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

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SamBoyle

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08 Aug 2017, 18:15

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Very tricky, I attached a color coded chart for it with an explanation of the exterior angle theorem .

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very difficult problem.png [ 74.76 KiB | Viewed 9517 times ]

kshitijrana37

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10 Aug 2017, 00:07

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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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10 Aug 2017, 00:23

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kshitijrana37

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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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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25 Feb 2018, 19:15

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Quote:

So,∠BAO+∠ACO=2∗∠BAO+∠BOA=3∗∠BAO=60º

I just don't get this step in the first statement. Please help.

Bunuel

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25 Feb 2018, 21:03

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erickrs

Quote:

So,∠BAO+∠ACO=2∗∠BAO+∠BOA=3∗∠BAO=60º

I just don't get this step in the first statement. Please help.

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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08 Nov 2019, 14:42

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burnttwinky

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.

Show Spoiler

Attachment:

The attachment Semicirlce.GIF is no longer available

Attachment:

The attachment Untitled.png is no longer available

Refer to Attached Picture:
We know that O is the center of the circle;
From that statement, we can infer that OB = OC, so their angles are also equal (a°)
We know that AB = OB, so their angles are also equal (c°).
In the picture, there are 5 angles to refer to (angle a°, b°, c°, d°, and e°). From that, we can infer:
c° + b° + e° = 180
a° + a° + b° = 180
c° + c° + d° = 180
a° + d° = 180
a° + a° = e° (exterior angle theory)
We have 5 unique equations so if we know one of those values, we can answer the prompt!

(1) The degree measure of angle COD is 60º.
Sufficient
(2) The degree measure of angle BCO is 40º.
Sufficient

The answer is D!!

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GMAT coiasda.png [ 17.8 KiB | Viewed 7335 times ]

harshbirajdar

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18 Jun 2021, 20:21

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This is not fair. We encounter trick questions all the time in GMAT Quant. It has taken consistent practice for a few months to realise that you cannot simply assume a few things - especially in DS. I don't understand why we are assuming ABC to be a straight line. I've read all of Bunuel's replies on the thread, but I am still not convinced that we can assume it to be a straight line.

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18 Jun 2021, 21:29

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harshbirajdar

Exactly. 😂 math in GMAT is simply forcing your to accept that.

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Bunuel

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19 Jun 2021, 03:54

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wushingling

harshbirajdar

Exactly. 😂 math in GMAT is simply forcing your to accept that.

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Not sure what is confusing there. Check what does the OG says about the sissue.

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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19 Sep 2021, 04:53

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It's given that AB = OC

OC = OB as it's the radius.So AB = OC = OB

If ∠BAO = x , then ∠ BOA = x
In the triangle BAO, the two sides AB and OB are equal so the angle opposite to equal sides are also equal.

∠CBO = 2x Using exterior angle theorem

In the triangle BOC, the two sides OC and OB are equal so the angle opposite to equal sides are also equal.
∠CBO = ∠BCO = 2x

∠BOC = 180 - 4x As the sum of the angles of the triangle BOC = 180º

∠COD = 3x (As ∠COD, ∠BOC , ∠ BOA forms a straight line and their sum is 180º )

We are asked to find ∠ BAO i.e value of x.

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

We know that ∠COD = 3x = 60º

Then ∠ BAO = x = 60/3 = 20º
Statement 1 alone is sufficient.

(2) The degree measure of angle BCO is 40º.

We know that ∠BCO = 2x = 40º
Then ∠ BAO = x = 40/2 = 20º
Statement 2 alone is sufficient.

Option D is the correct answer.

Thanks,
Clifin J Francis,
GMAT SME

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xcusemeplz2009

tashu

<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

how can we know that CAO is a triangle, its is not mentioned anywhere nor we can conclude, it can be a 4 sided polygon right?

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02 May 2022, 06:25

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fozzzy

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!

extremely simple & elegant!

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23 Nov 2022, 01:28

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I have often seen experts mentioning that we cannot assume anything not explicitly stated in the question stem so how can we assume over here that AC is a straight line ? we shall need to assume that AC is a straight line to arrive at the answers.

Please correct me if I am wrong

Bunuel

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23 Nov 2022, 03:21

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Anant87

It's not an assumption. It follows from the instructions you get before each test: https://gmatclub.com/forum/in-the-figur ... l#p2808015

P.S. Please read the whole discussion before posting. Thank you for cooperation.

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