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In the figure shown, point O is the center of the semicircle

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Joined: 13 Mar 2009
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In the figure shown, point O is the center of the semicircle [#permalink]

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04 Oct 2009, 23:22
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Difficulty:

55% (hard)

Question Stats:

64% (02:32) correct 36% (01:36) wrong based on 64 sessions

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OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-shown-point-o-is-the-center-of-the-semicircle-89662.html
[Reveal] Spoiler: OA

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Re: Geometry (Semicircle, Triangle) from GMATPrep [#permalink]

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05 Oct 2009, 06:26
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Editing: (statement says <COD=60 and in my calculation I mistaken it for <COB)

Write down everything you know from the Q:
BO=CO=r=AB --> BOC and ABO are isosceles.
<BAO=<BOA and <BCO=<OBC
<CBO=2<BAO

(1) <BAO +<ACO=<COD=60 degrees (Using exterior angle theorem)
<ACO = <CBO = 2<BAO
So, <BAO + <ACO = 2<BAO + <BOA = 3* (<BAO) = 60 degrees
<BAO = 20 degrees SUFFICIENT

(2) <BCO=40 degrees --> <BCO=<CBO=40 degrees=2<BAO --> <BAO=20 degrees. SUFFICIENT

D.
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Last edited by Bunuel on 05 Oct 2009, 12:28, edited 1 time in total.
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Re: Geometry (Semicircle, Triangle) from GMATPrep [#permalink]

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05 Oct 2009, 09:34
OB = OC = AB = OD.So each statement alone is suff.

I wud go with option D
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Re: Geometry (Semicircle, Triangle) from GMATPrep [#permalink]

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05 Oct 2009, 12:39
Thanks, edited my post, anyway answer is D. In such semicircle any angle is sufficient to determine <BAO.
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Last edited by Bunuel on 07 Oct 2009, 15:32, edited 1 time in total.
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Re: Geometry (Semicircle, Triangle) from GMATPrep [#permalink]

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07 Oct 2009, 14:48
deepakraam wrote:
OB = OC = AB = OD.So each statement alone is suff.

I wud go with option D

Can you please explain how you came up with option D just by stating OB = OC = AB = OD. I spent nearly 5 min to decide that ans is D.
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Re: Geometry (Semicircle, Triangle) from GMATPrep [#permalink]

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21 Apr 2010, 07:43
Imagining the shapes visually might prove to be another way of doing this.

The question translates into - Is there only ONE possible line that satisfies the given figure spatially.
(1) When $$\angle COD$$ is fixed, there can be many lines possibly drawn from C to A, but only one line would render length $$AB = radius$$
(2) When $$\angle BCO$$ is fixed, the same reasoning applies.

So now we know that in either of the cases, we CAN zero down to one particular line segment, wich means that $$\angle BAO$$ is fixed (whatever it be). So answer is D.
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Re: Geometry (Semicircle, Triangle) from GMATPrep [#permalink]

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17 Jan 2013, 07:03
Can some1 help me in explaining how do we get <CBO=2<BAO.I really dont uderstand how they derive this.Can someone shed some light on this?
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Re: Geometry (Semicircle, Triangle) from GMATPrep [#permalink]

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17 Jan 2013, 07:22
skamal7 wrote:
Can some1 help me in explaining how do we get <CBO=2<BAO.I really dont uderstand how they derive this.Can someone shed some light on this?

Check here: in-the-figure-shown-point-o-is-the-center-of-the-semicircle-89662.html#p664937

Hope it helps.
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Re: Geometry (Semicircle, Triangle) from GMATPrep   [#permalink] 17 Jan 2013, 07:22
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