∴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.