Question

△ABC中,A=,BC=3,则△ABC的周长为(    )

A.4sin(B+)+3                  B.4sin(B+)+3

C.6sin(B+)+3                        D.6sin(B+)+3

Answer 1

解析:A=,BC=3,设周长为x,

   由正弦定理知===2R,

    由合分比定理知

    =,

    即=.

    ∴2{+sinB+sin［π-(A+B)］}=x,即x=3+2［sinB+sin(B+)］

    =3+2(sinB+sinBcos+cosBsin)

    =3+2(sinB+sinB+cosB)

    =3+2(sinB+cosB)

    =3+6(sinB+cosB)

    =3+6sin(B+).

答案:D