由题意得∠B=180°-∠A-∠C=105°-∠A
设⊙O为△ABC外接圆
令⊙O半径为r,则OA=OB=OC=r
∠AOB=2∠C=150°,∠BOC=2∠A,∠COA=2∠B(圆心角等于二倍的圆周角)
S△AOB=(1/2)(OAxOBxSin∠AOB)=(1/4)(r^2)
S△BOC=(1/2)(OBxOCxSin∠BOC)=(1/2)(r^2)Sin(2∠A)
S△COA=(1/2)(OCxOAxSin∠COA)=(1/2)(r^2)Sin(2∠B)
则(1/4)(r^2)+(1/2)(r^2)Sin(2∠A)=√3x(1/2)(r^2)Sin(2∠B)
化简得1+2Sin2A=2√3Sin2B=2√3Sin(210°-2∠A)
=2√3(Sin210°Cos2A-Cos210°Sin2A)
=2√3(Sin210°Cos2A-Cos210°Sin2A)
=2√3(√3/2xSin2A-1/2xCos2A)
=3Sin2A-√3Cos2A
即1+√3Cos2A=Sin2A
两边平方1+2√3Cos2A+3(Cos2A)^2=(Sin2A)^2=1-(Cos2A)^2
2(Cos2A)^2+√3Cos2A=0
(Cos2A)(√3+2Cos2A)=0
Cos2A=0或者-√3/2
“△ABC为锐角三角形”可得0°