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 `angleBOF: angle BOC` is equal to-

Options

A. `1 : 2`
B. `2 : 3`
C. ` 3 : 4`
D. `4 : 5`

Answer

B

Explanation

`:. DeltaOAB` should be equilareral
Now `angleOBC=90^(@) & AB=BC`
`rArr angleBAC=angle BCA =15^(@)`
`:. Angle BOF =30^(@)&angle BOC=45^(@)" {Delta OBC" is right isosceles"}`
`:. (angle BOF)/(angleBOC)=(30^(@))/(45^(@))=2/3" "{angleBOF=2angleBAF}`