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GMATPrep Qn.

by pavithra » Thu Oct 15, 2009 9:28 pm
Image

In the figure shown, O is the centre of the semicircle and points B,C & D lie on the semicircle. If the length of line segment AB = length of line segment OC. what is the degree measure of angle BAO?

1) Angle COD=60 degrees
2) Angle BCO=40 degrees

Could someone explain the solution please? ( Answer is D)

Thanks
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by ershovici » Fri Oct 16, 2009 10:40 am
Here is the explanetion:

Statement 1 - CO=BO so angle BCO=CBO=40. CA is the lineand has 180 grades, so angle OBA is 140, finally since OB=BA BOA=BAO=(180-40)/2 = 20
Statement 2 - since BCO is 40 degrees and has equall legs so OBC is also 40 degrees. As a result ABO is 140 degrees and AB equals with radius, so AB=BO, AOB=BAO and = 20 (since all three angles sum is 180)
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by pavithra » Fri Oct 16, 2009 5:29 pm
I still don't get the explanation for statement 1. Given COD=60 degees, how did you arrive at OCB=OBC=40 degrees??

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by pavithra » Sun Oct 18, 2009 7:05 am
Yo ppl....this problem is killing me. could anyone please help.

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by happybruce » Sun Oct 18, 2009 8:50 am
what we know: cbo = bco (isoscles triangle).
bao = boa (isocles triangle).

(1) we know that cbo = 2bao b/c of the the external angle rule.
120 = bao + (180-2cbo) ; cbo = 2bao.
120 = bao + (180 -4bao)
3bao = 60; bao = 20. So (1) is sufficient.

(2) much easier to solve; bco = 2 bao
40 = 2bao; bao = 20
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by pavithra » Sun Oct 18, 2009 9:35 am
Thanks a lot...that was eally helpful.

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