题目内容
在△ABC中,A:B:C=3:1:2,则a:b:c=( )
| A.1:2:3 | B.3:2:1 | C.1:
| D.2:1:
|
在△ABC中,A:B:C=3:1:2,
设A=3k,B=k,C=2k,
可得A+B+C=3k+k+2k=π,即k=
,
∴A=
,B=
,C=
,
∴由正弦定理
=
=
,得:
=
=
,
则a:b:c=2:1:
.
故选D
设A=3k,B=k,C=2k,
可得A+B+C=3k+k+2k=π,即k=
| π |
| 6 |
∴A=
| π |
| 2 |
| π |
| 6 |
| π |
| 3 |
∴由正弦定理
| a |
| sinA |
| b |
| sinB |
| c |
| sinC |
| a |
| 1 |
| b | ||
|
| c | ||||
|
则a:b:c=2:1:
| 3 |
故选D
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