锐角三角形ABC,角A,B,C的对应边abc a=2bsina,1:求B的大小 ,2求COSA+SINC的取值范围!急用

(1)用正弦定理容易得sinB=1/2,B=π/6
（2）C=π-(A+B)=π-(A+π/6)
cosA+sinC=cosA+sin(A+π/6)
=3/2cosA+√3/2sinA
=√3(√3/2cosA+1/2sinA)
=√3 sin(A+π/3)
=√3 sin(A+π/2-π/6)
=√3 cos(A-π/6)
结合 A<π/2,得
1/2< cosA+sinC <= √3