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

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Re: In the figure above, point O is the center of the circle and [#permalink]

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New post 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 [#permalink]

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New post 29 Dec 2014, 21:14
I must have overlooked something because I get:

x+x+(180-2y)+y= 180
2x-y=0
2x-2x= 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+(180-2y)+y= 180
2x-y=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+(180-2y)+y= 180 and thought I would end up with 5x= 180 but instead kept getting 2x-2x= 0

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Re: In the figure above, point O is the center of the circle and [#permalink]

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New post 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 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."
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GMAT01 wrote:
I see what I did. Below was my thought process:

x+x+(180-2y)+y= 180
2x-y=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+(180-2y)+y= 180 and thought I would end up with 5x= 180 but instead kept getting 2x-2x= 0



From this equation: x+x+(180-2y)+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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In the figure above, point O is the center of the circle and [#permalink]

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New post 30 Dec 2014, 01:31
Algebra

x+x+180-y=180

2x+180-y=180 => 2x=y

so, x+2x+2x=180

5x=180 => x=36

Backsolving

take 34 (C)

34+34=68, y=68, 68*2=136, 180-136=44 not equal 34. We have to get higher value of X to make difference less (eliminate C,D,E)

go B (36)

36+36=72, y=72, 72*2=144, 180-144=36 (it is correct)


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In the figure above, point O is the center of the circle and [#permalink]

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New post 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 [#permalink]

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erikvm wrote:
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
Image


Hi erikvm,

Image
The angle which you have written as 180 - 2y = x because OA = OB (radii of the circle). So ∠OAB = ∠OBA.

∠(OAC + CAB) = y = 2x
x + ∠CAB = 2x which gives us ∠CAB = x.

Now summing up the angles in triangle OAB, we get x + 2x + 2x = 180 i.e. x = 36

Hope this helps :)

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In the figure above, point O is the center of the circle and [#permalink]

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DropBear wrote:

Geometry: What is the angle of x?



Attachment:
The attachment Capture.PNG is no longer available


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
D. 32
E. 30

I had to guess this one recently and even after reading the official answer and explanation there are still some inferences that I just don't understand. I am really looking forward to seeing a few different ways of solving. Personally this is one of the hardest questions I have faced... Maybe you will find it easy :)

The reason for the poll is I would like to see how difficult everyone finds this question. As I said, this one really beat me!

Would be great to receive my first Kudos if you find this useful too :-D


Straightforward question if you realize the additional constraint would come from the fact that in Triangle AOB, OA = OB = radius of the circle.

Now in triangle, ACO , \(\angle{COA} = \angle{OAC} = x\) (as OC = AC) and \(\angle{ACB} = 2x\) (external angle of a triangle)

Additionally, \(\angle{ACB}= \angle{ABC}\) = 2x (as AC = AB)

FInally, in triangle AOB, OA = OB = radius of the circle ---> \(\angle{OBA}=\angle{OAB}\) = 2x ---> \(\angle{CAB} =x\)

Thus , in triangle ACB,

\(\angle{ACB} + \angle{CBA} + \angle{BAC}\) =2x+2x+x = 180 ---> x = 36. B is the correct answer.

Please search for a question before posting.

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In the figure above, point O is the center of the circle and [#permalink]

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the measure of the exterior angle is equal to the sum of the two non-adjacent angles of the triangle --> Angle ACB=2x
OA=OB => 180-4x+x=2x; X=36°
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Re: In the figure above, point O is the center of the circle and [#permalink]

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New post 03 Oct 2015, 11:17
Something it's not quite right here.

I'm also stuck with the last calculation.

STEP 1
180-2x + y = 180
2x = y OK I KNOW HOW TO FIND THIS

STEP 2
Then I convert Y to become 2X
<AOB = x
<ABO = 2x
<BAO = x + (180-4x)-----Why I add x to 180-4x because that's how I think we get a straight line of 180 degree

therefore:
x + 2x + x + (180-4x) = 180
then I get 180 = 180

PLEASE HELP ME. TOTALLY STUCK

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Re: In the figure above, point O is the center of the circle and [#permalink]

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blendercroix wrote:
Something it's not quite right here.

