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

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

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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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very difficult problem.png [ 74.76 KiB | Viewed 2546 times ]

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

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

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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

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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 and point  [#permalink]

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25 Feb 2018, 18:15
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 and point  [#permalink]

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25 Feb 2018, 20:03
1
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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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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08 Nov 2019, 13:42
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

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

Attachments

GMAT coiasda.png [ 17.8 KiB | Viewed 528 times ]

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

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16 Dec 2019, 12:17
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

Hi Bunuel

I understand you used the triangle external angle theorem to determine $$\angle CBO = 2*\angle BAO$$

Question

Doesn't $$\angle CBO = 2*\angle BAO$$-- go against another theorem which says "Sides that are off the same length -- their corresponding opposite angles of these equal sides will also be the same"

Given ALL three sides (AB = BO = CO) are all the same -- shouldn't the opposite angles related to these sides ALL BE THE SAME ?
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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23 Feb 2020, 01:21
Bunuel wrote:
$$\angle ACO = \angle CBO = 2* \angle BAO$$

HI Bunuel and other experts, which property is this?
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Re: In the figure shown, point O is the center of the semicircle and point  [#permalink]

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23 Feb 2020, 02:50
OjhaShishir wrote:
Bunuel wrote:
$$\angle ACO = \angle CBO = 2* \angle BAO$$

HI Bunuel and other experts, which property is this?

https://gmatclub.com/forum/in-the-figur ... ml#p664724
https://gmatclub.com/forum/in-the-figur ... ml#p664937
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Re: In the figure shown, point O is the center of the semicircle and point   [#permalink] 23 Feb 2020, 02:50

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