Question

高中数学问题，在线急求需过程，望高手解答，马上给分

在三角形ABC中,角A,B,C分别为a,b,c,acosB+bcosA=2c(cosC)1求内角C2若a=3,c=根号7,求b

Answer 1

1、acosB+bcosA=2c(cosC)
用正弦定理：sinAcosB+sinBcosA=2sinCcosC
sin(A+B)=2sinCcosC
sinC=2sinCcosC
cosC=1/2
C=60°
2、根据余弦定理c^2=a^2+b^2-2abcosC得
7=9+b^2-3b
所以b=1或b=2

Answer 2

a:b:c=sinA:sinB:sinC
sinAcosB+sinBcosA=2sinC(cosC)
sin(A+B)=sin2C
sinC=2sinCcosC
sinC≠0
cosC=1/2
C=π/3
2.
c²=a²+b²-2abcosC
7=9+b²-3b
b=1或2

Answer 3

1、由正弦定理可得
a=k*sinA b=k*sinB c=k*sinc
acosB+bcosA=2c(cosC
sinAcosB+sinBcosA=2sinCcosC
sin(A+B)=2sinCcosC
cosC=1/2
C=60°
2、由余弦定理
c²=a²+b²-2abcosC
9+b²-3b=7
b=1或2