由正弦定理,有:
(2sinA+sinC)cosB+sinBcosC=0
移项
2sinAcosB=-(sinCcosB+sinBcosC)
即2sinAcosB=-sin(B+C)
2sinAcosB=-sin(π-A)=-sinA
故cosB=-1/2
B=2π/3
cosB=(a²+c²-b²)/2ac=-1/2
得a²+c²-13=-ac
(a+c)²-2ac-13=-ac
由a+c=4解得:ac=3
S△ABC=(1/2)acsinB=(3√3)/4
由正弦定理,有:
(2sinA+sinC)cosB+sinBcosC=0
移项
2sinAcosB=-(sinCcosB+sinBcosC)
即2sinAcosB=-sin(B+C)
2sinAcosB=-sin(π-A)=-sinA
故cosB=-1/2
B=2π/3
cosB=(a²+c²-b²)/2ac=-1/2
得a²+c²-13=-ac
(a+c)²-2ac-13=-ac
由a+c=4解得:ac=3
S△ABC=(1/2)acsinB=(3√3)/4