Last visit was: 12 Aug 2024, 22:06 It is currently 12 Aug 2024, 22:06
Toolkit
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Manager
Joined: 25 Jun 2009
Posts: 197
Own Kudos [?]: 1617 [378]
Given Kudos: 4
Director
Joined: 29 Nov 2012
Posts: 576
Own Kudos [?]: 6200 [116]
Given Kudos: 543
Math Expert
Joined: 02 Sep 2009
Posts: 94906
Own Kudos [?]: 649110 [110]
Given Kudos: 86922
Math Expert
Joined: 02 Sep 2009
Posts: 94906
Own Kudos [?]: 649110 [21]
Given Kudos: 86922
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
16
Kudos
4
Bookmarks
msunny wrote:
I still miss it. I saw the Circle Chapter from the Maths book as well.
Why is

<CBO=2<BAO

First of all note that ABO is isosceles triangle. Why? Given that AB=OC, OC is radius, but OB is also radius, hence AB=OC=OB=r --> two sides in triangle ABO namely AB and BO are equal. Which means that angles BAO and BOA are also equal.

So we have <BAO=<BOA.

Next step: angle <CBO is exterior angle for triangle BAO. According to the exterior angle theorem: An exterior angle of a triangle is equal to the sum of the opposite interior angles. --> <CBO=<BAO+<BOA, as <BAO=<BOA --> <CBO=<BAO+<BAO=2<BAO.

Hope it's clear.

General Discussion
Intern
Joined: 12 Oct 2008
Posts: 29
Own Kudos [?]: 14 [6]
Given Kudos: 3
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
6
Kudos
<CBO=2<BAO ?? Why
Manager
Joined: 09 May 2009
Posts: 110
Own Kudos [?]: 1068 [18]
Given Kudos: 13
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
16
Kudos
1
Bookmarks
tashu wrote:
<CBO=2<BAO ?? Why

actually <CBO=<BAO+<BOA........(RULE EXT ANGLE OF A TRIANGL = SUM OF OPPO INT ANG)
<BAO=<BOA(ANGLES OF EQUAL SIDE)=2<BAO

HENCE <CAO=2<BAO
Intern
Joined: 21 Nov 2009
Status:Applying Now
Posts: 46
Own Kudos [?]: 591 [1]
Given Kudos: 3
WE:Project Management (Manufacturing)
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
1
Kudos
I still miss it. I saw the Circle Chapter from the Maths book as well.
Why is

<CBO=2<BAO
Manager
Joined: 25 Aug 2009
Posts: 72
Own Kudos [?]: 360 [15]
Given Kudos: 3
Location: Streamwood IL
Concentration: Finance
Schools:Kellogg(Evening),Booth (Evening)
Q51  V34
GPA: 3.4
WE 1: 5 Years
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
13
Kudos
2
Bookmarks
Let angle BAO=x
since AB=BO we have angle BOA=x
SInce Angle CBO is an exterior angle to BAo and BOA it is equal to the sum of their individual angles
Angle CBO = x+x=2x

BO and CO are the two radii hence they subtend equal angles thus BCO = 2x
and BOC = 180-4x
We need x
Statement 1 gives COD
Since COD+BOC+AOB = 180
60+180-4x+x=180
We cans olve for x - sufficient

Statement 2 gives
BCO = 2x = 40

we can calculate x hence sufficient.

Answer is D (Hope this is clear.)
Intern
Joined: 21 Aug 2010
Posts: 25
Own Kudos [?]: 58 [5]
Given Kudos: 6
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
5
Kudos
Hi,

What have I missed? I too marked this as C. GMAT Prep says its D, so fine, agreed. But here is my reasoning.
The reason is that it is not mentioned in the question that ABC is one single line. (points A,B,C all lie on the same line).

Why do we need to consider it as one straight line? If we do not consider them on the straight line, we cannot apply Exterior angle sum rule.
If we cannot apply, we will not be able to solve this without both stmt1 and stmt2.

Cheers!
Math Expert
Joined: 02 Sep 2009
Posts: 94906
Own Kudos [?]: 649110 [3]
Given Kudos: 86922
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
2
Kudos
1
Bookmarks
gmatretake wrote:
Hi,

What have I missed? I too marked this as C. GMAT Prep says its D, so fine, agreed. But here is my reasoning.
The reason is that it is not mentioned in the question that ABC is one single line. (points A,B,C all lie on the same line).

