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In the figure above, point O is the center of the circle and

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In the figure above, point O is the center of the circle and [#permalink]  16 Jan 2011, 15:56
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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In the figure above, point O is the center of the circle and OC = AC = AB. What is the value of x ?

(A) 40
(B) 36
(C) 34
(0) 32
(E) 30

Problem Solving
Question: 162
Category: Arithmetic Statistics
Page: 83
Difficulty: 600

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Last edited by Bunuel on 16 Mar 2014, 04:42, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Quant Rev #161 Advanced Triangles Problem [#permalink]  16 Jan 2011, 18:07
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tonebeeze wrote:
In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

a. 40
b. 36
c. 34
d. 32
e. 30

It goes like this -

In triangle OAB , Angle O + Angle A + Angle B = 180 -------1
OA = OB (Radius) => Angle A = Angle B

In triangle ACB, Angle C = Angle O + Angle OAC (Sum of interior opposite angles)
=> Angle ACB = 2x;
Also, AC = AB => Angle ACB = Angle ABC = 2x each

Thus Angle A = Angle B = 2x each.

So, substituting in 1

5x = 180 => x = 36 deg.

This will help.
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Re: Quant Rev #161 Advanced Triangles Problem [#permalink]  17 Jan 2011, 18:38
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Amazing question...kinda a very good revision of my triagnometry
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Re: Quant Rev #161 Advanced Triangles Problem [#permalink]  18 Jan 2011, 19:51
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tonebeeze wrote:
In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

a. 40
b. 36
c. 34
d. 32
e. 30

A small diagram helps:
Attachment:

Ques1.jpg [ 9.23 KiB | Viewed 29666 times ]

