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The length of minor arc AB is twice the length of minor arc

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The length of minor arc AB is twice the length of minor arc [#permalink] New post 17 Dec 2007, 12:48
The length of minor arc AB is twice the length of minor arc BC and the length of minor arc AC is three times the length of minor arc AB. What is the measure of angle BCA?

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

I have no idea on where to begin. Please see below for the diagram. Thanks!!
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 [#permalink] New post 17 Dec 2007, 13:01
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I guess B. I just will try to draw...

....I've finished.

From question: AB=2BC and AC=3AB

1. Let x=BC, So, AB=2x and AC=6x.

2. Now, we can find angles for our minor arcs: 6x+2x+x=360 ==> x=40

3. angle AOC = 40+80=120. angle OAC=angle OCA=(180-120)/2=30

4. angle OCB=angle OBC=(180-40)/2=70

5. angle BCA= angle BCO - angle ACO=70-30=40
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 [#permalink] New post 17 Dec 2007, 15:19
There is a mistake in the question.

Quote:
length of minor arc AC is three times the length of minor arc AB


Since AC = 6x
AB=2x and
BC=x
Then 6x should be the MAJOR arc of AC and not the minor.
because the minor is equal to 3x

Walker's answer is correct as usual
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Re: Triangle and circle [#permalink] New post 17 Dec 2007, 16:09
pinal2 wrote:
The length of minor arc AB is twice the length of minor arc BC and the length of minor arc AC is three times the length of minor arc AB. What is the measure of angle BCA?

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

I have no idea on where to begin. Please see below for the diagram. Thanks!!


I guess the diagram in the question misleading. I am not sure if this was really attached with the question.
But any way the angles are in ratio 6:3:1 , and from that we can figure out the way Walker proceeded.
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 [#permalink] New post 17 Dec 2007, 16:15
The diagram is correct.
I believe the question has an error in it
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 [#permalink] New post 17 Dec 2007, 18:14
walker wrote:
I guess B. I just will try to draw...

....I've finished.

From question: AB=2BC and AC=3AB

1. Let x=BC, So, AB=2x and AC=6x.

2. Now, we can find angles for our minor arcs: 6x+2x+x=360 ==> x=40

3. angle AOC = 40+80=120. angle OAC=angle OCA=(180-120)/2=30

4. angle OCB=angle OBC=(180-40)/2=70

5. angle BCA= angle BCO - angle ACO=70-30=40


Same approach, but its not very hard to do. I just did it in my head. We have x+2x+6x=9x=180 x=20 We are looking for BCA or essentially AB

so 2*20=40.



Walker what program are you using to make those diagrams? is it possible that you could send it to me?
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 [#permalink] New post 17 Dec 2007, 18:53
I pasted the question and the diagram from MGMAT.

OA is indeed B, 40.

Great explanation guys.
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 [#permalink] New post 17 Dec 2007, 23:00
Expert's post
GMATBLACKBELT wrote:
Same approach, but its not very hard to do. I just did it in my head. We have x+2x+6x=9x=180 x=20 We are looking for BCA or essentially AB

so 2*20=40.


your approach is better!



GMATBLACKBELT wrote:
Walker what program are you using to make those diagrams? is it possible that you could send it to me?

Adobe PhotoShop
  [#permalink] 17 Dec 2007, 23:00
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