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In the figure above, point O is the center of the circle and OC = AC =
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Updated on: 04 Feb 2019, 04:59
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Originally posted by Bunuel on 16 Jan 2011, 16:56.
Last edited by Bunuel on 04 Feb 2019, 04:59, edited 5 times in total.
Renamed the topic and edited the question.




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Re: In the figure above, point O is the center of the circle and OC = AC =
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18 Jan 2011, 20:51
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 259351 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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Re: In the figure above, point O is the center of the circle and OC = AC =
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16 Jan 2011, 19:07
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: In the figure above, point O is the center of the circle and OC = AC =
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10 Mar 2011, 14:03



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Re: In the figure above, point O is the center of the circle and OC = AC =
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14 Dec 2013, 17:41
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 = 1802x. 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 = (1802x) So, the small angle C = 180  (1802x) > 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



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Re: In the figure above, point O is the center of the circle and OC = AC =
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25 May 2014, 14:31
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 1802y 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!



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Re: In the figure above, point O is the center of the circle and OC = AC =
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25 May 2014, 19:42
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 1802y 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 4040 or 5050 or 8080 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 5050, in the other triangle, the equal angles could be 7070. 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.
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Re: In the figure above, point O is the center of the circle and OC = AC =
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25 May 2014, 23:58
Answer = B = 36 Please refer diagram below:
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Re: In the figure above, point O is the center of the circle and OC = AC =
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27 May 2014, 16:15
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 1802y 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 4040 or 5050 or 8080 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 5050, in the other triangle, the equal angles could be 7070. 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: In the figure above, point O is the center of the circle and OC = AC =
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27 May 2014, 21:18
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 249647 times ]
Here one angle is 90 degrees and the other is acute.
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Re: In the figure above, point O is the center of the circle and OC = AC =
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29 Dec 2014, 14:43
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 Karishma  why is it that you did not have x+ x + (180  2y)+ y = 180 to include the entire triangle? I solved this equation by adding an additional line to form a diameter and then constructed another angle(titled z) and approached the solution that way but I am attempting to understand this method that you used to solve this problem.



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Re: In the figure above, point O is the center of the circle and OC = AC =
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29 Dec 2014, 20:07
GMAT01 wrote: Karishma  why is it that you did not have x+ x + (180  2y)+ y = 180 to include the entire triangle? I solved this equation by adding an additional line to form a diameter and then constructed another angle(titled z) and approached the solution that way but I am attempting to understand this method that you used to solve this problem. You need to find the relation between x and y. You can do it in any way you like; you will get the same result. Doing it your way: x+ x + (180  2y)+ y = 180 2x = y
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Re: In the figure above, point O is the center of the circle and OC = AC =
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29 Dec 2014, 20:38
VeritasPrepKarishma wrote: GMAT01 wrote: Karishma  why is it that you did not have x+ x + (180  2y)+ y = 180 to include the entire triangle? I solved this equation by adding an additional line to form a diameter and then constructed another angle(titled z) and approached the solution that way but I am attempting to understand this method that you used to solve this problem. You need to find the relation between x and y. You can do it in any way you like; you will get the same result. Doing it your way: x+ x + (180  2y)+ y = 180 2x = y Thank you. I must have overlooked something because I get: x+x+(1802y)+y= 180 2xy=0 2x2x= 0



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Re: In the figure above, point O is the center of the circle and OC = AC =
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29 Dec 2014, 20:41
GMAT01 wrote: VeritasPrepKarishma wrote: GMAT01 wrote: Karishma  why is it that you did not have x+ x + (180  2y)+ y = 180 to include the entire triangle? I solved this equation by adding an additional line to form a diameter and then constructed another angle(titled z) and approached the solution that way but I am attempting to understand this method that you used to solve this problem. You need to find the relation between x and y. You can do it in any way you like; you will get the same result. Doing it your way: x+ x + (180  2y)+ y = 180 2x = y Thank you. I must have overlooked something because I get: x+x+(1802y)+y= 180 2xy=0 2x2x= 0Ignore the highlighted step. after 2x  y = 0, when you take y to the other side, you get 2x = y Without doing this step, how did you substitute 2x for y in the highlighted step?
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Re: In the figure above, point O is the center of the circle and OC = AC =
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29 Dec 2014, 21:09
How are we getting different variables x and y when the sides are equal. Can you explain Krishna. Cause three sides are equal shouldn't their angles be noted with the same variable?



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Re: In the figure above, point O is the center of the circle and OC = AC =
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29 Dec 2014, 21:14
I must have overlooked something because I get:
x+x+(1802y)+y= 180 2xy=0 2x2x= 0[/quote]
Ignore the highlighted step. after 2x  y = 0, when you take y to the other side, you get 2x = y
Without doing this step, how did you substitute 2x for y in the highlighted step?[/quote] I see what I did. Below was my thought process:
x+x+(1802y)+y= 180 2xy=0 I then looked at the graph and applied the sum of two interior angles is equal to the opposite exterior which you labeled y. Once I got 2x= y I then went back to x+x+(1802y)+y= 180 and thought I would end up with 5x= 180 but instead kept getting 2x2x= 0



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Re: In the figure above, point O is the center of the circle and OC = AC =
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29 Dec 2014, 21:20
bankerboy30 wrote: How are we getting different variables x and y when the sides are equal. Can you explain Krishna. Cause three sides are equal shouldn't their angles be noted with the same variable? Note the sides that are equal OC = AC = AB OC and AC are sides if a triangle and the angles opposite to them are marked as x each (i.e. they are equal) AC and AB are equal sides of another triangle and angles opposite to them are marked as y each. Note that you cannot mark them as x too because they are equal angles in a different triangle. Their measure could be different from x. I have explained this in detail in a post given below. Giving the explanation here: "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 4040 or 5050 or 8080 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 5050, in the other triangle, the equal angles could be 7070. 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."
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Re: In the figure above, point O is the center of the circle and OC = AC =
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29 Dec 2014, 21:25
GMAT01 wrote: I see what I did. Below was my thought process:
x+x+(1802y)+y= 180 2xy=0 I then looked at the graph and applied the sum of two interior angles is equal to the opposite exterior which you labeled y. Once I got 2x= y I then went back to x+x+(1802y)+y= 180 and thought I would end up with 5x= 180 but instead kept getting 2x2x= 0 From this equation: x+x+(1802y)+y= 180, you derive that 2x = y. If you try to substitute 2x = y in this equation itself, you will just get 2x  2x = 0 which implies 0 = 0. This equation has 2 variables and you need two distinct equations to get the value of the two variables. If both equations are just 2x = y, you cannot get the value of x. You need another equation to get the value of x.
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Re: In the figure above, point O is the center of the circle and OC = AC =
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18 May 2015, 07:50
What step am I doing wrong here? I can't figure it out. I don't understand how to go to 5x = 180 I just get to 4x + 180  4x = 180 aka 4x  4x = 0.. edit: that last "8" kinda looks like a 5, but its supposed to be an 8



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Re: In the figure above, point O is the center of the circle and OC = AC =
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