∴C=180°-A-B=30°;
由正弦定理
| a |
| sinA |
| b |
| sinB |
| c |
| sinC |
a=
| bsinA |
| sinB |
5×
| ||||
|
5
| ||
| 3 |
| asinC |
| sinA |
| ||||||
|
5
| ||
| 3 |
(2)∵acosA+bcosB=ccosC,
∴sinAcosA+sinBcosB=sinCcosC,
∴sin2A+sin2B=sin2C,2sin(A+B)cos(A-B)=2sinCcosC,
∴cos(A-B)=-cos(A+B),2cosAcosB=0,
∴cosA=0或cosB=0,得 A=
| π |
| 2 |
| π |
| 2 |
∴△ABC是直角三角形.
