Question

正弦余弦定理题目求解，要过程，谢谢！！

Answer 1

1、
2abcosC(2sinA-sinB)=2accosBsinB
bcosC(2sinA-sinB)=csinBcosB
sinBcosC(2sinA-sinB)=sinCsinBcosB
cosC(2sinA-sinB)=sinCcosB
2sinAcosC-sinBcosC=sinCcosB
2sinAcosC=sinBcosC+sinCcosB
2sinAcosC=sin(B+C)
2sinAcosC=sinA
2cosC=1
cosC=0.5
0<C<π
C=π/3
2、 设角ADC＝θ　则 设角BDC＝π-θ
AC^2=CD^2+AD^2-2CD*ADcosθ　=6-4*2^2^0.5cosθ
BC^2=CD^2+BD^2-2CD*BDcos(π-θ) =6-4*2^2^0.5cosθ(π-θ) =6+4*2^2^0.5cosθ
AC^2+BC^2=12
AB^2=AC^2+BC^2-2AC*BCcosπ/3
8=12-AC*BC
AC*BC=4
S=0.5*AC*BC*sin角ACB =0.5*4*cosπ/3=根号3