Question

在三角形ABC中,B=60度,cosA=4/5,B=根号3.求sinc的值,求三角形abc的面积.

Answer 1

1.sinA=√(1-cos²A)=3/5

sinC=sin(180°-A-B)

=sinAcosB+sinBcosA

=(3+4√3)/10

2.正弦定理a/sinA=b/sinB

a=bsinA/sinB=6/5

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

=9(1+√3)/50