It is currently 28 Jun 2017, 09:35

# Live Now:

Open Admission Chat with MBA Expert, Donna Bauman. (Kellogg MBA Graduate and Former UNC Adcom)

### 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
Your Progress

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 semicircle and

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

VP
Joined: 08 Apr 2009
Posts: 1183
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
In the figure shown, point O is the center of semicircle and [#permalink]

### Show Tags

17 Jul 2009, 15:00
3
This post received
KUDOS
11
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

61% (01:22) correct 39% (00:47) wrong based on 182 sessions

### HideShow timer Statistics

In the figure shown, point O is the center of 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
(2) The degree measure of angle BCO is 40

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-shown-point-o-is-the-center-of-the-semicircle-88009.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jul 2012, 05:37, edited 2 times in total.
Edited the question and added the OA.
Current Student
Joined: 18 Jun 2009
Posts: 356
Location: San Francisco
Schools: Duke,Oxford,IMD,INSEAD
Re: semicircle and triangle [#permalink]

### Show Tags

17 Jul 2009, 15:37
1
This post received
KUDOS
1
This post was
BOOKMARKED
Assume angle OBC=y and angle BAO=z

Using (1)
=========
y = 2z ---(i)(exterior angle is sum of opp interior angles)

z+(180-2y)+60=180 ---(ii) (sum of angle AOB, BOC and COD is 180)
Solving (i) and (ii) we get z=20
Hence sufficient

Using (2)
=========
Angle BCO = y = 40
==> z=40/2 = 20
Hence sufficient

Answer (D).
Manager
Joined: 17 Dec 2007
Posts: 101
Re: semicircle and triangle [#permalink]

### Show Tags

17 Jul 2009, 15:38
1
This post received
KUDOS
asimov wrote:

In the figure shown, point O is the center of 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
(2) The degree measure of angle BCO is 40

I think 2 alone is sufficient

Given data : AB = OC which implies = BO.

So In triangle ABO, AB = BO , isoceles triangle hence opposite angles BOA =angle BAO
similarly triangle

We know BO=OC and hence triangle BOC is also isoceles. So if angle BCO = 40, then angle OBC =40 and hence angle ABO = 180-40 = 120

once angle ABO is known, we know angle BOA = BAO, hence 2 angle BAO+ angle ABO = 180
2 BAO = 60 and hence BAO is 30
VP
Joined: 08 Apr 2009
Posts: 1183
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
Re: semicircle and triangle [#permalink]

### Show Tags

18 Jul 2009, 07:32
hi gmanjesh, can you please help us understand why
y=2z
how do you know BAO and BOA have the same angle?

I'm assuming that's the same assumption you used to assume OBC and OCB are equal.
Manager
Joined: 17 Dec 2007
Posts: 101
Re: semicircle and triangle [#permalink]

### Show Tags

18 Jul 2009, 08:25
1
This post received
KUDOS
asimov wrote:
hi gmanjesh, can you please help us understand why
y=2z
how do you know BAO and BOA have the same angle?

I'm assuming that's the same assumption you used to assume OBC and OCB are equal.

Hi Asimov

triangle ABO is isoceles with OB and AB are equal sides, hence the opposite angles BAO and BOA are equal
Current Student
Joined: 18 Jun 2009
Posts: 356
Location: San Francisco
Schools: Duke,Oxford,IMD,INSEAD
Re: semicircle and triangle [#permalink]

### Show Tags

18 Jul 2009, 08:42
3
This post received
KUDOS
alwynjoseph wrote:
asimov wrote:
hi gmanjesh, can you please help us understand why
y=2z
how do you know BAO and BOA have the same angle?

I'm assuming that's the same assumption you used to assume OBC and OCB are equal.

