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钝角三角形ABC的外接圆半径为2,最长的边BC=23,求sinB+sinC的取值范围.

题目详情
BC=2
3
,求sinB+sinC的取值范围.
3
3
3
▼优质解答
答案和解析
∵钝角三角形ABC的外接圆半径为2,最长的边BC=2
3
,由正弦定理可得
BC
sinA
=2R=4,解得sinA=
3
2
,故A=
3

由于 sinB+sinC=sinB+sin(
π
3
-B)=
1
2
sinB+
3
2
cosB=sin(B+
π
3
 ),
 
π
3
<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
BC=2
3
3
33,由正弦定理可得
BC
sinA
=2R=4,解得sinA=
3
2
,故A=
3

由于 sinB+sinC=sinB+sin(
π
3
-B)=
1
2
sinB+
3
2
cosB=sin(B+
π
3
 ),
 
π
3
<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
BC
sinA
BCBCBCsinAsinAsinA=2R=4,解得sinA=
3
2
,故A=
3

由于 sinB+sinC=sinB+sin(
π
3
-B)=
1
2
sinB+
3
2
cosB=sin(B+
π
3
 ),
 
π
3
<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
3
2
3
3
3
3
33222,故A=
3

由于 sinB+sinC=sinB+sin(
π
3
-B)=
1
2
sinB+
3
2
cosB=sin(B+
π
3
 ),
 
π
3
<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
3
2π2π2π333.
由于 sinB+sinC=sinB+sin(
π
3
-B)=
1
2
sinB+
3
2
cosB=sin(B+
π
3
 ),
 
π
3
<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
π
3
πππ333-B)=
1
2
sinB+
3
2
cosB=sin(B+
π
3
 ),
 
π
3
<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
1
2
111222sinB+
3
2
cosB=sin(B+
π
3
 ),
 
π
3
<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
3
2
3
3
3
3
33222cosB=sin(B+
π
3
 ),
 
π
3
<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
π
3
πππ333 ),
 
π
3
<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
π
3
πππ333<B+
π
3
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
π
3
πππ333<
3
,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
3
2π2π2π333,∴
3
2
<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
3
2
3
3
3
3
33222<sin(B+
π
3
)≤1,故sinB+sinC的取值范围是 (
3
2
,1].
π
3
πππ333 )≤1,故sinB+sinC的取值范围是 (
3
2
,1].
3
2
3
3
3
3
33222,1].