Question

第16题求数学学霸帮忙。

Answer 1

2(cosA/2)^2=√3/3sinA ,
(cosA/2)^2=√3/3sinA/2*cosA/2
cosA/2(cosA/2-√3/3sinA/2)=0
cosA/2-√3/3sinA/2=0
sinA/2=√3cosA/2
tanA/2=√3
A/2=π/3
A=2π/3
B+C=π/3
sin(B-C)=4cosBsinC

sinBcosC-cosBsinC=4cosBsinC
sinBcosC=5cosBsinC
tanB=5tanC..............................(1)
tan(B+C)=(tanB+tanC)/(1-tanBtanC)=√3...................(2)
把(1)代入(2)得
6tanC/[1-(5tanC)^2]=√3.
5√3(tanC)^2+6tanC-√3=0
tanC=(2√6-3)/(5√3),或 tanC=(-2√6-3)/(5√3)<0 (舍去)
tanC=(2√6-3)/(5√3)
又sinBcosC=5cosBsinC
sinB/sinC=5cosB/cosC
sinB/sinC=5cos(π/3-C)/cosC=5[1/2cosC+√3/2sinC)/cosC
=5/2+(5√3/2*tanC)
=5/2+[5√3/2*(2√6-3)/(5√3)]
=5/2-3/2+√3
=1+√3