Question

On the circle with centre 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 sides of the line OB and AB = BC. The line 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

The correct option is B

2:3

1. $\Delta$ AOB is equilateral ($\angle$ AOB = $\angle$OAB = $\angle$OBA = ${60}^{\circ }$)

2. $\Delta$ OBC is right angled isosceles ($\angle$OBC = ${90}^{\circ }$)

3. $\Delta$ ABC is isosceles ($\angle$BAC = $\angle$BCA = ${15}^{\circ }$)

4. $\angle$ OAC = ${60}^{\circ }$ - $\angle$CAB = ${45}^{\circ }$

5. $\Delta$ AOF is right angled isosceles ($\angle$ AOF = ${90}^{\circ }$,$\angle$ OFA = ${45}^{\circ }$)

6. $\angle$ BOF = ${90}^{\circ }$ - $\angle$ AOB = ${30}^{\circ }$

7. $\Delta$ OBC is right angled isosceles ($\angle$BOC = ${45}^{\circ }$)

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