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

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This was a tough one...thanks to Bunuel as always for his genius. I figured a visual would help as well.

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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 3297 times ]

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

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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 2538 times ]

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

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

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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 1800 times ]

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

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@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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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:

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]

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Bunuel wrote:
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

Hope it's clear.

Hi, I thought we could not assume that the graphic is representative of the problem in a DS problem? But by using the Exterior Angle Theorem, we assume that AB is part of AC, don't we?

Thanks in advance, you have helped me a lot in my GMAT studies.
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Re: In the figure shown, point O is the center of the semicircle  [#permalink]

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tobnit wrote:

Hi, I thought we could not assume that the graphic is representative of the problem in a DS problem? But by using the Exterior Angle Theorem, we assume that AB is part of AC, don't we?

Thanks in advance, you have helped me a lot in my GMAT studies.

This is answered FOUR times on the previous page (last time in the post just above yours). I'm happy to answer again:

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]

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Quote:
So,∠BAO+∠ACO=2∗∠BAO+∠BOA=3∗∠BAO=60º

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

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erickrs wrote:
Quote:
So,∠BAO+∠ACO=2∗∠BAO+∠BOA=3∗∠BAO=60º

Not sure what to add to the below.

(1) The degree measure of angle COD is 60º:

$$\angle BAO +\angle ACO = \angle COD = 60º$$ degrees (Using exterior angle theorem)

$$\angle ACO = \angle CBO = 2* \angle BAO$$

$$So, \angle BAO + \angle ACO = 2* \angle BAO + \angle BOA = 3* \angle BAO = 60º$$

$$\angle BAO = 20º$$.
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In the figure shown, point O is the center of the semicircle  [#permalink]

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Bunuel wrote:
erickrs wrote:
Quote:
So,∠BAO+∠ACO=2∗∠BAO+∠BOA=3∗∠BAO=60º

Not sure what to add to the below.

(1) The degree measure of angle COD is 60º:

$$\angle BAO +\angle ACO = \angle COD = 60º$$ degrees (Using exterior angle theorem)

$$\angle ACO = \angle CBO = 2* \angle BAO$$

$$So, \angle BAO + \angle ACO = 2* \angle BAO + \angle BOA = 3* \angle BAO = 60º$$

$$\angle BAO = 20º$$.

Thank you for the reply Bunuel.

Now, I got the step $$\angle$$BAO + $$\angle$$ACO = 2* $$\angle$$BAO + $$\angle$$BOA

I see you're using the exterior angle theorem and at the same time substituting for the equal interior angle given it is isosceles:

$$\angle$$ACO = $$\angle$$BCO = $$\angle$$CBO = $$\angle$$OBC = 2* $$\angle$$BAO = 2* $$\angle$$BOA
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Re: In the figure shown, point O is the center of the semicircle  [#permalink]

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fozzzy wrote:
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!

Hi, where or how did you get 3x? as shown on the diagram, thanks
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Re: In the figure shown, point O is the center of the semicircle  [#permalink]

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[quote="Bunuel"] 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 $$\angle BCO = \angle CBO$$
$$\angle CBO = 2*\angle BAO$$ CA]] can someone explain this

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

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Hi Kp1703,
I hope you know the below property of triangles:-

An exterior angle of a triangle is equal to the sum of the opposite interior angles.

$$\angle{CBO}$$ is an external angle of the $$\triangle{ABO}$$. Opposite interior angles of $$\angle{CBO}$$ in the $$\triangle{ABO}$$ are $$\angle{BAO}$$ and $$\angle{BOA}$$.
Therefore, $$\angle{CBO}=\angle{BAO}+\angle{BOA}$$-----------------------(1)

Now, given AB=OC. As OC=OB=radius of circle. So, AB=OB.
Hence, the corresponding angles in the $$\triangle{BAO}$$ are equal. i.e,$$\angle{BAO}=\angle{BOA}$$------------(2)

So (1) becomes, $$\angle{CBO}=\angle{BAO}+\angle{BAO}$$=$$2*\angle{BAO}$$

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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PKN wrote:
Hi Kp1703,
I hope you know the below property of triangles:-

An exterior angle of a triangle is equal to the sum of the opposite interior angles.

$$\angle{CBO}$$ is an external angle of the $$\triangle{ABO}$$. Opposite interior angles of $$\angle{CBO}$$ in the $$\triangle{ABO}$$ are $$\angle{BAO}$$ and $$\angle{BOA}$$.
Therefore, $$\angle{CBO}=\angle{BAO}+\angle{BOA}$$-----------------------(1)

Now, given AB=OC. As OC=OB=radius of circle. So, AB=OB.
Hence, the corresponding angles in the $$\triangle{BAO}$$ are equal. i.e,$$\angle{BAO}=\angle{BOA}$$------------(2)

So (1) becomes, $$\angle{CBO}=\angle{BAO}+\angle{BAO}$$=$$2*\angle{BAO}$$

Hope it helps.

Thanks a lot

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

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Can someone explain why <CBO = 2* <BAO please?

Nevermind. Figured it out.
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 BCO is 40º:
$$\angle BCO=40º$$ --> $$\angle BCO = \angle CBO=40º = 2*\angle BAO$$ --> $$\angle BAO=20º$$.
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
http://gmatclub.com/forum/gmatprep-2-tr ... 76801.html
http://gmatclub.com:8080/forum/viewtopic.php?p=607910

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