| c |
| a+b |
| b |
| a+c |
∴a2+ab+ac+bc=c2+ac+b2+ab
∴b2+c2-a2=bc
∴2bccosA=ab
∴cosA=
| 1 |
| 2 |
∵0°＜∠A＜180°
∴∠A=60°
（2）∵
| c |
| b |
2+
| ||
| 4 |
∴令b=4t，c=（2+
| 3 |
cosA=
| b2+c2−a2 |
| 2bc |
16t2+(7+4
| ||
8(2+
|
| 1 |
| 2 |
解得t=1
∴b=4．
| c |
| a+b |
| b |
| a+c |
| c |
| b |
2+
| ||
| 4 |
| 15 |
| c |
| a+b |
| b |
| a+c |
| 1 |
| 2 |
| c |
| b |
2+
| ||
| 4 |
| 3 |
| b2+c2−a2 |
| 2bc |
16t2+(7+4
| ||
8(2+
|
| 1 |
| 2 |