Question

在△ABC中,角A,BC的对边长分别为abc,B=π/3,CosA=4/5,b=根号3,求sinC的值,求△ABC的面积

Answer 1

cosA=4/5 sinA=√[1-(4/5)²]=3/5

sinC=sin(π-A-B)=sin(A+B)

=sinAcosB+cosAsinB

=(3/5)cos(π/3)+(4/5)sin(π/3)

=(3/5)(1/2)+(4/5)(√3/2)

=(3+4√3)/10

由正弦定理,a/sinA=b/sinB

a=bsinA/sinB=(3/5)√3/(√3/2)=6/5

S△ABC=(1/2)absinC=(1/2)(6/5)(√3)(3+4√3)/10=9(4+√3)/50