1.

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

Answer»

`d sqrt3`
`d//sqrt3`
3d
`sqrt3d//2`

SOLUTION :
`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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