| AB |
| AC |
| AB |
| AB |
| BC |
| AB |
| AB |
| BC |
=
| AB |
∴|
| AB |
| AB |
（2）由（1）知2bcosA=1，2acosB=3
3bcosA=acosB
∴由正弦定理：3sinBcosA=sinAcosB…（8分）
| sin(A-B) |
| 3sinC |
| sinAcosB-sinBcosA |
| 3(sinAcosB+sinBcosA) |
| 1 |
| 6 |
| AB |
| AC |
| 1 |
| 3 |
| AB |
| BC |
| sin(A−B) |
| 3sinC |
| AB |
| AC |
| AB |
| AB |
| BC |
| AB |
| AB |
| BC |
| AB |
| AB |
| AB |
| sin(A-B) |
| 3sinC |
| sinAcosB-sinBcosA |
| 3(sinAcosB+sinBcosA) |
| 1 |
| 6 |