In the figure, you see (180 - 2x) + y = 180 (straight angle) so 2x = y

Also, the moment you see the circle and its two radii, mark them equal and the corresponding angles equal. (the red angle = blue angle)
x + 180 - 2y = y
5x = 180 (from above, y = 2x)
x = 36
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 08 Nov 2010 Posts: 417 WE 1: Business Development Followers: 7 Kudos [?]: 74 [0], given: 161 Re: Quant Rev #161 Advanced Triangles Problem [#permalink] 19 Jan 2011, 21:38 Cool, but i think its an ez 600 question. isnt it? _________________ Manager Joined: 11 Jul 2010 Posts: 227 Followers: 1 Kudos [?]: 63 [0], given: 20 Re: Quant Rev #161 Advanced Triangles Problem [#permalink] 19 Jan 2011, 22:57 in your original diagram OA and OB don't appear to be radii - the end points are off the circle... without getting them to be equal I dont think you can answer this question Senior Manager Joined: 08 Nov 2010 Posts: 417 WE 1: Business Development Followers: 7 Kudos [?]: 74 [0], given: 161 Re: Quant Rev #161 Advanced Triangles Problem [#permalink] 19 Jan 2011, 23:03 I agree. i assumed that A and B are on the circle. _________________ Director Status: No dream is too large, no dreamer is too small Joined: 14 Jul 2010 Posts: 649 Followers: 37 Kudos [?]: 529 [1] , given: 39 Re: Quant Rev #161 Advanced Triangles Problem [#permalink] 10 Mar 2011, 13:03 1 This post received KUDOS angle oac=x angle acb=2x=angle b=2x=a (as oa and ob are radii) so in triangle oab x+2x+3x=180 x=36 _________________ Collections:- PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html 100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html Manager Joined: 14 Feb 2011 Posts: 196 Followers: 4 Kudos [?]: 95 [0], given: 3 Re: Quant Rev #161 Advanced Triangles Problem [#permalink] 11 Mar 2011, 00:14 tonebeeze wrote: In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)? a. 40 b. 36 c. 34 d. 32 e. 30 The figure attached is ambiguous. A needs to be shown on the circle, or at least it should be mentioned that A is also a point on the circle. If A is any point inside the circle, you cant solve this question from given information. GMAT Club Legend Joined: 09 Sep 2013 Posts: 8175 Followers: 416 Kudos [?]: 111 [0], given: 0 Re: In the figure attached, point O is the center of the circle [#permalink] 14 Dec 2013, 02:35 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Senior Manager Joined: 13 May 2013 Posts: 472 Followers: 1 Kudos [?]: 115 [1] , given: 134 Re: In the figure attached, point O is the center of the circle [#permalink] 14 Dec 2013, 16:41 1 This post received KUDOS In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)? OC = AC = AB tells us that triangle OAC and BAC are isosceles. Also, as with any triangle on a straight line with an exterior angle, the exterior angle (or in this case, angle ACB) can be found by subtracting the interior angle from 180. We know that in triangle OAC (because it is an isosceles triangle) that angle o and angle a are equal to one another. Therefore, angle c = 180 - x - x --> angle c = 180-2x. Angle C (of the smaller isosceles triangle) also happens to share the same angle measurement as angle B. Furthermore, because OA and OB are radii, we know that they equal one another and that angle A = angle B. We know that angle c in the smaller isosceles triangle = 180 - the measure of the obtuse angle C. As shown above the obtuse angle C = (180-2x) So, the small angle C = 180 - (180-2x) --> angle C = 2x. So, angle C = B = 2x. If angle A = B then A = 2x and becaause the obtuse triangle has two x measurements, we know that the measure of A in the small isosceles triangle = x. Therefore, in the small isosceles triangle we have 2x+2x+x = 180. 5x = 180. x = 36. b. 36 Intern Joined: 09 Feb 2014 Posts: 1 Followers: 0 Kudos [?]: 0 [0], given: 21 Re: In the figure attached, point O is the center of the circle [#permalink] 18 Feb 2014, 01:13 Is there another way to do it without assuming that OA and OB are radii? We are unable to state that OA and OB are radii based on the ambiguous diagram, and it doesn't mention OA and OB being the radii anywhere in the problem. It just so happened to work out in this problem, but maybe next time assuming it to be the radii won't result in the correct answer. Math Expert Joined: 02 Sep 2009 Posts: 31286 Followers: 5345 Kudos [?]: 62155 [0], given: 9444 Re: In the figure attached, point O is the center of the circle [#permalink] 18 Feb 2014, 01:28 Expert's post chuanchiehlo wrote: Is there another way to do it without assuming that OA and OB are radii? We are unable to state that OA and OB are radii based on the ambiguous diagram, and it doesn't mention OA and OB being the radii anywhere in the problem. It just so happened to work out in this problem, but maybe next time assuming it to be the radii won't result in the correct answer. Actually the diagram in the source (OG) is more accurate: Edited the original post. Hope it helps. _________________ Senior Manager Joined: 15 Aug 2013 Posts: 328 Followers: 0 Kudos [?]: 34 [0], given: 23 Re: Quant Rev #161 Advanced Triangles Problem [#permalink] 25 May 2014, 13:31 VeritasPrepKarishma wrote: tonebeeze wrote: In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)? a. 40 b. 36 c. 34 d. 32 e. 30 A small diagram helps: Attachment: Ques1.jpg In the figure, you see (180 - 2x) + y = 180 (straight angle) so 2x = y Also, the moment you see the circle and its two radii, mark them equal and the corresponding angles equal. (the red angle = blue angle) x + 180 - 2y = y 5x = 180 (from above, y = 2x) x = 36 Hi Karishma, Thanks for the detailed explanation but I still have a nagging issue. I was having a hard time setting this up with the variables. For example, <OAC, you have it broken up as 180-2y and x, which i see why you did. When I started working on this problem, I labeled <ACB and <ABC as "x" because of the congruent lines and I labeled OAC and COA as "x" too because all of those lines were same according to the question stem. Doing so, conflicted with the fact that <OAB and <ABO are equal and that's when my whole setup went kaboom Why is that wrong? I guess I'm not following why you gave some variables "x" vs. some "y"? Thanks! Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6219 Location: Pune, India Followers: 1676 Kudos [?]: 9592 [1] , given: 197 Re: Quant Rev #161 Advanced Triangles Problem [#permalink] 25 May 2014, 18:42 1 This post received KUDOS Expert's post russ9 wrote: VeritasPrepKarishma wrote: tonebeeze wrote: In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)? a. 40 b. 36 c. 34 d. 32 e. 30 A small diagram helps: Attachment: Ques1.jpg In the figure, you see (180 - 2x) + y = 180 (straight angle) so 2x = y Also, the moment you see the circle and its two radii, mark them equal and the corresponding angles equal. (the red angle = blue angle) x + 180 - 2y = y 5x = 180 (from above, y = 2x) x = 36 Hi Karishma, Thanks for the detailed explanation but I still have a nagging issue. I was having a hard time setting this up with the variables. For example, <OAC, you have it broken up as 180-2y and x, which i see why you did. When I started working on this problem, I labeled <ACB and <ABC as "x" because of the congruent lines and I labeled OAC and COA as "x" too because all of those lines were same according to the question stem. Doing so, conflicted with the fact that <OAB and <ABO are equal and that's when my whole setup went kaboom Why is that wrong? I guess I'm not following why you gave some variables "x" vs. some "y"? Thanks! When two sides of a triangle are equal, the two opposite angles are equal. But can you say what the two angles are? No. Say a triangle has two sides of length 5 cm each. Do we know the measure of equal angles? No. They could be 40-40 or 50-50 or 80-80 etc. So if you have two different triangles with 2 sides of length 5 cm each, the equal angle could have different measures - in one triangle the equal angles could be 50-50, in the other triangle, the equal angles could be 70-70. In triangle OAC, since OC = AC, you have two equal angles as x each. The third angle here is 180 - 2x. In triangle ACB, since AC = AB, angle ACB = angle ABC but what makes you say that they must be x each too? This is a different triangle. Even if the sides have the same length as the sides of triangle OAC, there is no reason to believe that the equal angles need to be x each. So you call the angles y. The third angle here is 180 - 2y. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: In the figure above, point O is the center of the circle and [#permalink]  25 May 2014, 22:58
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Re: Quant Rev #161 Advanced Triangles Problem [#permalink]  27 May 2014, 15:15
VeritasPrepKarishma wrote:
russ9 wrote:

