Question

已知三角形ABC中，a，b，c分别是角A，B，C的对边，且(2a+c)cosB+bcosC=0

①求角B
②若b=根号13，a+c=4，求△ABC的面积

Answer 1

1.(2a+c)cosB+bcosC=0
2acosb+ccosb+bcosc=0 （ccosb+bcosc=a画个图做个高一看就知，相应3个结论）
b=120
2. b*b=a*a+c*c-2accos120
即13=a*a+c*c+ac=（a+c）方-ac ac=3
s=1/2*acsin120=3根3/4

Answer 2

解：①(2a+c)cosB+bcosC=0
2a*cosB+c*cosB+b*cosC=0
2a*cosB + a = 0
cosB = -1/2
B=120°
② b^2 = a^2+c^2-2ac*cosB
13 = a^2 + c^2 - 2ac * (-1/2)
a^2+c^2+ac = 13
(a+c)^2-ac =13
16-ac=13
ac = 3
S = 1/2ac*sinB = 1/2 ×3×√3/2 =3√3/4

Answer 3

①利用正弦定理可把(2a+c)cosB+bcosC=0转化为
（2sinA+sinC)cosB+sinBcosC=0
得2sinAcosB+sinCcosB+sinBcosC=0
得2sinAcosB+sin(C+B)=0
得2sinAcosB+sinA=0
得2cosB+1=0
得cosB=-1/2,得B=120
② b^2 = a^2+c^2-2ac*cosB
13 = a^2 + c^2 - 2ac * (-1/2)
a^2+c^2+ac = 13
(a+c)^2-ac =13
16-ac=13
ac = 3
S = 1/2ac*sinB = 1/2 ×3×√3/2 =3√3/4