It is currently 12 Dec 2017, 15:54

### 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:

### Hide Tags

Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 599

Kudos [?]: 649 [0], given: 298

Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

30 Jun 2015, 19:07
Bunuel wrote:
In the figure shown, point O is the center of the semicircle and B, C, 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 60º.
(2) The degree measure of angle BCD is 40º.

Write down everything you know from the stem:

BO=CO=r=AB --> BOC and ABO are isosceles.
<BAO=<BOA and <BCO=<OBC
<CBO=2<BAO

(1) <BAO +<ACO=<COD=60 degrees (Using exterior angle theorem)
<ACO = <CBO = 2<BAO
So, <BAO + <ACO = 2<BAO + <BOA = 3* (<BAO) = 60 degrees
<BAO = 20 degrees. SUFFICIENT

(2) <BCO=40 degrees --> <BCO=<CBO=40 degrees=2<BAO --> <BAO=20 degrees. SUFFICIENT

More solutions at:
http://gmatclub.com:8080/forum/viewtopic.php?p=464547
http://gmatclub.com:8080/forum/viewtopic.php?p=398461
http://gmatclub.com:8080/forum/viewtopic.php?p=581082
gmatprep-2-triangle-semicircle-76801.html
http://gmatclub.com:8080/forum/viewtopic.php?p=607910

What property have you used Bunuel -
<CBO=2<BAO
_________________

Like my post Send me a Kudos It is a Good manner.
My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

Kudos [?]: 649 [0], given: 298

Math Expert
Joined: 02 Sep 2009
Posts: 42571

Kudos [?]: 135392 [0], given: 12691

Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

01 Jul 2015, 00:24
honchos wrote:
Bunuel wrote:

In the figure shown, point O is the center of the semicircle and B, C, 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 ?

Write down everything you know from the stem:

$$BO=CO=radius=AB$$ --> triangles BOC and ABO are isosceles.
$$\angle BAO = \angle BOA$$ and $$\angle BCO = \angle CBO$$
$$\angle CBO = 2*\angle BAO$$

(1) The degree measure of angle COD is 60º:
$$\angle BAO +\angle ACO = \angle COD = 60º$$ degrees (Using exterior angle theorem)
$$\angle ACO = \angle CBO = 2* \angle BAO$$
$$So, \angle BAO + \angle ACO = 2* \angle BAO + \angle BOA = 3* \angle BAO = 60º$$
$$\angle BAO = 20º$$.
SUFFICIENT

(2) The degree measure of angle BCD is 40º:
$$\angle BCO=40º$$ --> $$\angle BCO = \angle CBO=40º = 2*\angle BAO$$ --> $$\angle BAO=20º$$.
SUFFICIENT

.

What property have you used Bunuel -
<CBO=2<BAO

Triangle Exterior Angle Theorem: An exterior angle of a triangle is equal to the sum of the opposite interior angles.
$$\angle CBO = \angle BAO + \angle BOA$$ and since we know that $$\angle BAO = \angle BOA$$, then $$\angle CBO = 2*\angle BAO$$.

Hope it's clear.
_________________

Kudos [?]: 135392 [0], given: 12691

Senior Manager
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 439

Kudos [?]: 517 [0], given: 58

Location: United States (CA)
Age: 38
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GRE 1: 337 Q168 V169
WE: Education (Education)
In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

28 Jul 2015, 10:05
This was a tough one...thanks to Bunuel as always for his genius. I figured a visual would help as well.

Attachments

File comment: Any questions, comments, suggestions or improvements, please let me know. On the actual GMAT I would probably forego the angle signs to save me some time, since they are not really necessary and it already takes long enough to get all the angle names right.

Screen Shot 2015-07-28 at 12.47.32 PM.png [ 207.34 KiB | Viewed 1779 times ]

_________________

Harvard grad and 770 GMAT scorer, offering high-quality private GMAT tutoring, both in-person and online via Skype, since 2002.

GMAT Action Plan - McElroy Tutoring

Kudos [?]: 517 [0], given: 58

Retired Moderator
Joined: 29 Apr 2015
Posts: 889

Kudos [?]: 1915 [0], given: 302

Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

12 Sep 2015, 07:26
Bunuel wrote:

In the figure shown, point O is the center of the semicircle and B, C, 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 ?

