Question

求解高二数学

Answer 1

追答

好评哦

Answer 2

cosB/cosC=-b/2a+c=-sinB/(2sinA+sinC)

2sinAcosB+sinCcosB+sinBcosC=0
2sinAcosB+sin(C+B)=0

sinA(2cosB+1)=0
cosB= - 1/2 ==>B=120度
由余弦定理得：
13=a^2+c^2-2accos120=(a+c)^2-ac=16-ac
ac=3
a+c=4
{a=1
{c=3
....................
{a=3
{c=1

Answer 3

a/sinA=b/sinB=c/sinC
cosB/cosC=-b/(2a+c)=-sinB/(2sinA+sinC)
cosB(2sinA+sinC)=-sinBcosC
2sinAcosB+sinCcosB+sinBcosC=0
2sinAcosB+sin(C+B)=0
2sinAcosB+sinA=0
因为sinA<>0, 2cosB+1=0, cosB=-1/2, B=2π/3

cosB=(a²+c²-b²)/(2ac)=-1/2
a²+c²-b²=-ac
(a+c)²-2ac-b²+ac=0
4²-ac-13=0
ac=3
S=1/2*ac*sinB=1/2*3*sin(2π/3)=3√3/4