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

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

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New post 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

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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New post 01 Jul 2015, 00:24
honchos wrote:
Bunuel wrote:
Image

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

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New post 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.

Additional video explanation: https://www.youtube.com/watch?v=V2Gl8crxN90
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
Screen Shot 2015-07-28 at 12.47.32 PM.png [ 207.34 KiB | Viewed 1779 times ]


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

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New post 12 Sep 2015, 07:26
Bunuel wrote:
Image

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.

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

For more about the geometry issues check the links below.


Why angle is ACO = 2*BAO?
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Re: In the figure shown, point O is the center of the semicircle [#permalink]

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New post 13 Sep 2015, 02:12
reto wrote:
Bunuel wrote:
Image

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.

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

For more about the geometry issues check the links below.


Why angle is ACO = 2*BAO?


Please read the whole thread: in-the-figure-shown-point-o-is-the-center-of-the-semicircle-89662-20.html#p1544308
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 13 Sep 2015, 05:17
reto wrote:
Bunuel wrote:
Image

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.

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

For more about the geometry issues check the links below.


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, \(\angle OCB = \angle OBC = 2*\angle BAO\)

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

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New post 24 Dec 2015, 05:14
burnttwinky wrote:
Image
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
gmatprep.png [ 5.11 KiB | Viewed 1293 times ]


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

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New post 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

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

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New post 23 Sep 2016, 07:28
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burnttwinky wrote:
Image
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:
Image

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

So, ∆ABO is an isosceles triangle.
Image

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:
Image

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

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

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

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

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

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

Finally, since ∆ABO is an isosceles triangle, the other two angles must each be 20º
Image
As we can see, ∠BAO = 20º
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer =
[Reveal] Spoiler:
D


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

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New post 08 Aug 2017, 17:15
Very tricky, I attached a color coded chart for it with an explanation of the exterior angle theorem .
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Re: In the figure shown, point O is the center of the semicircle [#permalink]

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New post 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

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

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New post 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

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

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