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# The length of arc AXB is twice the length of arc BZC, and

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The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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03 Dec 2010, 20:08
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The length of arc AXB is twice the length of arc BZC, and the length of arc AYC is three times the length of arc AXB. What is the measure of angle BCA?

A. 20
B. 40
C. 60
D. 80
E. 120

My reasoning was:

AXB=2BCZ
AYC=3AXB
AYC=6BCZ

2BCZ+BCZ+3BCZ=360 Degrees

BCZ=40 degrees

AXB=80 degrees

AYC=240 degrees

Since AXB=80 and is opposite of angle BCA will be half of that (40)
At the same time, since BCZ is 40 degrees, BAC will be 20, and finally, since AYC is 240, ABC will be 120. I am not sure if this rule exists, it just made sense to me, and my answer of 40 degrees was correct. Am i right here?

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Re: Can this MGMAT Geometry question be solved this way?  [#permalink]

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04 Dec 2010, 03:53
8
6
MisterEko wrote:
Following is from MGMAT CAT:

The length of arc AXB is twice the length of arc BZC, and the length of arc AYC is three times the length of arc AXB. What is the measure of angle BCA?

1. 20
2. 40
3. 60
4. 80
5. 120

My reasoning was:

AXB=2BCZ
AYC=3AXB
AYC=6BCZ

2BCZ+BCZ+3BCZ=360 Degrees

BCZ=40 degrees

AXB=80 degrees

AYC=240 degrees

Since AXB=80 and is opposite of angle BCA will be half of that (40)
At the same time, since BCZ is 40 degrees, BAC will be 20, and finally, since AYC is 240, ABC will be 120. I am not sure if this rule exists, it just made sense to me, and my answer of 40 degrees was correct. Am i right here?

The red part is not correct.

When we are told that "the length of arc AXB is twice the length of arc BZC" means that $$\angle{AOB}=2*\angle{BOC}$$, where O is the center of the circle. Similarly "the length of arc AYC is three times the length of arc AXB." means that $$\angle{YOC}=3*\angle{AOB}$$

So if $$\angle{BOC}=x$$ then $$x+2x+3*2x=360$$ --> $$x=40$$ --> $$\angle{AOB}=2*\angle{BOC}=2*40=80$$. Now, according to The Central Angle Theorem the measure of inscribed angle is always half the measure of the central angle so $$\angle{AOB}=80=2*\angle{BCA}$$ --> $$\angle{BCA}=40$$.

To elaborate more on angles and arcs:
You should know that when you measure the length of an arc in degrees then it's the measure of corresponding central angle. For example $$\angle{BOC}=40$$ means that arcBZC is 40/360=1/9 th of the circumference. Next, $$\angle{BAC}$$, according to the central angle theorem, will be half the measure of the central angle, so $$\angle{BAC}=\frac{\angle{BOC}}{2}=20$$. As for $$\angle{BZC}$$: it'll be supplementary angle to $$\angle{BAC}$$ (supplementary angles are two angles that add up to 180°), so it equals to $$\angle{BZC}=180-\angle{BAC}=160$$.

For more chek Circles chapter of Math Book: math-circles-87957.html

Hope it helps.
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Re: Can this MGMAT Geometry question be solved this way?  [#permalink]

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03 Dec 2010, 21:47
1
Quote:
Since AXB=80 and is opposite of angle BCA will be half of that (40)

It's not quite because of the fact that it's opposite to BCA. Any arc will subtend an angle at the center - that's the one you've found, where AXB = 80. If the same arc subtends any angle on the circumference, then that angle is half of the angle subtended at the center, and hence BCA is 40 (Central Angle Theorem)

For more information on the properties of the circle, check out the Math Book's section on Circles
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Re: Can this MGMAT Geometry question be solved this way?  [#permalink]

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03 Dec 2010, 23:39
2
2
It should be solved like following
length of AXB + BZC + AYC = Perimeter of the circle

also, length of AXB = 2 BZC, therefore ½ AXB = BZC
also, length of AYC = 3 AXB

Hence, length AXB + ½ AXB + 3 AXB = 2πr
Θ/360 2πr + ½ Θ/360 2πr + 3 Θ/360 2πr = 2πr
2πr Θ/360 ( 1 + ½ + 3 ) = 2πr
Θ = (2πr*360*2)/(2πr*9)
= 80
And ∠BCA = ½ m (arc AXB)
= ½ * 80
= 40
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Re: Can this MGMAT Geometry question be solved this way?  [#permalink]

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04 Dec 2010, 05:02
Hey friend Bunuel
as u have mention

Quote:
When we are told that "the length of arc AXB is twice the length of arc BZC" means that , where O is the center of the circle. Similarly "the length of arc AYC is three times the length of arc AXB." means that

So if then --> --> . Now, according to The Central Angle Theorem the measure of inscribed angle is always half the measure of the central angle so --> .

