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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2 : 3$

Vedclass · Answer

(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 $\frac{\angle BOF}{\angle BOC} = \frac{30^{\circ}}{45^{\circ}} = \frac{2}{3}$.

Similar Questions

If $\alpha \neq -4$ and $(2, \alpha)$ is the mid-point of a chord of the circle $x^2+y^2-4x+8y+6=0$,then the values of the $y$-intercept of the chord lie in the interval

The equation of the incircle of the triangle formed by the lines $x=0, y=0$ and $3x+4y-24=0$ is

If $(1,3)$ is the midpoint of a chord of the circle $x^2+y^2-4x-8y+16=0$,then the area of the triangle formed by that chord with the coordinate axes is

The number of common tangents to the circles $x^2+y^2-18x-15y+131=0$ and $x^2+y^2-6x-6y-7=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module