circle question

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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linkinpark: Master | Next Rank: 500 Posts; Posts: 124; Joined: Tue Dec 22, 2009 8:32 am; Location: Karjat; Thanked: 3 times

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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onedayi'll: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Wed Jan 06, 2010 8:16 pm; Location: Chandigarh, INDIA

by onedayi'll » Sat Jan 09, 2010 8:58 am

30 degree

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ace_gre: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Mon Nov 23, 2009 2:30 pm; Thanked: 26 times; Followed by:1 members

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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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Sat Jan 09, 2010 12:23 pm

Brent Hanneson - Creator of GMATPrepNow.com

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Lattefah84: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Mon Dec 14, 2009 4:15 am

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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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

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.

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Lattefah84: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Mon Dec 14, 2009 4:15 am

by Lattefah84 » Sat Jan 09, 2010 1:20 pm

Brent Hanneson wrote:

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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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Sat Jan 09, 2010 1:52 pm

Angle x is the same as angle ABD

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Jmx: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Fri Jan 08, 2010 7:01 am; Thanked: 1 times

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.