(1)由余弦定理,b²+c²-a²=2bccosA
又b²+c²-a²=bc ∴2cosA=1 cosA=1/2
A=60°
(2)∵a=√3,B=45°,A=60°
∴C=75°
由正弦定理,a/sinA=b/sinB=c/sinC
解得b=√2,c=(√6+√2)/2
所以△ABC的周长=√3+√2+(√6+√2)/2=√3+3√2/2+√6/2
(1)由余弦定理,b²+c²-a²=2bccosA
又b²+c²-a²=bc ∴2cosA=1 cosA=1/2
A=60°
(2)∵a=√3,B=45°,A=60°
∴C=75°
由正弦定理,a/sinA=b/sinB=c/sinC
解得b=√2,c=(√6+√2)/2
所以△ABC的周长=√3+√2+(√6+√2)/2=√3+3√2/2+√6/2