Question

设O使锐角三角形ABC的外心,若∠C=75°,且△AOB,△BOC,△COA的面积满足S△AOB+S△BOC=根号三S△

Answer 1

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)(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)

则(1/4)(r^2)+(1/2)(r^2)Sin(2∠A)=√3 x (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(√3/2 x Sin2A - 1/2 x Cos2A)

=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°