题目内容
在△ABC中,a,b,c分别是∠A,∠B,∠C的对边,∠A=60°,b=2,△ABC面积为
,则
=______.
| 3 |
| 2b+3c+4a |
| 4sin A+2sinB+3sinC |
∵∠A=60°,b=2,△ABC面积为
,
∴S=
bcsinA=
×2c×
=
,
解得:c=2,
∴a2=b2+c2-2bccosA=4+4-2×2×2×
=4,
解得:a=2,
∴由正弦定理得:
=
=
=
=
,
∴
=
=
=
=
=
.
故答案为:
| 3 |
∴S=
| 1 |
| 2 |
| 1 |
| 2 |
| ||
| 2 |
| 3 |
解得:c=2,
∴a2=b2+c2-2bccosA=4+4-2×2×2×
| 1 |
| 2 |
解得:a=2,
∴由正弦定理得:
| a |
| sinA |
| b |
| sinB |
| c |
| sinC |
| 2 | ||||
|
4
| ||
| 3 |
∴
| 4a |
| 4sinA |
| 2b |
| 2sinB |
| 3c |
| 3sinC |
| 2b+3c+4a |
| 4sin A+2sinB+3sinC |
| a |
| sinA |
4
| ||
| 3 |
故答案为:
4
| ||
| 3 |
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