Question

在三角形ABC中,角A,B,C,所对的边分别为a,b,c,B=派/3,cosA=4/5,b=根号3

Answer 1

cosA = 4/5 ＞ 0,A为锐角

sinA=√(1-cos^2A) = 3/5

B=π/3

sinC=sin(π-A-B)=sin(A+B)=sinAcosB+cosAsinB=3/5*1/2+4/5*√3/2 = (3+4√3)/10

b=√3

a/sinA=b/sinB

a=bsinA/sinB=√3*3/5 /(√3/2) = 6/5

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