(1)∵A=
,S△ABC=π 3
bcsinA=1 2
bc=
3
4
,
3
4
∴bc=1,
由余弦定理得:
=cosA=1 2
=
b2+c2?a2
2bc
,
b2+c2?1 2
整理得:b2+c2=2,
∴(b+c)2=b2+c2+2bc=4,
∴b+c=2;
(2)由正弦定理知
?sin(a b?c
-C)=π 3
?sin(sinA sinB?sinC
-C)π 3
=
=
sin(
3
2
?C)π 3 sin(
?C)?sinC2π 3