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　

(1)求sinC的值　(2)求三角形ABC的面积各位帮忙下　四点交卷　要过程

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

Answer 2

sinA=根号（1-cos²B)=3/5,sinB=sinπ/2=根号3/2，cosB=1/2
sinc=sin（180°-A-B)=sin(A+B)=sinA*cosB+sinB*cosA=3/5*1/2+根号3/2*4/5=（3+4根号3）/10
a/sinA=b/sinB, a=6/5
S△ABC=absinC/2=6/5*根号3sinC/2=（36+9根号3）/50