在三角形ABC中,a.b.c分别是角A,B,C的对边,且COSB分之COSC=-b分之 2a+c,1.求角B的大小

2. b=根下13,a+c=4.求三角形ABC的面积
2025-03-14 23:42:44
推荐回答(3个)
回答1:

cosC/cosB=-(2sinA+sinC)/sinB 化简整理得:
sinB×cosC=-cosB×(2sinA+sinC)
sinB ×cosC+cosB× sinC=-2cosB×sinA
sin(B+C)=-2cosB×sinA
sinA=-2cosB×sinA
cosB=-1/2
B=120°
2、根据余弦定理b^2=a^2+c^2-2ac×cosB=(a+c)^2-2ac-2ac×cosB
代入已知条件得:13=16-2ac(1+cosB)=16-ac, ac=3
三角形的面积为:1/2ac×sinB=1/2×3×√3/2=3√3/4

回答2:

1)据正弦定理:(2a+c):(-b)=(2sinA+sinC):(-sinB)
有题目条件知 可化简为 2sinA*cosB+sinA=0 解得cosB=(-0.5) 因为B是三角形的内角 可知
B=120°
2)三角形的面积S=0.5*ac*sinB
有余弦定理:b^2=a^2+c^2+2ac*cosB=13
因为a+c=4
可得ac=3
S=0.5*3*SINB=(三倍的根号三)除以(四)
由于计算机水平有限,看起来比较麻烦

回答3:

(1)120度
(2)4分之3倍根号3

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