Why do we need to consider it as one straight line? If we do not consider them on the straight line, we cannot apply Exterior angle sum rule.
If we cannot apply, we will not be able to solve this without both stmt1 and stmt2.

Cheers!

Ian Stewart:

"In general, you should not trust the scale of GMAT diagrams, either in Problem Solving or Data Sufficiency. It used to be true that Problem Solving diagrams were drawn to scale unless mentioned otherwise, but I've seen recent questions where that is clearly not the case. So I'd only trust a diagram I'd drawn myself. ...

Here I'm referring only to the scale of diagrams; the relative lengths of line segments in a triangle, for example. ... You can accept the relative ordering of points and their relative locations as given (if the vertices of a pentagon are labeled ABCDE clockwise around the shape, then you can take it as given that AB, BC, CD, DE and EA are the edges of the pentagon; if a line is labeled with four points in A, B, C, D in sequence, you can take it as given that AC is longer than both AB and BC; if a point C is drawn inside a circle, unless the question tells you otherwise, you can assume that C is actually within the circle; if what appears to be a straight line is labeled with three points A, B, C, you can assume the line is actually straight, and that B is a point on the line -- the GMAT would never include as a trick the possibility that ABC actually form a 179 degree angle that is imperceptible to the eye, to give a few examples).

So don't trust the lengths of lines, but do trust the sequence of points on a line, or the location of points within or outside figures in a drawing. "

Hope it helps.
Retired Moderator
Joined: 20 Dec 2010
Posts: 1107
Own Kudos [?]: 4767 [14]
Given Kudos: 376
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
11
Kudos
3
Bookmarks
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
D

Sol:

Attachment:

Semicircle_And_Triangle.PNG [ 12.23 KiB | Viewed 114116 times ]

From the stem:
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"
Intern
Joined: 10 Apr 2012
Posts: 15
Own Kudos [?]: 120 [11]
Given Kudos: 0
Concentration: Finance, Economics
GMAT 1: 760 Q50 V44
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
11
Kudos
Img always makes it pretty clear
Attachments

Note_20130401_173652_01.jpg [ 86.68 KiB | Viewed 200898 times ]

Manager
Joined: 18 Dec 2012
Posts: 64
Own Kudos [?]: 163 [0]
Given Kudos: 56
Location: India
Concentration: General Management, Strategy
GMAT 1: 530 Q37 V25
GMAT 2: 660 Q49 V32
GPA: 3.32
WE:Manufacturing and Production (Manufacturing)
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
Where does it say that Points A, B and C lie on the same line? Assumption?
Math Expert
Joined: 02 Sep 2009
Posts: 94906
Own Kudos [?]: 649110 [2]
Given Kudos: 86922
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
2
Bookmarks
Qoofi wrote:
Where does it say that Points A, B and C lie on the same line? Assumption?

OG13, page 272:
A figure accompanying a data sufficiency problem will 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 can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.

Hope it helps.
Math Expert
Joined: 02 Sep 2009
Posts: 94906
Own Kudos [?]: 649110 [12]
Given Kudos: 86922
In the figure shown, point O is the center of the semicircle and point [#permalink]
9
Kudos
3
Bookmarks
Bunuel 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.[/i]

Attachment:
The attachment GMAT1.jpg is no longer available

VERITAS PREP OFFICIAL SOLUTION:

That is a complicated-looking figure. Your instinct might be that you don’t have time to draw it, but these kinds of questions will be designed specifically to thwart our intuition if we attempt to do too much work in our heads. So the first thing to do is draw the figure on our scratch pad, and mark the relationships we’re given. We’re told that segment CO is equal to AB, so we’ll designate that relationship. We’ll also call angle BAO, which we’re asked about, ‘x.’ Now we have the following:
Attachment:

GMAT2.jpg [ 18.14 KiB | Viewed 64147 times ]

Fight the impulse to jump to the statements now. In a harder question like this, we’ll benefit from taking more time to derive additional relationships from the question stem. Psychologically, this is often a struggle for test-takers. You’re conscious of your time constraint. You want to work quickly. The trick is to trust that this pre-statement investment of time will allow you to evaluate the information provided in the statements more efficiently, ultimately saving time.

