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In the given figure A B is the diameter ...

In the given figure `A B` is the diameter of the circle, centred at `Odot` If `/_C O A=60^0,A B=2r ,A C=d ,` and `CD=l`

A

`d sqrt3`

B

`d//sqrt3`

C

3d

D

`sqrt3d//2`

Text Solution

Verified by Experts

The correct Answer is:
A

`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`
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