m |
n |
3 |
m |
n |
∴2sinB•
3 |
化简得
3 |
π |
6 |
又0<B<π,得-
π |
6 |
π |
6 |
11π |
6 |
∴2B-
π |
6 |
π |
2 |
π |
3 |
(2)∵a,b,c成等差数列,b=2,∴a+c=2b=4.
又∵b2=a2+c2-2ac•cosB,
∴4=a2+c2-2ac•cos
π |
3 |
将a+c=4代入,得a2-4a+4=0,得a=2,
从而c=2,三角形为等边三角形.…(12分)
因此,△ABC的面积S =
1 |
2 |
3 |
m |
n |
3 |
m |
n |
m |
n |
3 |
m |
n |
3 |
3 |
π |
6 |
π |
6 |
π |
6 |
11π |
6 |
π |
6 |
π |
2 |
π |
3 |
π |
3 |
1 |
2 |
3 |