三角ABC形中,内A,B,C角所对的边分别为a,b,c,己知cosB=13分子5sinC=5分孑3.求sinA的值

cosB=5/13 SinB=√1-cos^2 B=√1-25/169=12/13
SinC=3/5
CosC=±√1-Sin^2 C=±4/5
当CosC=4/5时
SinA=Sin(180-(B+C)=Sin(B+C)
=SinBCosC+SinCCosB
=12/13*4/5+3/5*5/13
=48/65+15/65=63/65
当CosC=-4/5时
SinA=Sin(180-(B+C)=Sin(B+C)
=SinBCosC+SinCCosB
=12/13*(-4/5)+3/5*5*13
=(-48+15)/65=-33/65
因为 0<A<180
所以SinA>0
所以SinA=63/65

sinB=(1-(5/13)^2)^(1/2)=12/13
cosC=+ -(1-(3/5)^2)^(1/2)=+ -4/5
sinA=sin(pi-(B+C))=sin(B+C)=sinBcosC+cosBsinC
=(12/13)(4/5)+(5/13)(3/5)=63/65

或sinA=(12/13)(-4/5)+(5/13)(3/5)=-33/65 (舍弃)

所以:sinA=63/65

因为：cosB=5/13
所以：由sinB的平方+cosB的平方=1
可得到：sinB=√1-25/169=12/13
同样求得cosC=±4/5
当cosC=4/5时
sinA=Sin(180-(B+C)）
=sin(B+C)
=sinBcosC+sinCcosB
=12/13*4/5+3/5*5/13
=48/65+15/65=63/65
当cosC=-4/5时
sinA=sin(180-(B+C)）
=sin(B+C)
=sinBcosC+sinCcosB
=12/13*(-4/5)+3/5*5*13
=(-48+15)/65=-33/65