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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

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 \(\angle CBO = 2*\angle BAO\).

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 1576 times ]

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

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12 Sep 2015, 08:26

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

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

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

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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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.

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 1090 times ]

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

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31 Aug 2016, 08:54

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

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

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23 Sep 2016, 08:28

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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 [#permalink]

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10 Aug 2017, 00:07

@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

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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