在△ABC中,内角A、B、C的对边长分别为a、b、c且sinAcosB=1/3,sinBcosA=1/6,△ABC的外接圆半径R=3.求a/b值

在△ABC中,内角A、B、C的对边长分别为a、b、c且sinAcosB=1/3,sinBcosA=1/6,△ABC的外接圆半径R=3.求a/b的值
sinAcosB+cosAsinB=sin(A+B)=1/3+1/6=1/2,故A+B=π/6或π-π/6=5π/6
sinAcosB-cosAsinB=sin(A-B)=1/3-1/6=1/6,故A-B=arcsin(1/6)或π-arcsin(1/6)
因此可能有四种组合：
(一)A+B=π/6.(1)；A-B=arcsin(1/6).(2)
(1)+(2)得2A=π/6+arcsin(1/6),故A=π/12+(1/2)arcsin(1/6)；
(1)-(2)得2B=π/6-arcsin(1/6),故B=π/12-(1/2)arcsin(1/6)；
∴a/b=sinA/sinB=sin[π/12+(1/2)arcsin(1/6)]/sin[π/12-(1/2)arcsin(1/6)]
=sin(15°+4.7970°)/sin(15°-4.7970°)=sin19.7970°/sin10.2030°=0.3387/0.1771=1.9125
(二)A+B=π/6.(1)；A-B=π-arcsin(1/6).(2)
(1)+(2)得2A=7π/6-arcsin(1/6),故A=π/12-(1/2)arcsin(1/6)；
(1)-(2)得2B=-5π/6+arcsin(1/6),故B=-5π/12+(1/2)arcsin(1/6)；
B为负角,舍去此情况.
(三)A+B=5π/6.(1)；A-B=arcsin(1/6).(2)
(1)+(2)得2A=5π/6+arcsin(1/6),故A=5π/12+(1/2)arcsin(1/6)
(1)-(2)得2B=5π/6-arcsin(1/6),故B=5π/12-（1/2）arcsin(1/6)
这时a/b=sin[5π/12+(1/2)arcsin(1/6)]/sin[5π/12-(1/2)arcsin(1/6)]
=sin(150°+4.7970°)/sin(150°-4.7970°)=sin154.7970°/sin145.203°=0.4258/0.5707=0.7461
(四)A+B=5π/6.(1)；A-B=π-arcsin(1/6).(2)
(1)+(2)得2A=11π/6-arcsin(1/6),故A=11π/12-(1/2)arcsin(1/6)；
(1)-(2)得2B=-π/6+arxsin(1/6),故B=-π/12+(1/2)arcsin(1/6)