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

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

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Updated on: 05 Oct 2019, 00:06
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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 [ 14.09 KiB | Viewed 105530 times ]

Attachment:

Untitled.png [ 1.77 KiB | Viewed 123986 times ]

Originally posted by burnttwinky on 15 Dec 2009, 13:09.
Last edited by Bunuel on 05 Oct 2019, 00:06, edited 4 times in total.
Edited the question
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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15 Dec 2009, 14:04
21
45

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 BCO is 40º:
$$\angle BCO=40º$$ --> $$\angle BCO = \angle CBO=40º = 2*\angle BAO$$ --> $$\angle BAO=20º$$.
SUFFICIENT

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

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05 Nov 2013, 06:59
36
10
You can solve this question quickly if you do everything up front. Look at the attached diagram

The portion marked in Red are equal => AB = OC ( given in the question stem)

Let Angle AOB = X

Statement 1 says COD = 60 = 3x ( as per diagram) => X=20

Statement 2 says BCO = 40 = 2x ( as per diagram) => X=20

Answer is D each statement is sufficient!
Attachments

circle triangle.jpg [ 17.29 KiB | Viewed 115695 times ]

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

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18 Dec 2009, 18:33
2
Given OC = AB
OC = OB since both are radii.
Therefore OB = AB.
Since angles opposite to equal sides are equal. Therefore,
Angle OCB = Angle OBC and Angle OAB = Angle BOA

Statement 1 Angle COD = 60 degrees. Not sufficient. Since we cannot determine angle COB or angle OCB.
Statement 2 Angle OCB = 40 degrees. Not sufficient Since we cannot determine angle COB

Both together helps us determine Angle COB and angle BOA = 180 - angle COD - angle COB
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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18 Dec 2009, 20:46
3
<CBO=2<BAO ?? Why
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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18 Dec 2009, 21:14
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1
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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
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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19 Dec 2009, 12:33
I still miss it. I saw the Circle Chapter from the Maths book as well.
Why is

<CBO=2<BAO
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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19 Dec 2009, 12:58
13
1
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.

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

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19 Dec 2009, 13:01
12
1
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.)
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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28 Oct 2010, 10:16
2
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!
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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02 Nov 2010, 06:09
1
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.
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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04 May 2011, 09:25
10
2
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 32548 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"
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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01 Apr 2013, 15:40
11
Img always makes it pretty clear
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Note_20130401_173652_01.jpg [ 86.68 KiB | Viewed 119717 times ]

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

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30 Jul 2013, 04:38
Where does it say that Points A, B and C lie on the same line? Assumption?
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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30 Jul 2013, 04:42
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.
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In the figure shown, point O is the center of the semicircle and point  [#permalink]

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30 Jun 2015, 20:07
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

What property have you used Bunuel -
<CBO=2<BAO
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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01 Jul 2015, 01:24
2
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.
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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28 Jul 2015, 11:05
1
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 10489 times ]

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

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24 Dec 2015, 06:14
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.

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 9707 times ]

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

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Updated on: 16 Apr 2018, 12:49
2
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

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Originally posted by GMATPrepNow on 23 Sep 2016, 08:28.
Last edited by GMATPrepNow on 16 Apr 2018, 12:49, edited 1 time in total.
Re: In the figure shown, point O is the center of the semicircle and point   [#permalink] 23 Sep 2016, 08:28

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