AB |
AC |
AB |
AB |
BC |
=
AB |
AB |
AB |
BC |
AB |
∴|
AB |
(2)由已知及(1)有:2bcosA=1,2acos(π-B)=-3,
∴acosB=3bcosA(8分)
由正弦定理得:sinAcosB=3sinBcosA(10分)
∴
sin(A-B) |
sinC |
sin(A-B) |
sin(A+B) |
sinAcosB-cosAsinB |
sinAcosB+cosAsinB |
1 |
2 |
AB |
AC |
AB |
BC |
sin(A?B) |
sinC |
AB |
AC |
AB |
AB |
BC |
AB |
AB |
AB |
BC |
AB |
AB |
sin(A-B) |
sinC |
sin(A-B) |
sin(A+B) |
sinAcosB-cosAsinB |
sinAcosB+cosAsinB |
1 |
2 |