Question

在三角形中,角A.B.C的对边分别为a.b.c,B=三分之派,cosA=五分之四,b=根号三. (1)求a的值 (2)求三角形ABC的

Answer 1

(1)
cosA=五分之四
sinA=√(1-cos^2A)=3/5
a/sinA=b/sinB
a=sinA*b/sinB=3/5*√3/√3/2=10/3

(2)
sinC=sin(180-(A+B)=sin(A+B)=sinAcosB+cosAsinB=3/5*1/2+4/5*√3/2=(4√3+3)/10
S△ABC=1/2*a*bsinA
=1/2*10/3*√3*(4√3+3)/10
=(4+√3)/2

Answer 2

1)根据正弦定理
a/sinA=b/sinB
因为cosA=4/5, B=π/3
所以sinA=3/5，sinB=根号3/2
代入数值为
a=bsinA/sinB=根号3*3/5/根号3/2=6/5

Answer 3

sinA=根号(1-cos^2A)=根号(1-16/25)=3/5
a/sinA=b/sinB
a=bsinA/sinB=根号3*3/5 / sinπ/3 = 根号3*3/5 / (根号3/2) = 6/5

sinC=sin(180-A-B)=sin(A+B)=sinAcosB+cosAsinB=3/5cosπ/3+4/5sinπ/3=3/5*1/2+4/5*根号3/2=(3+4根号3)/10
S△ABC=1/2absinC=1/2*6/5*根号3*(3+4根号3)/10=(9根号3+36)/50