Write down everything you know from the stem:

$$BO=CO=radius=AB$$ --> triangles BOC and ABO are isosceles.
$$\angle BAO = \angle BOA$$ and $$\angle BCO = \angle CBO$$
$$\angle CBO = 2*\angle BAO$$

(1) The degree measure of angle COD is 60º:
$$\angle BAO +\angle ACO = \angle COD = 60º$$ degrees (Using exterior angle theorem)
$$\angle ACO = \angle CBO = 2* \angle BAO$$
$$So, \angle BAO + \angle ACO = 2* \angle BAO + \angle BOA = 3* \angle BAO = 60º$$
$$\angle BAO = 20º$$.
SUFFICIENT

(2) The degree measure of angle BCD is 40º:
$$\angle BCO=40º$$ --> $$\angle BCO = \angle CBO=40º = 2*\angle BAO$$ --> $$\angle BAO=20º$$.
SUFFICIENT

More solutions at:
viewtopic.php?p=464547
viewtopic.php?p=398461
viewtopic.php?p=581082
gmatprep-2-triangle-semicircle-76801.html
viewtopic.php?p=607910

Why angle is ACO = 2*BAO?
_________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

Kudos [?]: 1915 [0], given: 302

Math Expert
Joined: 02 Sep 2009
Posts: 42571

Kudos [?]: 135392 [0], given: 12691

Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

13 Sep 2015, 02:12
reto wrote:
Bunuel wrote:

In the figure shown, point O is the center of the semicircle and B, C, 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 ?

Write down everything you know from the stem:

$$BO=CO=radius=AB$$ --> triangles BOC and ABO are isosceles.
$$\angle BAO = \angle BOA$$ and $$\angle BCO = \angle CBO$$
$$\angle CBO = 2*\angle BAO$$

(1) The degree measure of angle COD is 60º:
$$\angle BAO +\angle ACO = \angle COD = 60º$$ degrees (Using exterior angle theorem)
$$\angle ACO = \angle CBO = 2* \angle BAO$$
$$So, \angle BAO + \angle ACO = 2* \angle BAO + \angle BOA = 3* \angle BAO = 60º$$
$$\angle BAO = 20º$$.
SUFFICIENT

(2) The degree measure of angle BCD is 40º:
$$\angle BCO=40º$$ --> $$\angle BCO = \angle CBO=40º = 2*\angle BAO$$ --> $$\angle BAO=20º$$.
SUFFICIENT

More solutions at:
viewtopic.php?p=464547
viewtopic.php?p=398461
viewtopic.php?p=581082
gmatprep-2-triangle-semicircle-76801.html
viewtopic.php?p=607910

Why angle is ACO = 2*BAO?

_________________

Kudos [?]: 135392 [0], given: 12691

Current Student
Joined: 20 Mar 2014
Posts: 2671

Kudos [?]: 1789 [0], given: 796

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

13 Sep 2015, 05:17
reto wrote:
Bunuel wrote:

In the figure shown, point O is the center of the semicircle and B, C, 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 ?

Write down everything you know from the stem:

$$BO=CO=radius=AB$$ --> triangles BOC and ABO are isosceles.
$$\angle BAO = \angle BOA$$ and $$\angle BCO = \angle CBO$$
$$\angle CBO = 2*\angle BAO$$

(1) The degree measure of angle COD is 60º:
$$\angle BAO +\angle ACO = \angle COD = 60º$$ degrees (Using exterior angle theorem)
$$\angle ACO = \angle CBO = 2* \angle BAO$$
$$So, \angle BAO + \angle ACO = 2* \angle BAO + \angle BOA = 3* \angle BAO = 60º$$
$$\angle BAO = 20º$$.
SUFFICIENT

(2) The degree measure of angle BCD is 40º:
$$\angle BCO=40º$$ --> $$\angle BCO = \angle CBO=40º = 2*\angle BAO$$ --> $$\angle BAO=20º$$.
SUFFICIENT

More solutions at:
viewtopic.php?p=464547
viewtopic.php?p=398461
viewtopic.php?p=581082
gmatprep-2-triangle-semicircle-76801.html
viewtopic.php?p=607910

Why angle is ACO = 2*BAO?

Consider triangle BOC, OB =OC (both are radii of the semicircle).

You are given that AB = OC=OB ---> In triangle ABO, $$\angle AOB = \angle BAO$$ and $$\angle OBC$$ is the external angle ---> $$\angle OBC = \angle BAO + \angle BOA= 2 \angle BAO$$

Also, as OB = OC, in triangle BOC, $$\angle OCB = \angle OBC = 2*\angle BAO$$

Kudos [?]: 1789 [0], given: 796

Director
Joined: 10 Mar 2013
Posts: 589

Kudos [?]: 492 [0], given: 200

Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

24 Dec 2015, 05:14
burnttwinky wrote:

In the figure shown, point O is the center of the semicircle and points B, C, 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 60º.
(2) The degree measure of angle BCO is 40º.