Now the name of the game is to try to label as much of this figure as we can without introducing a new variable. Notice that segments CO and BO are both radii of the circle, so we know those are equal. Our diagram now looks like this:
Attachment:

GMAT3.jpg [ 18.4 KiB | Viewed 64139 times ]

Next, look at triangle ABO. Notice that segments AB and BO are equal. If angles opposite equal angles are equal to each other, we can then designate angle AOB as ‘x’ because it must be equal to angle BAO, as those two angles are opposite sides that are of equal length. Moreover, if the three interior angles of a triangle will sum to 180, the remaining angle, ABO, can be designated 180-2x. This gives us the following.
Attachment:

GMAT4.jpg [ 18.91 KiB | Viewed 64101 times ]

No reason to stop here. Notice that angles ABO and CBO lie on a line. Angles that lie on a line must sum to 180. If angle ABO is 180-2x, then angle CBO must be 2x. Now we have this:
Attachment:

GMAT5.jpg [ 19.62 KiB | Viewed 64099 times ]

Analyzing triangle CBO, we see that sides BO and CO are equal, meaning that the angles opposite those sides must be equal. So now we can label angle BCO as ‘2x.’ If angles CBO and BOC sum to 4x, the remaining angle, BOC, must then be 180-4x, so that the interior angles of the triangle will sum to 180.
Attachment:

GMAT6.jpg [ 19.59 KiB | Viewed 64064 times ]

We’ve got enough at this point that we can very quickly evaluate our statements, However, there is one last interesting relationship. Notice that angle COD is an exterior angle of triangle CAO. An exterior angle, by definition, must be equal to the sum of the two remote interior angles. So, in this case, Angle COD is equal to the sum of angles BCO and BAO. Therefore COD = 2x + x = 3x, which I’ve circled in the figure. (Triangle CAO is outlined in blue in the figure below to more clearly demarcate the exterior angle.)

Attachment:

GMAT7.jpg [ 20.77 KiB | Viewed 64063 times ]

That’s a lot of work. Determining all of these relationships will likely take close to two minutes. But watch how quickly we can evaluate our statements if we’ve done all of this preemptive groundwork:

Statement 1: Angle COD = 60. We’ve designated angle COD as 3x, so 3x = 60. Clearly we can solve for x. Sufficient. Eliminate BCE.

Statement 2: Angle BCO = 40. We’ve designated angle BCO as 2x, so 2x = 40. Clearly we can solve for x. Sufficient. Answer is D.

Notice, all of the heavy lifting for this question came before we even so much as glanced at our statements.

Takeaway: For a challenging Data Sufficiency question in which you’re given a lot of information in the question stem, the best approach is to spend some time taming the complexity of the problem before examining the statements. When you work out these relationships, try to minimize the number of variables you use when doing so, as this will simplify your calculations once you’re ready to go to the statements. Most importantly, don’t do too much work in your head. There’s no need to rely on the limited bandwidth of your working memory if you have the option of putting everything into a concrete form on your scratch pad.
Senior Manager
Joined: 17 Apr 2013
Status:Verbal Forum Moderator
Posts: 360
Own Kudos [?]: 2250 [0]
Given Kudos: 298
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
In the figure shown, point O is the center of the semicircle and point [#permalink]
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

What property have you used Bunuel -
<CBO=2<BAO
Math Expert
Joined: 02 Sep 2009
Posts: 94906
Own Kudos [?]: 649110 [2]
Given Kudos: 86922
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
2
Kudos
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.
Tutor
Joined: 10 Jul 2015
Status:Expert GMAT, GRE, and LSAT Tutor / Coach
Affiliations: Harvard University, A.B. with honors in Government, 2002
Posts: 1182
Own Kudos [?]: 2444 [1]
Given Kudos: 274
Location: United States (CO)
Age: 44
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GMAT 4: 730 Q48 V42 (Online)
GRE 1: Q168 V169

GRE 2: Q170 V170
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
1
Kudos
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 91015 times ]

Current Student
Joined: 10 Mar 2013
Posts: 355
Own Kudos [?]: 2751 [0]
Given Kudos: 200
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.7
WE:Marketing (Telecommunications)
Re: In the figure shown, point O is the center of the semicircle and point [#permalink]
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.

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 89981 times ]

GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6804
Own Kudos [?]: 30998 [11]
Given Kudos: 799
In the figure shown, point O is the center of the semicircle and point [#permalink]
6
Kudos
5
Bookmarks
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.

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

RELATED VIDEO

Originally posted by BrentGMATPrepNow on 23 Sep 2016, 08:28.
Last edited by BrentGMATPrepNow on 29 Jun 2021, 06:51, edited 2 times in total.
In the figure shown, point O is the center of the semicircle and point [#permalink]
1   2
Moderator:
Math Expert
94906 posts