Hi Asimov

triangle ABO is isoceles with OB and AB are equal sides, hence the opposite angles BAO and BOA are equal

thanks alwynjoseph for explaining it

Asimov OB and OC are the radius and hence they are euqal and as per the question AB=OC therefore AB=OC=OB
VP
Joined: 08 Apr 2009
Posts: 1183
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
Re: semicircle and triangle [#permalink]

### Show Tags

18 Jul 2009, 09:17
thanks a lot. I didn't catch on the radius -> isoceles triangle. it makes sense to me now.

the OA is D
Intern
Joined: 20 Jul 2009
Posts: 1
Re: semicircle and triangle [#permalink]

### Show Tags

20 Jul 2009, 05:05
Hi, guys. If I follow your explanation correctly, I think you're assuming that AC is a straight line and that ACO is a triangle.

If that is what you're doing, is there anything in the question setup that says we can assume that AC is a straight line?

Thanks.
Manager
Joined: 21 Jul 2009
Posts: 249
Location: New York, NY
Re: semicircle and triangle [#permalink]

### Show Tags

27 Aug 2009, 13:24
gmanjesh wrote:

y = 2z ---(i)(exterior angle is sum of opp interior angles)

I don't understand this part. Is anyone able to illustrate please?
Intern
Joined: 14 Apr 2010
Posts: 5
Re: semicircle and triangle [#permalink]

### Show Tags

21 Apr 2010, 19:09
I understand how statement II is sufficient, but I am having a hard time with statement I. Can someone please help me out?

One thing that seems unexplained with the statement I work done thus far:

I agree that angle BAO = BOA, but in superman's equation he assumed that angle BOC = BOA. How do we know that?
Intern
Joined: 23 May 2010
Posts: 2
Re: semicircle and triangle [#permalink]

### Show Tags

23 May 2010, 16:25
3
This post received
KUDOS
ok I found out why y = 2z.

If angle BAO = x, then angle BOA = x (because triangle BAO is Isosceles)

Therefore, angle ABO = 180 - 2x (because the sum of all the angles of the triangle is 180)

Since line segment AC is a straight line that means ABO + CBO = 180.

If you substitute

ABO = 180 - 2x

into

ABO + CBO = 180

CBO = 2x. Then BOC = 180 - 4x. (because triangle BOC is isosceles)
Manhattan GMAT Instructor
Joined: 28 Aug 2009
Posts: 152
Location: St. Louis, MO
Schools: Cornell (Bach. of Sci.), UCLA Anderson (MBA)
Re: semicircle and triangle [#permalink]

### Show Tags

19 Jun 2010, 10:03
grifter000 wrote:
Hi, guys. If I follow your explanation correctly, I think you're assuming that AC is a straight line and that ACO is a triangle.

If that is what you're doing, is there anything in the question setup that says we can assume that AC is a straight line?

Thanks.

Good general question. You should almost assume that DS diagrams are not drawn to scale. But there are things about the picture that you can take as fact:

(1) If two lines look like they meet, they do. (i.e. no microscopic gaps)
(2) If a line looks straight, it is. (i.e. no trick angles of 179.99 degrees)
_________________

Emily Sledge | Manhattan GMAT Instructor | St. Louis

Manhattan GMAT Discount | Manhattan GMAT Course Reviews | Manhattan GMAT Reviews

Director
Joined: 25 Aug 2007
Posts: 937
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Re: semicircle and triangle [#permalink]

### Show Tags

22 Jun 2010, 11:33
IMO D.

This is simple test of triangle basics. Thanks for sharing this question.
_________________

Want to improve your CR: http://gmatclub.com/forum/cr-methods-an-approach-to-find-the-best-answers-93146.html
Tricky Quant problems: http://gmatclub.com/forum/50-tricky-questions-92834.html
Important Grammer Fundamentals: http://gmatclub.com/forum/key-fundamentals-of-grammer-our-crucial-learnings-on-sc-93659.html

Manager
Joined: 30 Mar 2010
Posts: 83
GMAT 1: 730 Q48 V42
Re: semicircle and triangle [#permalink]

### Show Tags

26 Nov 2010, 08:15
Ok so I understand that BOA = BAO and if BOA = 2z then OBC = z. But I thought that given that OC = AB = OB then BOA = BAO = OBC?

