On the circle with center O,points A and B ar | vedclass

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(B) Given that $OA = OB = OF$ are radii of the circle. $BC$ is tangent to the circle at $B$,so $\angle OBC = 90^{\circ}$.In $	riangle OAB$,we have $O

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$2 : 3$

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(B) Given that $OA = OB = OF$ are radii of the circle. $BC$ is tangent to the circle at $B$,so $\angle OBC = 90^{\circ}$.In $	riangle OAB$,we have $OA = OB = AB$ (radii and given condition),so $	riangle OAB$ is an equilateral triangle.Thus,$\angle AOB = 60^{\circ}$ and $\angle OAB = 60^{\circ}$.In $	riangle ABC$,we have $AB = BC$. Since $\angle ABC = \angle ABO + \angle OBC = 60^{\circ} + 90^{\circ} = 150^{\circ}$,and $	riangle ABC$ is isosceles with $AB = BC$,the base angles are $\angle BAC = \angle BCA = (180^{\circ} - 150^{\circ}) / 2 = 15^{\circ}$.Now,consider $	riangle OAC$. In $	riangle OAB$,$\angle OAB = 60^{\circ}$. In $	riangle ABC$,$\angle BAC = 15^{\circ}$. Thus,$\angle OAC = \angle OAB + \angle BAC = 60^{\circ} + 15^{\circ} = 75^{\circ}$.Since $OA = OF$,$	riangle OAF$ is isosceles,so $\angle OFA = \angle OAF = 75^{\circ}$.Then $\angle AOF = 180^{\circ} - (75^{\circ} + 75^{\circ}) = 30^{\circ}$.Since $\angle AOB = 60^{\circ}$,we have $\angle BOF = \angle AOB - \angle AOF = 60^{\circ} - 30^{\circ} = 30^{\circ}$.In $	riangle OBC$,$\angle BOC = 180^{\circ} - (90^{\circ} + \angle OCB) = 180^{\circ} - (90^{\circ} + 15^{\circ}) = 45^{\circ}$.Therefore,the ratio $\frac{\angle BOF}{\angle BOC} = \frac{30^{\circ}}{45^{\circ}} = \frac{2}{3}$.

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