Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 03 May 2015, 17:16

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
VP
Joined: 22 Nov 2007
Posts: 1105
Followers: 7

Kudos [?]: 193 [1] , given: 0

In the figure shown, point O is the center of the semicircle [#permalink]  10 Mar 2008, 10:43
1
KUDOS
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (00:00) wrong based on 5 sessions
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.

see file
Attachments

figure.doc [27 KiB]

Director
Joined: 10 Sep 2007
Posts: 949
Followers: 7

Kudos [?]: 205 [1] , given: 0

Re: gmatprep geo [#permalink]  10 Mar 2008, 12:39
1
KUDOS
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.
VP
Joined: 22 Nov 2007
Posts: 1105
Followers: 7

Kudos [?]: 193 [0], given: 0

Re: gmatprep geo [#permalink]  15 Mar 2008, 01:52
elmagnifico wrote:
D

OA is D but explain your choice always
Manager
Joined: 02 Mar 2008
Posts: 211
Concentration: Finance, Strategy
Followers: 1

Kudos [?]: 40 [1] , given: 1

Re: gmatprep geo [#permalink]  15 Mar 2008, 05:16
1
KUDOS
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
Intern
Joined: 02 Apr 2008
Posts: 37
Followers: 0

Kudos [?]: 11 [1] , given: 0

Re: gmatprep geo [#permalink]  06 Apr 2008, 01:16
1
KUDOS
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.
Intern
Joined: 02 Apr 2009
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

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

Thanks
Intern
Joined: 02 Mar 2009
Posts: 45
Location: Austin
Followers: 0

Kudos [?]: 2 [1] , given: 0

Re: gmatprep geo [#permalink]  18 May 2009, 05:20
1
KUDOS
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.
Intern
Joined: 04 Dec 2009
Posts: 24
Followers: 0

Kudos [?]: 10 [0], given: 13

Re: gmatprep geo [#permalink]  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
Intern
Joined: 04 Dec 2009
Posts: 24
Followers: 0

Kudos [?]: 10 [0], given: 13

Re: gmatprep geo [#permalink]  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
Similar topics Replies Last post
Similar
Topics:
41 In the figure shown, point O is the center of the semicircle 19 19 Dec 2009, 07:28
11 In the figure shown, point O is the center of the semicircle 8 15 Dec 2009, 12:09
3 In the figure shown, point O is the center of the semicircle 7 04 Oct 2009, 22:22
17 In the figure shown, point O is the center of semicircle and 17 17 Jul 2009, 14:00
1 In the figure shown, point O is the center of the semicircle 2 01 Jun 2006, 21:17
Display posts from previous: Sort by