(Ⅰ)若f(1)=0且B=C+,试求角A、B、C的大小;
(Ⅱ)若f(2)=0,求角C的取值范围.
(Ⅰ)若f(1)=0且B=C+,试求角A、B、C的大小;
(Ⅱ)若f(2)=0,求角C的取值范围.
解:(Ⅰ)由f(1)=0得b2-4c2=0,即b=2c
由正弦定理得:sinB=2sinC
又B=C+
∴sin(C+)=2sinC
即sinCcos+cosCsin
=2sinC
∴3sinC=cosC
∴tanC=∴C=
,B=C+
=
,A=
.
(Ⅱ)由f(2)=0得a2+b2=2c2由余弦定理得:
cosC==
≥
∴
≤cosC<1,
故0<C≤.