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

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

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Updated on: 01 Jul 2015, 01:03
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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 91511 times ]

Attachment:

Untitled.png [ 1.77 KiB | Viewed 107479 times ]

Originally posted by burnttwinky on 15 Dec 2009, 13:09.
Last edited by Bunuel on 01 Jul 2015, 01:03, edited 3 times in total.
Edited the question
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In the figure shown, point O is the center of the semicircle  [#permalink]

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15 Dec 2009, 14:04
19
38

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

Answer: D.

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
http://gmatclub.com/forum/gmatprep-2-tr ... 76801.html
http://gmatclub.com:8080/forum/viewtopic.php?p=607910

For more about the geometry issues check the links below.
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Re: In the figure shown, point O is the center of the semicircle  [#permalink]

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05 Nov 2013, 06:59
22
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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)

OB = OC ( radius)

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 101725 times ]

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

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18 Dec 2009, 20:46
1
<CBO=2<BAO ?? Why
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Re: In the figure shown, point O is the center of the semicircle  [#permalink]

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

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Updated on: 23 Oct 2013, 03:41
3
10
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 BCO is 40º.

Attachment:

exteriorAngle.GIF [ 14.09 KiB | Viewed 126241 times ]

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Originally posted by msunny on 19 Dec 2009, 08:28.
Last edited by Bunuel on 23 Oct 2013, 03:41, edited 3 times in total.
Edited the question
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In the figure shown, point O is the center of the semicircle  [#permalink]

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19 Dec 2009, 10:25
12
9

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

Answer: D.

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

For more about the geometry issues check the links below.
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Re: Exterior Angle in a Semi Circle  [#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: Exterior Angle in a Semi Circle  [#permalink]

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19 Dec 2009, 12:58
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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.

For more about the triangles check the link abot triangles below.
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Re: Exterior Angle in a Semi Circle  [#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: Exterior Angle in a Semi Circle  [#permalink]

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28 Oct 2010, 10:16
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: Exterior Angle in a Semi Circle  [#permalink]

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02 Nov 2010, 06:09
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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  [#permalink]

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26 Mar 2012, 05:43
Hi,
how did we know that points A,B, and C are collinear, and hence apply the theorem of the exterior angle ?
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Re: In the figure shown, point O is the center of the semicircle  [#permalink]

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26 Mar 2012, 06:02
imadkho wrote:
Hi,
how did we know that points A,B, and C are collinear, and hence apply the theorem of the exterior angle ?

Please read explanations given above: in-the-figure-shown-point-o-is-the-center-of-the-semicircle-89662.html#p811512

So, 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 these points are actually on the line in the order given on the diagram.
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In the figure shown, point O is the center of the semicircle and  [#permalink]

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01 Apr 2013, 15:40
10
Img always makes it pretty clear
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Re: In the figure shown, point O is the center of the semicircle  [#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  [#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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Re: In the figure shown, point O is the center of the semicircle  [#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

Answer: D.

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

For more about the geometry issues check the links below.

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  [#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

Answer: D.
.

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   [#permalink] 01 Jul 2015, 01:24

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