Math, asked by saronsunny2633, 1 year ago

in the figure, chord AB of circle with centre O, is produced to C such that BC = OB . CO is joined and produced to meet the circle at D.If angle ACD = y and angle AOD = x,show that x = 3y.

Answers

Answered by 9419258724
39
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Answered by shaista16lm
3

Answer:

The relation between ∠AOD and ∠ACD is x = 3y

Step-by-step explanation:

According the given question, the figure can be drawn as(refer to the attachment)

given,

AB is the chord of the circle (center O)

BC=OB

Since the sides BC and OB is equal, we know that, angles opposite to equal sides are equal.

  hence ∠OCB=∠BOC=y°

From the figure, ∠OBA is the exterior angle of the triangle ΔBCO.

With the exterior angle property of the triangle i.e. sum of opposite interior angels is equal to exterior angle

∴ ∠OBA =∠BOC+∠OCB

              = y°+y° = 2y°

   ∠OBA=2y°

OA=OB (the radii of the circle)

Hence, ∠OAB=∠OBA (angles opposite to equal sides are equal. ).

∠OAB= 2y°

we need to find, ∠AOD, which is exterior angle of ΔAOC

∠AOD=OAB+∠OCA

   x°= 2y°+y°

 ∴  x°= 3y°

Hence Proved

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