m |
n |
m |
n |
π |
4 |
B |
2 |
即2sinB•[1-cos2(
π |
4 |
B |
2 |
即2sinB+2sin2B-2+1-2sinB2=0,
解得sinB=
1 |
2 |
由于0<B<π,所以B=
π |
6 |
5π |
6 |
(2)由a>b,得到A>B,即B=
π |
6 |
由余弦定理得:b2=a2+c2-2accosB,
代入得:1=3+c2-2
3 |
| ||
2 |
3 |
| ||
2 |
即c2+3c+2=0(无解)或c2-3c+2=0,
解得c=1或c=2.(12分)
m |
n |
π |
4 |
B |
2 |
m |
. |
n |
3 |
m |
n |
m |
n |
π |
4 |
B |
2 |
π |
4 |
B |
2 |
1 |
2 |
π |
6 |
5π |
6 |
π |
6 |
3 |
| ||
2 |
3 |
| ||
2 |