设锐角三角形ABC的内角A，B，C的对边分别为a，b，c，a=2bsinA （Ⅰ_爱下问搜题

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问题解答题

设锐角三角形ABC的内角A，B，C的对边分别为a，b，c，a=2bsinA

（Ⅰ）求B的大小；

（Ⅱ）求cosA+sinC的取值范围．

答案

（Ⅰ）由a=2bsinA，根据正弦定理得sinA=2sinBsinA，所以sinB=

1

2

，

由△ABC为锐角三角形得B=

π

6

．

（Ⅱ）cosA+sinC=cosA+sin(π-

π

6

-A)=cosA+sin(

π

6

+A)=cosA+

1

2

cosA+

3

2

sinA=

3

sin(A+

π

3

)．

由△ABC为锐角三角形知，

π

3

＜A＜

π

2

．

2π

3

＜A+

π

3

＜

5π

6

，

所以

1

2

＜sin(A+

π

3

)＜

3

2

．

由此有

3

2

＜

3

sin(A+

π

3

)＜

3

2

×

3

=

3

2

，

所以，cosA+sinC的取值范围为(

3

2

，

3

2

)．

完形填空

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