Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club

 It is currently 26 Mar 2017, 18:01

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# geometry 9

Author Message
Manager
Joined: 19 Aug 2009
Posts: 79
Followers: 1

Kudos [?]: 235 [0], given: 46

### Show Tags

02 Oct 2009, 11:29
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Attachments

geometry 9.JPG [ 15.42 KiB | Viewed 893 times ]

Manager
Joined: 08 Nov 2005
Posts: 182
Followers: 2

Kudos [?]: 20 [0], given: 10

### Show Tags

05 Oct 2009, 12:26
Where did you see this problem? It looks extremely difficult.
I hope someone post the answer and the explanation. I am curious now.
Manager
Joined: 02 Jan 2009
Posts: 96
Location: India
Schools: LBS
Followers: 2

Kudos [?]: 84 [0], given: 6

### Show Tags

07 Oct 2009, 14:01
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,

Also triangles AOO' and BOO' are identical.

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

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

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.
_________________

The Legion dies, it does not surrender.

Re: geometry 9   [#permalink] 07 Oct 2009, 14:01
Display posts from previous: Sort by