∴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.
