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Luci
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Post subject: In the figure shown, point O is the center of  Posted: Wed Aug 15, 2007 3:13 pm |
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I understand why B is sufficient. While both triangles are isosceles and we know BCO is 40º using the exterior angle rule BAO must be 20º
But how do we know A is sufficient? Using the rule again if we know that COD is 60º we know the oposite angles will sum 60º as well, but since ACO is not isosceles, how do we know the proportion between them?
Any help?
Thanks
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GMAT 2007
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Post subject: Posted: Thu Aug 16, 2007 11:21 am |
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We know AOD is a straight line -
Angle(AOB) + Angle(BOC) + Angle(COD) = 180---(1)
We know Angle(AOB) = Angle(BAO)---(2) Because(AB = OB)
Substitute in (1)
Angle(BAO) + Angle(BOC) + Angle(COD) = 180
Also, Angle(BOC) = 180-2(Angle(OBC)) Because OB = OC (Radii of Circle)
Substitute in (1)
Angle(BAO) + 180 - 2(Angle(OBC)) + Angle(COD) = 180
Also, in Triangle BAO, Angle OBC is the exterior angle so Angle(OBC) = Angle(BAO) + Angle(BOA) (Exterior angle is equal to the sum of the opposite angles)
So Angle(OBC) = 2(Angle(BAO)) ---from (2)
substitute it back in (1)
Angle(BAO) + 180 - 4Angle(BAO) + Angle(COD) = 180
We know Angle (COD) already so, (A) is also sufficient.
Hope it helps
GMAT 2007
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StaceyKoprince
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Post subject: Posted: Sat Aug 18, 2007 3:38 pm |
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| ManhattanGMAT Staff |
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Location: San Francisco
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This one's tough to show without the ability to draw. Try following through GMAT 2007's approach, above, or mine, below, while drawing at every step so you can see what's going on.
Re-draw just the straight lines - ignore the arc of the semi-circle. Label COD 60. Label BAO and BOA with a variable (I'll use "a") and label CBO and BCO with another variable (I'll use "b").
Use that info to label ABO. I know that AC = 180, so ABO = 180-b. That's part of a small triangle, which I can write: a+a+(180-b) = 180 which simplifies to 2a - b = 0
I know using the exterior angle rule that COD = BAO + BCO (look at the big triangle only). Given my labels, I can re-write that equation as: a+b = 60.
I can now combine the two equations.
2a - b = 0
a+b = 60
Substitute however you want, but solve for a, not b (since that's what you are asked to find).
2a = b and b = 60-a, so 2a = 60-a. 3a=60. a = 20.
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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Luci
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Post subject: Posted: Sat Aug 18, 2007 4:41 pm |
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Thank you guys, great explanations. I think Stacey´s is easier, because it is more direct. Although I think GMAT 2007 is going to perform terrific in the quant part because he/she solves the toughest problems :-)
Thanks again.
Luci
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rschunti
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Post subject: Pls clarify Posted: Sun Mar 02, 2008 12:26 pm |
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In this figure what is the reason of beleiving that "Line aegment "ABC" is a straight line?It could be possible that Line segments "AB" and "BC" may not be colinear as nothing is mentioned in this question that proves this.?Pls can you clarify the reason why we are assuming that ABC is a straight line?
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RonPurewal
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Post subject: Re: Pls clarify Posted: Mon Mar 03, 2008 5:54 am |
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| ManhattanGMAT Staff |
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rschunti wrote: In this figure what is the reason of beleiving that "Line aegment "ABC" is a straight line?It could be possible that Line segments "AB" and "BC" may not be colinear as nothing is mentioned in this question that proves this.?Pls can you clarify the reason why we are assuming that ABC is a straight line?
collinearity, and for that matter linearity in general, is one of the few things that you are allowed to assume on the gmat. the reason is pretty utilitarian: if you couldn't assume that things that look like straight lines are actually straight lines, then no diagram would ever be useful! (imagine if that cute little triangle in your diagram might turn out to be a heptagon... you get the picture)
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mclaren7
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Post subject: Posted: Sat Mar 29, 2008 1:29 pm |
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Dear friends
AB = OC = OB
Angle BAO = Angle BOA = X
Angle ABO = 180 - 2X, therefore angle CBO = 2x.
since OC = OB angle CBO = angle BCO = 2X, angle BOC = 180 - 4X
Therefore angle COD = 180 - (180-4X) - X = 3X
1: angle COD = 3x = 60 , X = 20 angle BAO = 20 <== sufficient
2: angle BCO = 2x = 40 , X = 20 angle BAO = 20 <== sufficient
KH
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iharden
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Post subject: Re: In the figure shown, point O is the center of Posted: Sat Aug 20, 2011 6:31 pm |
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Another Way!
1)
BAO = BOA = x
OBC = BCO = 2x
OAC + ACO = COD
x + 2x = 60
x = 20
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RonPurewal
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Post subject: Re: In the figure shown, point O is the center of Posted: Thu Aug 25, 2011 4:08 am |
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| ManhattanGMAT Staff |
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iharden wrote: Another Way!
1)
BAO = BOA = x
OBC = BCO = 2x
OAC + ACO = COD
x + 2x = 60
x = 20 this is correct, although, if you're going to go to the trouble of posting it on the forum, a few words of explanation would go a long way. i.e., right now there's no explanation for these steps, so this solution will only make sense to people who already know how to solve the problem. -- nice three-and-a-half-year thread bump, by the way. you must be a champion at searching the forum.
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hifunda_88
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Post subject: Re: In the figure shown, point O is the center of Posted: Sat May 19, 2012 2:04 pm |
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Ron, come on! You must know we come to MGMAT forums all the time when searching for GMATPrep explanations from Google right? (Whether it's 2008 or 2012)
:)
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RonPurewal
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Post subject: Re: In the figure shown, point O is the center of Posted: Sun May 20, 2012 3:15 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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hifunda_88 wrote: Ron, come on! You must know we come to MGMAT forums all the time when searching for GMATPrep explanations from Google right? (Whether it's 2008 or 2012)
:) right, but it's also a surprise that google would turn up a thread that hasn't had a response in years. unless you guys are actually scrolling past the first page on google (which hardly anyone ever does, pretty much ever).
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