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geometry 9

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Manager
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Joined: 19 Aug 2009
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geometry 9 [#permalink] New post 02 Oct 2009, 10:29
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Q.
2 circles C(O,r) and C( O',r') intersect at 2 points A and B and O lies on C(O',r'). A tangent CD is drawn to the circle C(O',r') at A.Then

a. angle(OAC)= OAB
b. angle( OAB)=AO'O
c. angle (AO'B)=AOB
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Manager
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Joined: 08 Nov 2005
Posts: 182
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Re: geometry 9 [#permalink] New post 05 Oct 2009, 11:26
Where did you see this problem? It looks extremely difficult.
I hope someone post the answer and the explanation. I am curious now.
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Schools: LBS
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Re: geometry 9 [#permalink] New post 07 Oct 2009, 13:01
virtualanimosity wrote:
Q.
2 circles C(O,r) and C( O',r') intersect at 2 points A and B and O lies on C(O',r'). A tangent CD is drawn to the circle C(O',r') at A.Then

a. angle(OAC)= OAB
b. angle( OAB)=AO'O
c. angle (AO'B)=AOB


Here,

O'A=OO'=O'B (radiuses)

Also triangles AOO' and BOO' are identical.

so angle O'AO=AOO'=OBO'=BOO' = x

So OO'A = OO'B = 180-2x

angle O'AC=90 (tangent + radius)

So OAB= 90-x

Let AB and OO' intersect at E.

In triangle OAE, angle OEA=90.
Since <EOA = x, angle OAE=OAB=90-x

so OAB=OAC.
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Re: geometry 9   [#permalink] 07 Oct 2009, 13:01
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