You are taking Length of an arc and measure angle of an arc as same
Length of an arc is a part of the circumference and it is calculated with formula Θ/360 2πr
where Θ = the measure angle of the arc

Measure angle of the arc is the central angle, which arc makes with the center.

What mentioned here in the question is not the measure angle, which will total to 360; But rather, what mentioned is length of the arcs and that will be equal to circumference i.e. 2πr.
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Re: Can this MGMAT Geometry question be solved this way?  [#permalink]

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04 Dec 2010, 05:18
3
vyassaptarashi wrote:
Hey friend Bunuel
as u have mention

Quote:
When we are told that "the length of arc AXB is twice the length of arc BZC" means that , where O is the center of the circle. Similarly "the length of arc AYC is three times the length of arc AXB." means that

So if then --> --> . Now, according to The Central Angle Theorem the measure of inscribed angle is always half the measure of the central angle so --> .

You are taking Length of an arc and measure angle of an arc as same
Length of an arc is a part of the circumference and it is calculated with formula Θ/360 2πr
where Θ = the measure angle of the arc

Measure angle of the arc is the central angle, which arc makes with the center.

What mentioned here in the question is not the measure angle, which will total to 360; But rather, what mentioned is length of the arcs and that will be equal to circumference i.e. 2πr.

I had a little typo, which I corrected (initially I calculated x to be equal to 60 not to 40). But else is correct in my solution.

You can transform the arc length into the central angle measure. So $$arcAXB+arcBZC+arcAYC=2\pi{r}$$ can be written as $$\angle{AOB}+\angle{BOC}+\angle{AOC}=360$$ which I did.

For more chek Circles chapter of Math Book: math-circles-87957.html

Hope it's clear.
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Re: Can this MGMAT Geometry question be solved this way?  [#permalink]

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04 Dec 2010, 06:22
2
hey Bunuel

Quote:
You can transform the arc length into the central angle measure. So can be written as which I did.

For more chek Circles chapter of Math Book: math-circles-87957.html

Hope it's clear.

You are right, it can be converted into angle form. I actually solve it and then found u r right.
Thank you for making my concepts clearer to this next limit.

+1 for you
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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04 Mar 2013, 12:07
simply tooooooo gooodd Bunuel
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Re: Angle Measure from Length of Arcs  [#permalink]

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04 May 2013, 12:25
bharatdasaka wrote:
The length of arc AXB is twice the length of arc BZC, and the length of arc AYC is three times the length of arc AXB. What is the measure of angle BCA?

a) 20
b) 40
c) 60
d) 80
e) 120

Can Some one Please Explain this

Total length of the arcs = 2pi*r = AXB+BZC+AYC =$$\frac{9}{2}*AXB$$.Thus, as for arc length of 2pi*r the central angle measure is 360 degrees, thus for arc AXB, the angle measure = $$\frac{2}{9}*360$$ = 80 degrees. Thus, from the central angle theorem, the required angle measure is half of this = 40 degrees.

B.
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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09 May 2013, 21:04
1
Hi Buneul, Is my approach correct?
As question said Arc AXB is twice Arc BZC and Arc AYC is three times AXB.
Let,
BAC=x
ABC=6x
BCA=2x

Now, x+2x+6x=180 ie x=20.

Hence, Required angle BCA=2x ie 2*20= 40 degree.

Sorry for the poor writing....
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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10 May 2013, 01:07
atalpanditgmat wrote:
Hi Buneul, Is my approach correct?
As question said Arc AXB is twice Arc BZC and Arc AYC is three times AXB.
Let,
BAC=x
ABC=6x
BCA=2x

Now, x+2x+6x=180 ie x=20.

Hence, Required angle BCA=2x ie 2*20= 40 degree.

Sorry for the poor writing....

Yes, that's correct.
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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10 Jun 2013, 08:09
Hi Bunuel,

I would like you to clarify one below doubt.

It has been deduced above that :
$$\angle {AOB} = 80 \angle{BOC}= 40 and \angle {AOC}=240$$
Now, here I'm getting confused, how can angle {AOC} be 240, If I join, AO and OC, then in this case, it becomes a triangle, and the sum of all the angles of triangle is equal to 180, which is contradictory to the above finding.

Regards,
imhimanshu

Quote:

You can transform the arc length into the central angle measure. So $$arcAXB+arcBZC+arcAYC=2\pi{r}$$ can be written as $$\angle{AOB}+\angle{BOC}+\angle{AOC}=360$$ which I did.

