a |
c |
sinA |
sinC |
sin2C |
sinC |
a |
2c |
由余弦定理得cosC=
a2+b2−c2 |
2ab |
(a+c)(a−c)+b2 |
2ab |
∵a+c=2b,
∴cosC=
2b(a−c)+b•
| ||
2ab |
2(a−c)+
| ||
2a |
∴
a |
2c |
2(a−c)+
| ||
2a |
整理得2a2-5ac+3c2=0,解得a=
3 |
2 |
所以a:b=6:5.所以a:b:c=6:5:4
三角形的三边之比为:6:5:4.
a |
c |
sinA |
sinC |
sin2C |
sinC |
a |
2c |
a2+b2−c2 |
2ab |
(a+c)(a−c)+b2 |
2ab |
2b(a−c)+b•
| ||
2ab |
2(a−c)+
| ||
2a |
a |
2c |
2(a−c)+
| ||
2a |
3 |
2 |