Geometry Problem : Quant Question Archive [LOCKED]
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Geometry Problem

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20 Nov 2007, 07:33
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

This data suff. question is based on the diagram, which I've added in the attachment. Here is the question:

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

(1) The degree measure of angle COD is 60 degrees.
(2) The degree measure of angle BCO is 40 degrees.

I got really stuck with this question. Would you guys please explain each step that you have taken when working on this problem?
thanks!
Attachments

File comment: Semicircle
Semi Circle.doc [24.5 KiB]

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20 Nov 2007, 07:41
the OA is D, but how??????? i'm lost
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20 Nov 2007, 07:47
SEGEMNTS OC,OD,OB,AB are equal in lenght
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20 Nov 2007, 08:21
tarek99 wrote:
the OA is D, but how??????? i'm lost

But i am able to find the answer with 2nd statement alone.

So it should be B. Here is the solution

If angle BCO is 40 then angle OBC is also 40. The angle ABO is 180-40= 140

then in triangle ABO 140+angle BAO + angle BOA= 180

Since AB=BO-OC=OD
then angle BAO = angle BOA

So angle BAO = 20, what do u say....
is it B or not??
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20 Nov 2007, 08:26
no, the OA is D. this was from the gmatprep. i'm also lost and can't seem to figure out how exactly. if you can, i would love to know how.

regards,
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20 Nov 2007, 11:06
yezz, would you please explain how A is suff? I understand how B is, but i'm completely clueless with A. if COD is 60, that means that COA is 120. then where did you go from there? how can you determine BOA before reaching to BAO??
regards,
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21 Nov 2007, 04:37
anyone knows how option 1 is also suff.?
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21 Nov 2007, 05:39
Let angle OAB = x.

AB = CO = BO (both radii of semi-circle). Therefore Triangle ABO is isosceles.

Therefore AOB = x; ABO = 180-2x; OBC = 2x; BCO = 2x (Triangle BCO is isosceles); BOC = 180-4x

AOB + BOC + COD = 180

Therefore x + (180 - 4x) + 60 = 180; therefore x = 20.
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21 Nov 2007, 06:38
mantymooney, THANK YOU SO MUCH! i now understand this very well. i appreciate it!

regards
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21 Nov 2007, 09:55
tarek99 wrote:
anyone knows how option 1 is also suff.?

I dont think I would have initially guessed D. But spending about 7min on this in detail, I can see why D is the answer.

AB=OC=BO=OX=AX X is the midpoint of OA

So now we have two iscocoles triangles.

Lets name the angles with variables: OCB=y OBC=y BCO=z

BAO=x BOA=x ABO=w

(its obviously much easier to draw these)

S1: COD=60*

We need to write the appropriate eqautions out. 2y+z=180, 2x+w=180
y+w=180, Since we know COD=60* We know that x+z=120*

We have 4 equations and 4 unknowns. Should be enough. Lets solve anyway. We want to know what x=?

z=120-x 2y+120-x=180 --> 2y-x=60 x=2y-60.

w=180-2x --> y+180-2x=180--> y-2x=0 --> y=2x So x=2(2x)-60

-3x=-60 ---> x=20.

S2:
This one is very easy: we still have the same equations from above.

y+z=180, 2x+w=180, y+w=180

Since y=40* w=140* --> 2x=40* x=20*
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23 Nov 2007, 23:08
mantymooney wrote:
Let angle OAB = x.

AB = CO = BO (both radii of semi-circle). Therefore Triangle ABO is isosceles.

Therefore AOB = x; ABO = 180-2x; OBC = 2x; BCO = 2x (Triangle BCO is isosceles); BOC = 180-4x

AOB + BOC + COD = 180

Therefore x + (180 - 4x) + 60 = 180; therefore x = 20.

mantymooney - good approach!!!
23 Nov 2007, 23:08
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