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In the figure shown, point O is the center of the semicircle

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In the figure shown, point O is the center of the semicircle [#permalink] New post 10 Mar 2008, 10:43
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In the figure shown, point O is the center of the semicircle and point points B, C
and D lie on the semicircle. If the length of line segment AB = OC, what is the
degree measure of angle BAO?

(1) angle COD = 60 degrees.
(2) angle BCO = 40 degrees.


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Re: gmatprep geo [#permalink] New post 10 Mar 2008, 12:39
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Sum of all the internal angles of a triangle is 180

So For triangle ACO we can write that
Angle CAO + Angle COA + Angle ACO = 180..(1)

Angle on a stright line is also 180 degree. So we can also write that
Angle COA + Angle COD = 180..(2)

Also Angle ACO is same as Angle BCO.

Since equation 1 and 2 have same RHS, so we can equate LHS. We have
Angle CAO + Angle COA + Angle ACO = Angle COA + Angle COD
As you can see Angle COA cancels out in both side, so we are left with
Angle CAO + Angle ACO = Angle COD => Angle CAO = Angle COD - Angle ACO = 60 - 40 = 20

Angle CAO is same as Angle BAO so it is also 20 degrees.

As both options are required, so I will go with C.

Last edited by abhijit_sen on 10 Mar 2008, 16:42, edited 1 time in total.
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Re: gmatprep geo [#permalink] New post 15 Mar 2008, 01:52
elmagnifico wrote:
D



OA is D but explain your choice always
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Re: gmatprep geo [#permalink] New post 15 Mar 2008, 05:16
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ABO and BOC are isosceles triangles (AB=BO=r, BO=OC=r)
Can derive CBO=BCO=2BAO
(2) is enough
COD=BCO+BAO=3BAO -> (1) is enough
=> D
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Re: gmatprep geo [#permalink] New post 06 Apr 2008, 01:16
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AlbertNTN wrote:
ABO and BOC are isosceles triangles (AB=BO=r, BO=OC=r)
Can derive CBO=BCO=2BAO


Could you please elaborate? Thank you.
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Re: gmatprep geo [#permalink] New post 11 May 2009, 22:03
Yes, can you please explain how you got CBO=BCO-2BAO?

Thanks
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Re: gmatprep geo [#permalink] New post 18 May 2009, 05:20
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Here is how I approached it:

bao = boa = x degrees (angles opposite equal sides in a triangle are equal)
obc = ocb = a degrees (angles opposite equal sides in a triangle are equal)

from triangle ABO
a +a+ (180-x) = 180 (sum of angles in a triangle is 180)
=> 2a = x

From triangle BCO:
x+x+[180-60-a] = 180
=> 2x + 120 -a = 180; 2x -a = 60; 4a-a = 60; a = 20.

So, statement I is suff.

Similarly, you can approach statement II.
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Re: gmatprep geo [#permalink] New post 25 Jan 2010, 20:34
I am still not very clear.

Here is what i infer

If radius of semicircle = x then OC=AB=OB=x

BOC and BAO are isoceles triangles

So Angle OBC = Angle OCB = y So OBC+OCB+COB=180 => 2y+COB=180

and Angle BAO = Angle BOA = z So 2z+ABO=180

We need to find BAO = z

From 1) COD =60 => COA=120 => y+z+120=180 => y+z=60 => Not sure how this is sufficient

From 2) BCO = 40 => BCO=y=40 => COB=100 => BOA =20 => BAO =z= 20 => Sufficient
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Re: gmatprep geo [#permalink] New post 25 Jan 2010, 20:46
gautamsubrahmanyam wrote:
I am still not very clear.

Here is what i infer

If radius of semicircle = x then OC=AB=OB=x

BOC and BAO are isoceles triangles

So Angle OBC = Angle OCB = y So OBC+OCB+COB=180 => 2y+COB=180

and Angle BAO = Angle BOA = z So 2z+ABO=180

We need to find BAO = z

From 1) COD =60 => COA=120 => y+z+120=180 => y+z=60 => Not sure how this is sufficient

From 2) BCO = 40 => BCO=y=40 => COB=100 => BOA =20 => BAO =z= 20 => Sufficient


I am clear now after re - reading the explanation given by "packem ".Thanks a lot for the explanation
Re: gmatprep geo   [#permalink] 25 Jan 2010, 20:46
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