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circle question

Expert replies
Source: — Problem Solving |

by Lattefah84 » Sat Jan 09, 2010 1:10 am
As I look at this picture, I would say x is 15. But I wouldn't know what is official method to solve this. This angle just looks too tiny for 30.

And what is OA?
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by linkinpark » Sat Jan 09, 2010 8:25 am
30 is my take
AB = BC = CD if you notice these are radii of circle so traingle ABC and ACD are equilaterals hence angle BAC = CAD = 60 and their sum = 120, now triangle ABD is iscosceles coz two sides are radii hence other angle will be x and 2x=60 so x=30

hope its clear
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by onedayi'll » Sat Jan 09, 2010 8:58 am
30 degree
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by ace_gre » Sat Jan 09, 2010 11:10 am
Thanks for all your responses..

@ linkinpark, one question... How do we know that BD is the bisector? From your explanation I see that ABC=60 degree, but how do we know that it is 2x? Thanks!
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by Brent@GMATPrepNow » Sat Jan 09, 2010 12:23 pm
Image
Brent Hanneson - Creator of GMATPrepNow.com
Image
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by Lattefah84 » Sat Jan 09, 2010 12:33 pm
ace_gre wrote:Can anyone help solve this question ? Cannot seem to figure this out! Thanks

OA is B
but the x arc at your picture is not the same as on Brent Hanneson's picture. OA is really 30 degrees???
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by Brent@GMATPrepNow » Sat Jan 09, 2010 12:35 pm
Lattefah84 wrote:
ace_gre wrote:Can anyone help solve this question ? Cannot seem to figure this out! Thanks

OA is B
but the x arc at your picture is not the same as on Brent Hanneson's picture. OA is really 30 degrees???
I'm not sure what you mean. I used the same diagram in my solution.
Brent Hanneson - Creator of GMATPrepNow.com
Image
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by Lattefah84 » Sat Jan 09, 2010 1:20 pm
Brent Hanneson wrote:Image
I think the area where arc x was marked is not the same where you marked it. I think of the arc contained of straight black line and full red line on the left from straight black line. If x is the arc as marked, it's too narrow for 30 degrees.
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by Brent@GMATPrepNow » Sat Jan 09, 2010 1:52 pm
Angle x is the same as angle ABD
Brent Hanneson - Creator of GMATPrepNow.com
Image
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by Jmx » Sat Jan 09, 2010 6:10 pm
I think Brent's pictures explain it greatly, but they can be confusing as they are not drawn to scale (same in the problem)... The bisector in an equilateral triangle should intersect the Base at its middle.
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by blaster » Mon Jan 11, 2010 5:00 am
x=30
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