Question

在△ABC中，角A.B.C的对边分别为a，b，c，B=3分之π，cosA=5分之4，b=根号3 1求sinC的值 2求△ABC的面积

Answer 1

1. 已知cosA=5分之4
则sinA=√(1-cos²A)=3/5
sinC=sin(180°-A-B)=sin(A+B)=sinAcosB+sinBcosA
=(3/5)*cos(π/3)+sin(π/3)*(4/5)
=(3/5)*(1/2)+(4/5)*(√3/2)
=(3+4√3)/10
2. 由正弦定理a/sinA=b/sinB
a=bsinA/sinB=√3*(3/5)/(√3/2)=6/5
△ABC的面积=(1/2)absinC=(1/2)*(6/5)*(√3)*(3+4√3)/10
=9(1+√3)/50