In the given figure, AB is the diameter of the circle, centered at O. If `angle COA =60^(@), AB=2r, Ac=d and CD=l,` then l is equal to

Options

A. `d sqrt3`
B. `d//sqrt3`
C. 3d
D. `sqrt3d//2`

Answer

A

Explanation

`AC = d, OA = OB r, CD = BD = I, angleCOA = (pi)/(3)`
Clearly `DeltaAOC` is equilateral
`:. d = r`
Also, `angleBOD = angleCOD = (2pi)/(3xx2) = (pi)/(3)`
or `tan.(pi)/(3) = (BD)/(OB) = (l)/(r) rArr l = r sqrt3 = d sqrt3`