Where is my mistake??
Intern
Joined: 03 Jun 2009
Posts: 49
Re: semicircle and triangle [#permalink]

### Show Tags

27 Nov 2010, 08:42
superman wrote:
Assume angle OBC=y and angle BAO=z

Using (1)
=========
y = 2z ---(i)(exterior angle is sum of opp interior angles)

z+(180-2y)+60=180 ---(ii) (sum of angle AOB, BOC and COD is 180)
Solving (i) and (ii) we get z=20
Hence sufficient

Using (2)
=========
Angle BCO = y = 40
==> z=40/2 = 20
Hence sufficient

Answer (D).

Gud Xplanation Superman!!! Kudos +1
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010
Re: semicircle and triangle [#permalink]

### Show Tags

04 May 2011, 09:25
7
This post received
KUDOS
2
This post was
BOOKMARKED
asimov wrote:
In the figure shown, point O is the center of 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
(2) The degree measure of angle BCO is 40
[Reveal] Spoiler: Official Answer
D

Sol:

Please find image herewith attached.

Attachment:

Semicircle_And_Triangle.PNG [ 12.23 KiB | Viewed 19253 times ]

From the stem:
AB=OC=OB=Radius
Thus, ABO and BOC are both isosceles triangles.

$$m\angle{BAO}=m\angle{BOA}=x^{\circ}$$ [Note: OB=AB]

$$m\angle{OBC}=m\angle{OCB}=y^{\circ}$$ [Note: OB=OC]

Theorem:
An exterior angle of a triangle is equal to the sum of its interior opposite angles.

$$m\angle{CBO}=m\angle{BAO}+m\angle{BOA}$$

$$y=x+x=2x$$----------------------1

Q: What is x?

1.
$$\angle{COD}=60^{\circ}=t$$

Theorem:
Sum of three angles of a triangle is $$180^{\circ}$$

$$m\angle{OCB}+m\angle{OBC}+m\angle{COB}=180$$

$$y+y+z=180$$

$$z=180-2y$$---------------------2

Using 1 and 2:

$$z=180-2(2x)$$

$$z=180-4x$$-----------------3

Theorem:
Angles on one side of a straight line will always add to $$180^{\circ}$$

$$m\angle{COD}+m\angle{BOA}+m\angle{COB}=180^{\circ}$$

$$60+x+z=180$$

$$z=120-x$$----------------4

Using equations 3 and 4:

$$180-4x=120-x$$

$$3x=60$$

$$x=20^{\circ}$$

Sufficient.

2.

$$m\angle{OCB}=40^{\circ}=y=m\angle{CBO}$$

$$y=40^{\circ}$$ ----------------------5

Using 1 and 5:

$$2x=40$$

$$x=20$$

Sufficient.

Ans: "D"

_________________
Manager
Joined: 18 Sep 2010
Posts: 53
Re: semicircle and triangle [#permalink]

### Show Tags

04 May 2011, 09:46
woooow!
fluke, your solution is very helpful
thank you so much!
_________________

(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Manager
Joined: 11 Jul 2009
Posts: 167
WE: Design (Computer Software)
Re: semicircle and triangle [#permalink]

### Show Tags

06 May 2011, 09:00
each statement alone is sufficient, so... D
_________________

Kaustubh

Re: semicircle and triangle   [#permalink] 06 May 2011, 09:00
Similar topics Replies Last post
Similar
Topics:
14 In the figure shown, point O is the center of the circle and points B, 9 10 Dec 2016, 16:01
1 In the Figure above, O is the center of a semicircle, and p 3 13 May 2016, 05:21
140 In the figure shown, point O is the center of the semicircle 28 23 Sep 2016, 08:28
5 In the figure shown, point O is the center of the semicircle 7 17 Jan 2013, 07:22
8 In the figure shown, point O is the center of the semicircle and B, C, 6 02 Oct 2014, 03:26
Display posts from previous: Sort by

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.