三角函数题目已知锐角三角形ABC中,cos(A+B)=4/5；sin(A-B)=1/5.c=3,求:三角形三边的长

cos(A+B)=4/5>0,sin(A+B)=3/5
sinAcosB+cosAsinB=3/5
sinAcosB-cosAsinB=1/5
sinAcosB=2/5
cosAsinB=1/5
tanA*cotB=2
tanA=2tanB
设c的高为h=CD
h/AD=tanA
h/BD=tanB
h/tanA+h/tanB=3
3h/tanA=3
h=tanA
2/1=AD/BD,AD+BD=3
AD=2BD
BD=1
AD=2
sin(A+B)=sinC=3/5
设AE⊥BC于E,AE/AC=3/5
设BF⊥AC于F,BF/BC=3/5
5AE=3AC,5BF=3BC
据面积相等,BC*AE=3BC*AC/5=3*h
BC*AC=5h
又,1+h^2=AC^2,4+h^2=BC^2
(1+h^2)(4+h^2)=BC^2*AC^2=25h^2
4+5h^2+h^4-25h^2=0
h^4-20h^2+4=0
h^2=10±√[(400-16)/4]
=10±4√6
h1=√6+2（A+B为锐角,舍去）
h2=√6-2
AC=√(11-4√6)
BC=√(14-4√6)
AB=3

cos(A+B)=4/5>0,sin(A+B)=3/5
sinAcosB+cosAsinB=3/5