Question

设锐角三角形ABC的内角A，B，C的对边分别为a，b，c，且a=2bsinA.求B的大小。 请给出

设锐角三角形ABC的内角A，B，C的对边分别为a，b，c，且a=2bsinA.求B的大小。
请给出非常详细的解题过程，一定采纳！

Answer 1

解:∵a=2bsina,∴a/sina=2b
又sinB=b/(a/sina)=b/2b=1/2,∴B=30°.
cosA+sinC=cos[180°-(B+C)]+sinC=cos(150°-c)+sinc
=cos[90°+(60°-c)]+sinC
=-sin(60°-c)+sinC
=-(sin60°cosc-cos60°sinc)+sinc
=-(√3/2)cosc+(3/2)sinC
=(3/2)[sinC-(√3/3)cosC]
=(3/2)[sinC-tan30°cosc]
=(3/2)(1/cos30°)(sinccos30°-coscsin30°)
=(√3)sin(c-30°)
∵△abc是锐角三角形,∴0°<C<90°,故-30°<C-30°<60°
∴-1/2<sin(C-30°)<√3/2
∴-√3/2<cosA+sinC<3/2.

Answer 2

解：根据正弦定理
a/sinA=b/sinB
sinB=bsinA/a
sinB=bsinA/(2bsinA)=1/2
B=30°

Answer 3

Answer 4

解:
∵a=2bsinA,
∴a/sinA=2b
又a/sinA=b/sinB
sinB=b/(a/sinA)=b/2b=1/2,
∴B=30°.

Answer 5