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DS

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TriColor
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DS [#permalink] New post   04 Oct 2007, 11:42
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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Please, explain your answer...



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Re: DS [#permalink] New post   04 Oct 2007, 12:09
You have the right answer. But can you explain you you arrived to D?
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Re: DS [#permalink] New post   04 Oct 2007, 12:21
TriColor wrote:
You have the right answer. But can you explain you you arrived to D?

A bit lengthy explanation...
So please draw it into a page or it'll be very confusing.

Let OB=OC=AB=R ;
Let OAB =y= BOA ;
So ABO = 180-2y
So OBC = 2y ; OBC is equilateral tringle so BCO=2y
IN OBC Angle BOC = 180-4y

Now from 1st COD = 60
So COA =180-60 =120
Or COA=BOC + BOA = 180-4y + y = 180 -3y = 120
3y =60
y =20


From 2nd BCO =40
We have find out that BCO =2y
So y =20

Thats what we need to know..
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anonymousegmat
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Re: DS [#permalink] New post   04 Oct 2007, 23:41
rainbow wrote:
TriColor wrote:
You have the right answer. But can you explain you you arrived to D?

A bit lengthy explanation...
So please draw it into a page or it'll be very confusing.

Let OB=OC=AB=R ;
Let OAB =y= BOA ;
So ABO = 180-2y
So OBC = 2y ; OBC is equilateral tringle so BCO=2y
IN OBC Angle BOC = 180-4y

Now from 1st COD = 60
So COA =180-60 =120
Or COA=BOC + BOA = 180-4y + y = 180 -3y = 120
3y =60
y =20


From 2nd BCO =40
We have find out that BCO =2y
So y =20

Thats what we need to know..


your explanation is perfect. this is exactly what you should be posting!

my only question is how do you know triangle BOC is equilateral. I solved it b/c it is isoceles; how are you making the jump that chord BC=r
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Re: DS [#permalink] New post   05 Oct 2007, 00:09
anonymousegmat wrote:
rainbow wrote:
TriColor wrote:
You have the right answer. But can you explain you you arrived to D?

A bit lengthy explanation...
So please draw it into a page or it'll be very confusing.

Let OB=OC=AB=R ;
Let OAB =y= BOA ;
So ABO = 180-2y
So OBC = 2y ; OBC is equilateral tringle so BCO=2y
IN OBC Angle BOC = 180-4y

Now from 1st COD = 60
So COA =180-60 =120
Or COA=BOC + BOA = 180-4y + y = 180 -3y = 120
3y =60
y =20


From 2nd BCO =40
We have find out that BCO =2y
So y =20

Thats what we need to know..


your explanation is perfect. this is exactly what you should be posting!

my only question is how do you know triangle BOC is equilateral. I solved it b/c it is isoceles; how are you making the jump that chord BC=r


It is not........it is issoceles triangle 100..40..40
The explanation is ok just that the triangle is issoceles
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Re: DS [#permalink] New post   05 Oct 2007, 00:12
anonymousegmat wrote:
rainbow wrote:
TriColor wrote:
You have the right answer. But can you explain you you arrived to D?

A bit lengthy explanation...
So please draw it into a page or it'll be very confusing.

Let OB=OC=AB=R ;
Let OAB =y= BOA ;
So ABO = 180-2y
So OBC = 2y ; OBC is equilateral tringle so BCO=2y
IN OBC Angle BOC = 180-4y

Now from 1st COD = 60
So COA =180-60 =120
Or COA=BOC + BOA = 180-4y + y = 180 -3y = 120
3y =60
y =20


From 2nd BCO =40
We have find out that BCO =2y
So y =20

Thats what we need to know..


your explanation is perfect. this is exactly what you should be posting!

my only question is how do you know triangle BOC is equilateral. I solved it b/c it is isoceles; how are you making the jump that chord BC=r


It is not........it is issoceles triangle 100..40..40
The explanation is ok just that the triangle is issoceles
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Re: DS [#permalink] New post   05 Oct 2007, 07:35
At this risk of muddying the waters and furthermore, just being flat-out wrong...
Can't we just do this one by falling back on the exterior angle rule, ie, the measure of an exterior angle is = to the measure of the two opposite interior angles? Lines BO, CO and DO are radii and thus the same length. AB is = to OC, so it's the same length as a radii. O to the left edge of the semicircle (point not named) is also a radii. So once you have the measure of any one of the angles within the semicircle, you can back it out through the figure to give you the proper angle measurements...or so I thought. I picked D immediately on this question and thought it was a "gimme". :oops:
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Re: DS [#permalink] New post   05 Oct 2007, 07:53
TriColor wrote:
Please, explain your answer...


D.

Given AB=OC
You know that BO = OC
This means AB=BO=OC
Using the isso triangle property, you can say that
Angle BAO = Angle BOA

Using Line property, you also know that
(#2) Angle OBC = 2*BAO

Using isso triangle property, you can say that
Angle BCO = Angle OBC

This means that Angle COB = 180- 2*2*BAO
The final equation is:
BOA + 180 - 4*BOA + COD = 180
(#1) 180 -3*BOA + COD = 180

(1) is SUFFICIENT using (#1) equation
(2) is SUFFICIENT using (#2) equation
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rainbow
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Re: DS [#permalink] New post   05 Oct 2007, 16:47
anonymousegmat wrote:
rainbow wrote:
TriColor wrote:
You have the right answer. But can you explain you you arrived to D?

A bit lengthy explanation...
So please draw it into a page or it'll be very confusing.

Let OB=OC=AB=R ;
Let OAB =y= BOA ;
So ABO = 180-2y
So OBC = 2y ; OBC is equilateral tringle so BCO=2y
IN OBC Angle BOC = 180-4y

Now from 1st COD = 60
So COA =180-60 =120
Or COA=BOC + BOA = 180-4y + y = 180 -3y = 120
3y =60
y =20


From 2nd BCO =40
We have find out that BCO =2y
So y =20

Thats what we need to know..


your explanation is perfect. this is exactly what you should be posting!

my only question is how do you know triangle BOC is equilateral. I solved it b/c it is isoceles; how are you making the jump that chord BC=r


BOC is isoceles not equilateral. Sorry it was typo..



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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gmatclubot
Re: DS [#permalink]
05 Oct 2007, 16:47
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