Question

On the circle with center O,points A,B are such that OA=AB . A point C is located on the tangent at B to the circle such that A and C are on the opposite side of the line OB and AB =BC.The segment AC intersects the circle again at F.Then the ratio $\angle BOF:\angle BOC$ is equal to :

• A. $1:2$
• B. $2:3$
• C. $3:4$
• D. $4:5$

Answer 1

Answer: (B)

$2:3$

1.$△AOB$ is equilatrual$(\angle AOB=\angle OAB=\angle OBA=60)$
2.$△OBC$ is right angled isosceles $(\angle OBC=90)$
3.$△ABC$ is isosceles $(\angle BAC=\angle BCA=15)$
4.$\angle OAC=60-\angle CAB=45$
5.$△AOF$ is right angled isosceles $(\angle AOF=90,\angle OFA=45)$
6.$\angle BOF=90-\angle AOB=30$
7.$△OBC$ is right angled isosceles $(\angle BOC=45)$
$\therefore \frac{\angle BOF}{\angle BOC}=\frac{30}{45}=\frac{2}{3}$