判断三角形形状,求详细过程 

由正弦定理得
a=2RsinA,b=2RsinB,c=2RsinC(R为△ABC外接圆半径)
∴acosA+bcosB=ccosC
→2RsinAcosA+2RsinBcosB=2RsinCcosC
→sin2A+sin2B=sin2C
→2sin(A+B)cos(A-B)=2sinCcosC
→2sin(π-C)cos(A-B)=2sinCcosC
→cos(A-B)=cosC
→A-B=C
→A=B+C
→2A=A+B+C=π
→A=π/2.
即△ABC是以A为直角的直角三角形
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