Can anyone help solve this question ? Cannot seem to figure this out! Thanks
OA is B
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Lattefah84
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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?
And what is OA?
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linkinpark
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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
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
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ace_gre
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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!
@ 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
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Lattefah84
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but the x arc at your picture is not the same as on Brent Hanneson's picture. OA is really 30 degrees???ace_gre wrote:Can anyone help solve this question ? Cannot seem to figure this out! Thanks
OA is B
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Brent@GMATPrepNow
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I'm not sure what you mean. I used the same diagram in my solution.Lattefah84 wrote:but the x arc at your picture is not the same as on Brent Hanneson's picture. OA is really 30 degrees???ace_gre wrote:Can anyone help solve this question ? Cannot seem to figure this out! Thanks
OA is B
Brent Hanneson - Creator of GMATPrepNow.com
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Lattefah84
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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.Brent Hanneson wrote:
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