题目内容
已知角A,B,C为△ABC的三个内角,其对边分别为a,b,c,若m=(-cos
,sin
),n=(cos
,sin
),a=2
,且m•n=
.
(1)求角A的值.
(2)求b+c的取值范围.
| A |
| 2 |
| A |
| 2 |
| A |
| 2 |
| A |
| 2 |
| 3 |
| 1 |
| 2 |
(1)求角A的值.
(2)求b+c的取值范围.
(1)m=(-cos
,sin
),
n=(cos
,sin
),且m•n=
.
∴-cos2
+sin2
=
,即-cosA=
,
又A∈(0,π),∴A=
;
(2)由正弦定理得:
=
=
=
=4,
又B+C=π-A=
,
∴b+c=4sinB+4sinC=4sinB+4sin(
-B)=4sin(B+
)(8分)
∵0<B<
,则
<B+
<
.
则
<sin(B+
)≤1,即b+c的取值范围是(2
,4].(10分)
| A |
| 2 |
| A |
| 2 |
n=(cos
| A |
| 2 |
| A |
| 2 |
| 1 |
| 2 |
∴-cos2
| A |
| 2 |
| A |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
又A∈(0,π),∴A=
| 2π |
| 3 |
(2)由正弦定理得:
| b |
| sinB |
| c |
| sinC |
| a |
| sinA |
2
| ||
sin
|
又B+C=π-A=
| π |
| 3 |
∴b+c=4sinB+4sinC=4sinB+4sin(
| π |
| 3 |
| π |
| 3 |
∵0<B<
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| 2π |
| 3 |
则
| ||
| 2 |
| π |
| 3 |
| 3 |
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