For more chek Circles chapter of Math Book: math-circles-87957.html

Hope it's clear.
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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12 Jun 2013, 05:18
Bump. Can someone please respond to the above doubt.
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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12 Jun 2013, 05:27
imhimanshu wrote:
Bump. Can someone please respond to the above doubt.

O is the center of the circle.

It has been deduced above that :
$$\angle {AOB} = 80\angle{BOC}= 40and \angle {AOC}=240$$
Now, here I'm getting confused, how can angle {AOC} be 240, If I join, AO and OC, then in this case, it becomes a triangle, and the sum of all the angles of triangle is equal to 180, which is contradictory to the above finding.

I do not understand the red part: the sum of the angles of a triangle will always be 180, and this does not contradict the above finding, which refers to the angles that are made with the center O. I attached an image, I think that you are probably not considering the correct drawing of the thing...
Attachments

1479_TriangleinCircleQ(edit).gif [ 6.68 KiB | Viewed 11449 times ]

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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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12 Jun 2013, 05:43
However, I feel, I still have the same doubt. Let me try to explain again.

It has been deduced above that Angle AOC is 240 degrees.
Now, If I join AO and OC and AC; then it will become a Triangle with Angle AOC is 240, which is against the rules of Triangles.

Can you please tell me where I'm going wrong. I have no idea..

Regards,
imhimanshu
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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12 Jun 2013, 06:14
1
imhimanshu wrote:
However, I feel, I still have the same doubt. Let me try to explain again.

It has been deduced above that Angle AOC is 240 degrees.
Now, If I join AO and OC and AC; then it will become a Triangle with Angle AOC is 240, which is against the rules of Triangles.

Can you please tell me where I'm going wrong. I have no idea..

Regards,
imhimanshu

I see, below there is a correct picture of the circle.

In the above one, the proportions are not correct and you get a wrong idea of what sides you are connecting.
Attachments

Untitled.png [ 5.61 KiB | Viewed 11431 times ]

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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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12 Jun 2013, 07:48
Zarrolou wrote:

I see, below there is a correct picture of the circle.

In the above one, the proportions are not correct and you get a wrong idea of what sides you are connecting.

Hi Zarrolou,
Thanks for the explanation.
I have another follow up question, so does that mean that we should not trust the diagrams in PS as well. I know that trusting a figure in DS is a sin, but Is the same goes for PS as well.
In addition, can we do such reshuffle of the points as well, as we did in the this problem, as long as logic is permitting us to do so.

Regards,
imhimanshu
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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12 Jun 2013, 10:06
1
imhimanshu wrote:
Zarrolou wrote:

I see, below there is a correct picture of the circle.

In the above one, the proportions are not correct and you get a wrong idea of what sides you are connecting.

Hi Zarrolou,
Thanks for the explanation.
I have another follow up question, so does that mean that we should not trust the diagrams in PS as well. I know that trusting a figure in DS is a sin, but Is the same goes for PS as well.
In addition, can we do such reshuffle of the points as well, as we did in the this problem, as long as logic is permitting us to do so.

Regards,
imhimanshu

Actual figure looks as shown below:
Attachment:

Arcs.png [ 5.75 KiB | Viewed 11396 times ]

OG13, page150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.

So, the question is not designed according to the GMAT rules, it should have been mentioned that the figure is not drawn accurately.

Hope it helps.
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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12 Jun 2013, 23:17
imhimanshu wrote:
Zarrolou wrote:

I see, below there is a correct picture of the circle.

In the above one, the proportions are not correct and you get a wrong idea of what sides you are connecting.

Hi Zarrolou,
Thanks for the explanation.
I have another follow up question, so does that mean that we should not trust the diagrams in PS as well. I know that trusting a figure in DS is a sin, but Is the same goes for PS as well.
In addition, can we do such reshuffle of the points as well, as we did in the this problem, as long as logic is permitting us to do so.

Regards,
imhimanshu

Responding to a pm:

In addition, note that when you read the question, you can make out that the diagram is not to scale. AXB certainly doesn't look like twice of BZC. The moment you notice this, realize that you need to make your own diagram. As I have pointed out before, if they are mutilating the actual diagram, it means the actual diagram can make the question easier for you. So make the diagram on your own according to scale. That said, the actual GMAT will probably have more dependable diagrams.
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Re: The length of arc AXB is twice the length of arc BZC, and  [#permalink]

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27 Sep 2014, 04:58

Simple : mark all the sides related to all the arcs

AB=2x , BC=x,AC=6x

So angle opposite to AB is BCA = 40 (180/9x=20)= 20*2

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Re: The length of arc AXB is twice the length of arc BZC, and &nbs [#permalink] 27 Sep 2014, 04:58

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