| tanA |
| tanB |
| 2c |
| b |
| sinAcosB |
| cosAsinB |
| 2c |
| b |
由正弦定理可得，1+
| sinAcosB |
| cosAsinB |
| 2sinC |
| sinB |
| sinAcosB+sinBcosA |
| sinBcosA |
| 2sinC |
| sinB |
∴sin（A+B）=2sinCcosA，cosA＝
| 1 |
| 2 |
∵0＜A＜π∴A＝
| π |
| 3 |
故答案为：
| π |
| 3 |
| tanA |
| tanB |
| 2c |
| b |
| tanA |
| tanB |
| 2c |
| b |
| sinAcosB |
| cosAsinB |
| 2c |
| b |
| sinAcosB |
| cosAsinB |
| 2sinC |
| sinB |
| sinAcosB+sinBcosA |
| sinBcosA |
| 2sinC |
| sinB |
| 1 |
| 2 |
| π |
| 3 |
| π |
| 3 |