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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 18 May 2015, 10:19
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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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Re: In the figure above, point O is the center of the circle and  [#permalink]

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New post 25 Jul 2015, 18:40
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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.

Merged the posts.
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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 08 Sep 2015, 00:40
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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, 10:17
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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New post 04 Oct 2015, 22:37
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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, 00:09
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, 09:36
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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New post 09 Nov 2016, 19:36
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Attached is a visual that should help.
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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 20 Nov 2016, 07:43
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, 06:42
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, 02:10
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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Re: In the figure above, point O is the center of the circle and  [#permalink]

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New post 18 May 2017, 04:03
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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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Re: In the figure above, point O is the center of the circle and  [#permalink]

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New post 25 May 2017, 10:35
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, 08:10
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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In the figure above, point O is the center of the circle and  [#permalink]

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New post Updated on: 16 Apr 2018, 11:27
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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


Since OC = AC, ∆AOC is an isosceles triangle, which means ∠OAC is also x°
Image

Since all 3 angles in ∆AOC must add to 180°, we can conclude that ∠OCA = (180-2x)°
Image

Since angles on a LINE must add to 180°, we can conclude that ∠ACB = 2x°
Image

Since AC = AB, ∆ACB is an isosceles triangle, which means ∠CBA is also 2x°
Image

Finally, since OA and OB are radii of the same circle, we know that ∆OAB is an isosceles triangle, which means ∠OABis also 2x°
Image

At this point, we can see that the 3 angles ∆OAB are x°, 2x° and 2x°
Since the angles in a triangle must add to 180°, we can write: x° + 2x° + 2x° = 180°
Simplify: 5x = 180
Solve: x = 36

Answer: B

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Originally posted by GMATPrepNow on 12 Oct 2017, 10:56.
Last edited by GMATPrepNow on 16 Apr 2018, 11:27, edited 1 time in total.
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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 20 Oct 2017, 16:58
[quote="VeritasPrepKarishma"]
Hi,

if OC = AC = AB, then how come the three angles are not equal? (it's illustrated as x, y, and y in your diagram versus all y's)

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

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

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New post 19 Mar 2018, 20:32
modernx wrote:
VeritasPrepKarishma wrote:
Hi,

if OC = AC = AB, then how come the three angles are not equal? (it's illustrated as x, y, and y in your diagram versus all y's)

Thanks



When two sides of a triangle are equal, the angles opposite them are equal. OC and AC form triangle OAC. The angles opposite to these two will be equal angle AOC = OAC. But we can't say what the measure of the equal angles is.

While AC and BC form triangle ABC. Angles opposite these will be equal angle BAC = CBA. But we can't say what the measure of the equal angles is.

Note that just because the triangles share a side AC, it doesn't mean the opposite angles will all be of the same measure. They are angles in different triangles. The length of the side does not give the angle.
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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 24 Apr 2018, 06:21
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

Image
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


OC=AC so its an isosceles triangle angle OAC = x. Now sum of angles AOC + OAC = ACB = 2x (An exterior angle of a triangle is equal to the sum of the opposite interior angles.). AC=AB so ABC=2x. Important point OA=OB radius of circle so OAB=OBA=2x. Equation of sum of angles of triangle OAB becomes -> x+2x+2x=180. 5x=180. x=36.

Answer: (B).
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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 28 Apr 2018, 04:20
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



VeritasPrepKarishma hello :)

cant understand the logic behind this (180 - 2x) + y = 180 (straight angle) so 2x = y

if you are deducting 2x from 180 then why are you adding y ? there are 2x and 2y right and how 2x = y :?

why triangle OAB an isosceles and not an equilateral :?

please explain:)

have a great wekkend :)
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Re: In the figure above, point O is the center of the circle and &nbs [#permalink] 28 Apr 2018, 04:20

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