题目内容
△ABC的三内角A,B,C所对边的长分别为a,b,c,
=(a+b,c),
=(b-a,c-b),若
⊥
,则sinB+sinC的取值范围是( )
m |
n |
m |
n |
A.(-
| B.(
| C.[
| D.[
|
∵
=(a+b,c),
=(b-a,c-b),
⊥
,
∴(a+b)(b-a)+c(c-b)=0,
∴a2=b2+c2-bc,
由余弦定理知,a2=b2+c2-2bccosA,
∴cosA=
,而A为△ABC的内角,
∴A=
.
∵△ABC中,A+B+C=π,
∴B+C=π-A=
,
∴sinB+sinC
=sin(
-C)+sinC
=
cosC-(-
)sinC+sinC
=
sinC+
cosC
=
sin(C+
).
∵0<C<
,故
<C+
<
.
∴
<sin(C+
)≤1.
∴
<
sin(C+
)≤
.即
<sinB+sinC≤
.
故选B.
m |
n |
m |
n |
∴(a+b)(b-a)+c(c-b)=0,
∴a2=b2+c2-bc,
由余弦定理知,a2=b2+c2-2bccosA,
∴cosA=
1 |
2 |
∴A=
π |
3 |
∵△ABC中,A+B+C=π,
∴B+C=π-A=
2π |
3 |
∴sinB+sinC
=sin(
2π |
3 |
=
| ||
2 |
1 |
2 |
=
3 |
2 |
| ||
2 |
=
3 |
π |
6 |
∵0<C<
2π |
3 |
π |
6 |
π |
6 |
5π |
6 |
∴
1 |
2 |
π |
6 |
∴
| ||
2 |
3 |
π |
6 |
3 |
| ||
2 |
3 |
故选B.
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