Question

高中数学 求详细解答！！！

Answer 1

2bsin(C+π/6)=a+c
2sinBsin(C+π/6)=sinA+sinC
2sinB(√3/2sinC+1/2cosC)=sinA+sinC
√3sinBsinC+sinBcosC=sin(B+C)+sinC
√3sinBsinC+sinBcosC=sinBcosC+sinCcosB+sinC
√3sinBsinC=sinCcosB+sinC
√3sinB=cosB+1
√3/2sinB-1/2cosB=1/2
sin(B-π/6)=1/2
B-π/6=π/6 或B-π/6=5π/6
B=π/3 或 B=π(舍去)
B=π/3
2)AM=AC ,B=π/3,作AD垂直于BC于D
所以,AD=√3/2*AB=√3/2*c ,
AB=2BD
BD=1/2*c
BD=3/4BC=3/4*a=1/2*c
3a=2c
cosC=CD/AC=1/4*a/b=a/4b
c^2=a^2+b^2-2abcosC
c^2=a^2+b^2-2ab*a/4b
c^2=a^2+b^2-1/2*a^2
c^2=1/2a^2+b^2
c^2=1/2*4/9c^2+b^2
b^2=7/9*c^2
b=√7/3*c
cosA=(b^2+c^2-a^2)/2bc=(7/9*c^2+c^2-4/9*c^2)/2√7/3*c^2=2√7/7
A=arccos2√7/7