题目内容
△ABC的内角A,B,C的对边分别为a,b,c,已知 b=2,B=
,C=
,则△ABC的面积为( )
| π |
| 6 |
| π |
| 4 |
A.2
| B.
| C.2
| D.
|
∵b=2,B=
,C=
,
∴由正弦定理
=
得:c=
=
=2
,A=
,
∴sinA=sin(
+
)=cos
=
,
则S△ABC=
bcsinA=
×2×2
×
=
+1.
故选B
| π |
| 6 |
| π |
| 4 |
∴由正弦定理
| b |
| sinB |
| c |
| sinC |
| bsinC |
| sinB |
2×
| ||||
|
| 2 |
| 7π |
| 12 |
∴sinA=sin(
| π |
| 2 |
| π |
| 12 |
| π |
| 12 |
| ||||
| 4 |
则S△ABC=
| 1 |
| 2 |
| 1 |
| 2 |
| 2 |
| ||||
| 4 |
| 3 |
故选B
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