Hi Karishma,

Thanks for the detailed explanation but I still have a nagging issue. I was having a hard time setting this up with the variables. For example, <OAC, you have it broken up as 180-2y and x, which i see why you did. When I started working on this problem, I labeled <ACB and <ABC as "x" because of the congruent lines and I labeled OAC and COA as "x" too because all of those lines were same according to the question stem. Doing so, conflicted with the fact that <OAB and <ABO are equal and that's when my whole setup went kaboom Why is that wrong?

I guess I'm not following why you gave some variables "x" vs. some "y"?

Thanks!

When two sides of a triangle are equal, the two opposite angles are equal. But can you say what the two angles are? No. Say a triangle has two sides of length 5 cm each. Do we know the measure of equal angles? No. They could be 40-40 or 50-50 or 80-80 etc. So if you have two different triangles with 2 sides of length 5 cm each, the equal angle could have different measures - in one triangle the equal angles could be 50-50, in the other triangle, the equal angles could be 70-70.

In triangle OAC, since OC = AC, you have two equal angles as x each. The third angle here is 180 - 2x.

In triangle ACB, since AC = AB, angle ACB = angle ABC but what makes you say that they must be x each too? This is a different triangle. Even if the sides have the same length as the sides of triangle OAC, there is no reason to believe that the equal angles need to be x each. So you call the angles y. The third angle here is 180 - 2y.

Hi Karishma,

This is news to me -- you can't assume that because length AC is shared, that implies that the corresponding angle in Triangle ACO will equal the corresponding angle in ACB?

Hmm -- learn something new everyday! Thanks!
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Re: Quant Rev #161 Advanced Triangles Problem [#permalink]  27 May 2014, 20:18
Expert's post
russ9 wrote:

Hi Karishma,

This is news to me -- you can't assume that because length AC is shared, that implies that the corresponding angle in Triangle ACO will equal the corresponding angle in ACB?

Hmm -- learn something new everyday! Thanks!

Corresponding angles are equal only when you have parallel lines with a common transversal.
Make two triangles with a common side. Can you make the angles made on the common side such that the angles have very different measures? Sure!

Attachment:

Ques3.jpg [ 6.81 KiB | Viewed 20725 times ]

Here one angle is 90 degrees and the other is acute.
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Re: Quant Rev #161 Advanced Triangles Problem [#permalink]  28 May 2014, 08:15
VeritasPrepKarishma wrote:
russ9 wrote:

Hi Karishma,

This is news to me -- you can't assume that because length AC is shared, that implies that the corresponding angle in Triangle ACO will equal the corresponding angle in ACB?

Hmm -- learn something new everyday! Thanks!

Corresponding angles are equal only when you have parallel lines with a common transversal.
Make two triangles with a common side. Can you make the angles made on the common side such that the angles have very different measures? Sure!

Attachment:
Ques3.jpg

Here one angle is 90 degrees and the other is acute.

Makes total sense. Thanks!
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Re: In the figure above, point O is the center of the circle and [#permalink]  28 May 2014, 08:46
x + 2x + 2x = 180 => x = 36
Re: In the figure above, point O is the center of the circle and   [#permalink] 28 May 2014, 08:46

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