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Geometry

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Geometry [#permalink] New post 26 Mar 2009, 15:17
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
DS in Geometry
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Re: Geometry [#permalink] New post 26 Mar 2009, 17:19
ANS:D
recently discussed in forum.
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Re: Geometry [#permalink] New post 27 Mar 2009, 13:47
Can you please post the link which has the explanation,Probably last time i missed it.
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Re: Geometry [#permalink] New post 27 Mar 2009, 14:42
IMO D.

I tired solving this problem using one of the basic rules of the triable.

Exterior angle of a triable = sum of the opposite interior angles.

It is given AB = OC.

Following can also be derived. OB = AB.
Let <BAO = <BOA = x (Because of the Isoceles triable proerty)
Let <BCO = <CBO = y. (Becuase of the isoceles triangle property)

From stmt1 : <COD = 60 Then x + y = 60 (By the rule, stated above)
Again <BAO + <BOA = <CBO ==> y + y = x ==> 2y = x(By the rule stated above)

Solving these two, we can get the value of y. Hence sufficient.

From Stmt 2: Given <BCO = 40. ==> < CBO = 40.

Again, <BAO + <BOA = <CBO ==> y + y = x ==> 2y = x(By the rule stated above)
Here we know x = 40.

Solving it, we can get the value of Y. Hence sufficient.

Other solutions can be found in the following link.

gmatprep-2-triangle-semicircle-76801.html
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Re: Geometry [#permalink] New post 27 Mar 2009, 14:43
nitindas wrote:
Can you please post the link which has the explanation,Probably last time i missed it.


Don't know the post but, here is the solution:

from the stem , AB = OC = OB (Since OC and OB are radii of the same circle) --- (a)

from (1) , we have <COD = 60 . Let's solve:

Consider triangle CBO.
<CBO + <COB + <BCO = 180 (sum of angles in a triangle) --- (b)
<CBO = <BCO (since OB = OC, its an isosceles triangle)

equation "b" becomes:
2<CBO + <COB = 180
Now, <CBO = <BAO + <BOA (external angle in a triangle is equal to the sum of two internal angles)
also, <BAO = <BOA (from equation a above)

which implies, 4<BAO + <COB = 180 --> 4<BAO + (120 - <BOA) = 180

or, 4<BAO = 120 - <BOA = 180
or, 3<BAO = 60 (since <BAO = <BOA )
or, <BAO = 20

Hence (1) is sufficient.

Now from (2) we have, <BCO = 40 --> <CBO = 40 --> 2<BAO = 40 --> <BAO = 20. Hence (2) is sufficient.

I have not explained here but if you understood (1) above, I think this is easy to stand.

Therefore the answer is (D).
_________________

kris

Re: Geometry   [#permalink] 27 Mar 2009, 14:43
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