由題意得∠B=180°-∠A-∠C=105°-∠A
設⊙O為△ABC外接圓
令⊙O半徑為r,則OA=OB=OC=r
∠AOB=2∠C=150°,∠BOC=2∠A,∠COA=2∠B （圓心角等于二倍的圓周角）
S△AOB=(1/2)(OA x OB x Sin∠AOB)=(1/4)(r^2)
S△BOC=(1/2)(OB x OC x Sin∠BOC)=(1/2)(r^2)Sin(2∠A)
S△COA=(1/2)(OC x OA x Sin∠COA)=(1/2)(r^2)Sin(2∠B)
則(1/4)(r^2)+(1/2)(r^2)Sin(2∠A)=√3 x (1/2)(r^2)Sin(2∠B)
化簡得1+2Sin2A=2√3Sin2B=2√3Sin(210°-2∠A)
=2√3(Sin210°Cos2A-Cos210°Sin2A)
=2√3(Sin210°Cos2A-Cos210°Sin2A)
=2√3(√3/2 x Sin2A - 1/2 x Cos2A)
=3Sin2A-√3Cos2A
即1+√3Cos2A=Sin2A
兩邊平方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為銳角三角形”可得0°