、c=2acosB得
sinB=sinCcosA
sinC=2sinAcosB
于是
sinC=sin(π-C)
=sin(A+B)
=sinAcosB+cosAsinB
=2sinAcosB

sinAcosB-cosAsinB=0
sin(A-B)=0·········①
sinB=sin(π-B)
=sin(A+C)
=sinAcosC+cosAsinC
=sinCcosA

sinAcosC=0·······②
又0<A、B、C<π,由②得
C=π/2
且0<A、B<π/2,由①得
A-B=0
A=B
综上,△ABC为等腰直角△.

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