I'm also stuck with the last calculation.

STEP 1
180-2x + y = 180
2x = y OK I KNOW HOW TO FIND THIS

STEP 2
Then I convert Y to become 2X
<AOB = x
<ABO = 2x
<BAO = x + (180-4x)-----Why I add x to 180-4x because that's how I think we get a straight line of 180 degree

therefore:
x + 2x + x + (180-4x) = 180
then I get 180 = 180

PLEASE HELP ME. TOTALLY STUCK


After step 2,
<BAO = <ABO
x + 180 - 4x = 2x
x = 36

The reason your third step doesn't work is because you used the property of total sum of triangle = 180 to get the relations. Now you are putting the relations back in sum of triangles is 180. You cannot get a value for x in this case. You need to put the relations in another property to arrive at a new conclusion (value of x).
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Re: In the figure above, point O is the center of the circle and [#permalink]

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New post 09 Feb 2016, 01:09
Here, since OC=CA, angle OAC=x. Hence angle ACB=2x and ABC=2x
If we extend BO to Q, then we get an inscribed angle AQB which is half the central angle x.
and the angle QAB=90
so, x/2+2x+90=180
5x/2=90
x=36
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Re: In the figure above, point O is the center of the circle and [#permalink]

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New post 18 Jul 2016, 10:36
OA=OB (radius)

Therefore angle oab = angle oba

oc=ac(given)
therefore angle aoc = angle cao

therefore, angle oac = x degrees

angle acb = 2x degrees (because exterior angle =sum of remote interior angles)

ac=ab
therefore angle abc = 2x degrees(isosceles triangle rule)

ao=pb therefore, angle oab = 2x(proved before that angle abo =2x --->isosceles triangle rule)

triangle aob =180 = 2x+2x+x= 5x=180
therefore x= 36

therefore answer is B

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Re: In the figure above, point O is the center of the circle and [#permalink]

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Attached is a visual that should help.
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In the figure above, point O is the center of the circle and [#permalink]

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New post 20 Nov 2016, 08:43
Hi Engr2012,

I was about to post the same logic as you described.

what I like about this method is how easy you get to the answer.

For anyone else trying to answer this question and in search for optimization. Take a look at Engr2012 post and my graph.
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Re: In the figure above, point O is the center of the circle and [#permalink]

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New post 23 Mar 2017, 07:42
I loved this question!

take a look at the triangle properties: if 2 sides are equal-then angles too
so as oc=ac means that AOC=OAC
x or AOB=180-2y
as AC=AB -> ABC=ACB->CAB is x

this means that angles OAB is made up of x + x=y
so x=180-2(2x)->5x=180, x=36

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Re: In the figure above, point O is the center of the circle and [#permalink]

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New post 18 May 2017, 03:10
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


good explanation but I am not understand (the red angle=blue angle )?

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mkumar26 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


good explanation but I am not understand (the red angle=blue angle )?


OA and OB are radii of the same circle so they will be equal. So in triangle OAB, angle OAB = OBA ie. red angle = blue angle (in my diagram)
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New post 25 May 2017, 11:35
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Attachment:
Untitled.png
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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We know that OB and OA are radii of the circle, so their lengths are equal. And since OC = AC, we can denote their respective angles with x. For the supplementary angle of AC, we can use y and determine that 180-2x + y = 180 or 2x = y.

We also know that angle OAB = y, so x + 180 - 2y = y and thus x + 180 - 4x = 2x

-3x + 180 = 2x
180 = 5x

x = 36.

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Re: In the figure above, point O is the center of the circle and [#permalink]

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New post 30 Jul 2017, 09:10
Very tough question. A key is recognizing that since OB is a radius and OA is a radius therefore angle OAB = angle OBA. The rest is also very tricky. Attached is a diagram.
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Re: In the figure above, point O is the center of the circle and   [#permalink] 30 Jul 2017, 09:10

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