Question

在△ABC中,内角A,B,C的对边分别为a,b,c,已知sinB/(sinA+sinC)=1-si

在△ABC中,内角A,B,C的对边分别为a,b,c,已知sinB/(sinA+sinC)=1-sinC/(sinA
＋sinB)，且b＝5,acosC＝-1.
【1】求角A
【2】求△ABC的面积

Answer 1

sinB/(sinA+sinC)=1-sinC/(sinA＋sinB)，
b/(a+c)=1-[c/(a+b)]
整理得
b^2+c^2-a^2=bc
cosA=(b^2+c^2-a^2)/2bc=bc/2bc=1/2
cosA=1/2
A=π/3
2)acosC＝-1.

a^2[1-(sinC)^2]=1
(sinC)^2]=(a^2-1)/a^2
a/sinA=c/sinC
a^2/(sinA)^2=c^2/(sinC0^2
4a^2/3=c^2a^2/(a^2-1)
3以^2=4a^2-4...................(1)
c^2=b^2+a^2-2abcosC=25+a^2-2*5*(acosC)=a^2+25+10

c^2=a^2+35......................(1)
解(1)(2)得
a^2=109
c^2=a^2+35=109+35=144
c=12
S=1/2bcsinA=1/2*12*5*√3/2=15√3
S=15√3