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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.

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Re: In the figure shown, point O is the center of the semicircle
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24 Dec 2015, 05:14

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.

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

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In the figure shown, point O is the center of the semicircle
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Updated on: 16 Apr 2018, 11:49

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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.

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

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

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
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30 Jan 2018, 09:37

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

Answer: D. .

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.

Re: In the figure shown, point O is the center of the semicircle
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30 Jan 2018, 09:49

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:

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
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06 Aug 2018, 06:57

[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

WE: Supply Chain Management (Energy and Utilities)

In the figure shown, point O is the center of the semicircle
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06 Aug 2018, 07:35

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
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06 Aug 2018, 07:37

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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Re: In the figure shown, point O is the center of the semicircle &nbs
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