(1)∵2cos(B+C)=1
∴cos(B+C)=1/2
又A+B+C=π
∴cosA=-cos(B+C)=-1/2
∴A=2π/3
(2)由余弦定理,有
cosA=(b²+c²-a²)/2bc
则 a²=b²+c²-2bc*cosA
=(b+c)²-2bc-2bc*cosA
=(3√3)²-2×4-2×4×(-1/2)
=23
故 a=√23
所以,边BC的长度为√23
(3)△ABC的面积
S=(1/2)*bc*sinA
=(1/2)×4×sin(2π/3)
=√3