A+B |
2 |
7 |
2 |
得4×
1-cos(A+B) |
2 |
7 |
2 |
即2+cosC-cos2C=
7 |
2 |
即4cos2C-4cosC+1=0,
解得cosC=
1 |
2 |
π |
3 |
(2)由C=
π |
3 |
得49=a2+b2-2abcos
π |
3 |
即ab=40,解得a=5,b=8或a=8或b=5,
则三角形的面积S=
1 |
2 |
1 |
2 |
| ||
2 |
3 |
则△ABC的内切圆的半径r=
2S |
a+b+c |
2×10
| ||
5+8+7 |
3 |
则△ABC的内切圆面积S=π•r2=3π
A+B |
2 |
7 |
2 |
A+B |
2 |
7 |
2 |
1-cos(A+B) |
2 |
7 |
2 |
7 |
2 |
1 |
2 |
π |
3 |
π |
3 |
π |
3 |
1 |
2 |
1 |
2 |
| ||
2 |
3 |
2S |
a+b+c |
2×10
| ||
5+8+7 |
3 |