题目内容
在△ABC中,若
=
=
,则△ABC的形状是( )
| a | ||
cos
|
| b | ||
cos
|
| c | ||
cos
|
分析:由已知结合正弦定理以及二倍角公式可得sin
=sin
=sin
,再结合角的范围和正弦函数的单调性可得A=B=C=
,即得答案.
| A |
| 2 |
| B |
| 2 |
| C |
| 2 |
| π |
| 3 |
解答:解:由
=
=
结合正弦定理可得
=
=
,即
=
=
,
化简得sin
=sin
=sin
,又
,
,
∈(0,
)此时正弦函数单调递增,
故
=
=
,又A+B+C=π,故A=B=C=
,即△ABC为等边三角形
故选B.
| a | ||
cos
|
| b | ||
cos
|
| c | ||
cos
|
| sinA | ||
cos
|
| sinB | ||
cos
|
| sinC | ||
cos
|
2sin
| ||||
cos
|
2sin
| ||||
cos
|
2sin
| ||||
cos
|
化简得sin
| A |
| 2 |
| B |
| 2 |
| C |
| 2 |
| A |
| 2 |
| B |
| 2 |
| C |
| 2 |
| π |
| 2 |
故
| A |
| 2 |
| B |
| 2 |
| C |
| 2 |
| π |
| 3 |
故选B.
点评:本题考查三角形形状的判断,涉及正弦定理和正弦函数的单调性以及二倍角公式,属中档题.
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