[Reveal] Spoiler:
Attachment:
The attachment Semicirlce.GIF is no longer available

Attachment:
The attachment Untitled.png is no longer available

Hi Guys, here's an alternative approach. The graph is for statement 1, as statement 2 can be calculated easily.
Forget about the circle, the only information we need from it, is that the 3 sides as marked in my graph are equal and we have two isosceles triangles. Just draw a perpendicular line to the diameter and we get a 90° angle and using the external angle property we can derive that the angles of a small traingle are 0.5y (see attachement)
Now, we need to add all the given angles that are equal to 180°: $$90+30+y+0.5y=180$$, we get $$y=40$$ and the angle we're looking for is equal $$0.5y=20$$
Attachments

gmatprep.png [ 5.11 KiB | Viewed 1293 times ]

_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Kudos [?]: 492 [0], given: 200

Intern
Joined: 08 Apr 2016
Posts: 2

Kudos [?]: [0], given: 12

Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

31 Aug 2016, 07:54
we can solve the same question by joining CD and applying the property of a semi-circle... Angle inscribed in a semi circle is a right angle ..then in that case <ACD=90 ..and by using the either of the conditions ,we can get the value of <BAO

Kudos [?]: [0], given: 12

SVP
Joined: 11 Sep 2015
Posts: 1904

Kudos [?]: 2740 [2], given: 364

Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

23 Sep 2016, 07:28
2
KUDOS
Expert's post
Top Contributor
burnttwinky wrote:

In the figure shown, point O is the center of the semicircle and points B, C, 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 60º.
(2) The degree measure of angle BCO is 40º.

[Reveal] Spoiler:
Attachment:
Semicirlce.GIF

Attachment:
Untitled.png

Target question: What is the degree measure of ∠BAO?

Given: The length of line segment AB is equal to the length of line sement OC

Statement 1: The degree measure of angle COD is 60º
So, we have the following:

Since the radii must have equal lengths, we can see that OB = OC

So, ∆ABO is an isosceles triangle.

If we let ∠BAO = x degrees, then we can use the facts that ∆ABO is isosceles and that angles must add to 180º to get the following:

Since angles on a LINE must add to 180º, we know that ∠OBC = 2x

Now, we can use the facts that ∆BCO is isosceles and that the angles must add to 180º to get the following:

Finally, we can see that the 3 angles with blue circles around them are on a line.

So, they must add to 180 degrees.
We get: x + (180-4x) + 60 = 180
Simplify: 240 - 3x = 180
Solve to get: x = 20
In other words, ∠BAO = 20º
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The degree measure of angle BCO is 40º
So, we have the following:

Since the radii must have equal lengths, we can see that OB = OC

So, ∆BCO is an isosceles triangle, which means OBC is also 40º

Since angles on a line must add to 180 degrees, ∠ABO = 140º

Finally, since ∆ABO is an isosceles triangle, the other two angles must each be 20º

As we can see, ∠BAO = 20º
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

[Reveal] Spoiler:
D

RELATED VIDEO

_________________

Brent Hanneson – Founder of gmatprepnow.com

Kudos [?]: 2740 [2], given: 364

Manager
Joined: 31 Dec 2016
Posts: 91

Kudos [?]: 12 [0], given: 22

In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

08 Aug 2017, 17:15
Very tricky, I attached a color coded chart for it with an explanation of the exterior angle theorem .
Attachments

very difficult problem.png [ 74.76 KiB | Viewed 371 times ]

Kudos [?]: 12 [0], given: 22

Intern
Joined: 23 Jan 2017
Posts: 4

Kudos [?]: 5 [0], given: 5

Concentration: Operations, Finance
GMAT 1: 710 Q49 V38
GMAT 2: 730 Q50 V38
GPA: 3.54
Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

09 Aug 2017, 23:07
@bunuel
Are we not assuming ABC to be a straight line. This is not specified anywhere. If ABC is not a straight line then exterior angle theorem is invalid for triangle ABO

Sent from my SM-G900H using GMAT Club Forum mobile app

Kudos [?]: 5 [0], given: 5

Math Expert
Joined: 02 Sep 2009
Posts: 42571

Kudos [?]: 135392 [0], given: 12691

Re: In the figure shown, point O is the center of the semicircle [#permalink]

### Show Tags

09 Aug 2017, 23:23
kshitijrana37 wrote:
Bunuel
Are we not assuming ABC to be a straight line. This is not specified anywhere. If ABC is not a straight line then exterior angle theorem is invalid for triangle ABO

Sent from my SM-G900H using GMAT Club Forum mobile app

This is answered thrice on the previous page. I'm happy to answer again:

OFFICIAL GUIDE:

Problem Solving
Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

Data Sufficiency:
Figures:
• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2).
• Lines shown as straight are straight, and lines that appear jagged are also straight.
• The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero.
• All figures lie in a plane unless otherwise indicated.
_________________

Kudos [?]: 135392 [0], given: 12691

Re: In the figure shown, point O is the center of the semicircle   [#permalink] 09 Aug 2017, 23:23

Go to page   Previous    1   2   [ 32 posts ]

Display posts from previous: Sort by