（1）因BC對應于∠A,AB對應于∠C.
應用正弦定理得：
BC/sinA=AB/sinC
AB=BCsinC/sinA=BC2sinA/sinA=2BC
故,AB=2√5.
(2) sin(2A-π/4)=sin2Acos(π/4)-cos2Asin(π/4)
=[(√2)/2](sin2A-cos2A)
利用余弦定理求角A:
cosA=(AB²+AC²-BC²)/2AB*AC
=[(2√5)²+3²-(√5)²]/2×(2√5)×3
=(20+9-5)/12(√5）
故,cosA=(2√5)/5
sinA=√[1-cos²A]=(√5)/5
sin(2A-π/4)=[(√2)/2][2sinAcosA-(2cos²A-1)]
=[(√2)/2]{2×(√5/5)×(2√5/5)-[2×(2√5/5)²-1]}
整理后得：
sin(2A-π/4)=(√2)/10