Question

在△ABC中,已知三内角A:B:C=3:4:5且三边a+b+c=3根号2+2根号3+根号6,求a,b,c

及面积S

Answer 1

△ABC中,已知三内角A:B:C=3:4:5

因为A+B+C=π
则A=π/4,B=π/3,C=5π/12

根据正弦定理得
a:b:c=sinA:sinB:sinC
=sinπ/4:sinπ/3:sin5π/12
=根号2/2:根号3/2:(根号2+根号6)/4
=2根号2/4:2根号3/4:(根号2+根号6)/4

a+b+c=3根号2+2根号3+根号6

则,a=2根号2
b=2根号3
c=根号2+根号6

S△=1/2×（acsinB）=[2根号2×(根号2+根号6)×根号3/2]÷2=根号3+3