1)sin?A=sin(π/3+B)sin(π/3-B)+sin?B
=(sinπ/3cosB+cosπ/3sinB)(sinπ/3cosB-cosπ/3sinB)+sin?B
=(sinπ/3cosB)?-(cosπ/3sinB)?+sin?B
=3cos?B/4-sin?B/4+sin?B
=(3cos?B+3sin?B)/4
=3/4
A為銳角
故sinA=√3/2
A=π/3
(2)AB*AC=bc=12，a=2√7
根據(jù)余弦定理有：
cosA=(b^+c^ -a^)/2bc
b^+c^ -28=2*12*1/2
b^+c^ =40(1)
即(b+c)^-2bc=40
(b+c)^=64
b+c=8(2)
b<c,聯(lián)立（1）（2）解得：
b=2,c=6