题目内容：

A=60° 三角形abc面积为二分之根号3 a=根号3 c>b 求角B

优质解答

S = 1/2 * bc sinA
1/2 bc √3/2 = √3/2
bc =2
余弦定理 cosA = (b^2 +c^2 -a^2)/2bc
1/2 = (b^2 +c^2 -3)/4
b^2 +c^2 =5
又c > b 所以c =2 b =1
c^2 =a^2 +b^2 是直角三角形
B =30°