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

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Senior Manager
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In the figure shown, point O is the center of the semicircle [#permalink]  15 Dec 2009, 12:09
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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.

[Reveal] Spoiler:
Attachment:

Semicirlce.GIF [ 14.09 KiB | Viewed 2725 times ]

Attachment:

Untitled.png [ 1.77 KiB | Viewed 373 times ]
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 Jul 2015, 00:03, edited 3 times in total.
Edited the question
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In the figure shown, point O is the center of the semicircle [#permalink]  15 Dec 2009, 13:04
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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

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
gmatprep-2-triangle-semicircle-76801.html
http://gmatclub.com:8080/forum/viewtopic.php?p=607910

For more about the geometry issues check the links below.
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Re: In the figure shown, point O is the center of the semicircle [#permalink]  18 Dec 2009, 12:44
Took me 15 min to figure this one out.

Bunuel I followed the same approach albeit from the 14th min

thanks for the lovely explanation
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Re: In the figure shown, point O is the center of the semicircle [#permalink]  18 Dec 2009, 17:33
Given OC = AB
OC = OB since both are radii.
Therefore OB = AB.
Since angles opposite to equal sides are equal. Therefore,
Angle OCB = Angle OBC and Angle OAB = Angle BOA

Statement 1 Angle COD = 60 degrees. Not sufficient. Since we cannot determine angle COB or angle OCB.
Statement 2 Angle OCB = 40 degrees. Not sufficient Since we cannot determine angle COB

Both together helps us determine Angle COB and angle BOA = 180 - angle COD - angle COB
Therefore, answer is C
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Re: In the figure shown, point O is the center of the semicircle [#permalink]  18 Dec 2009, 19:46
<CBO=2<BAO ?? Why
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Re: In the figure shown, point O is the center of the semicircle [#permalink]  18 Dec 2009, 20:14
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tashu wrote:
<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
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Re: In the figure shown, point O is the center of the semicircle [#permalink]  23 Jul 2014, 01:13
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Re: In the figure shown, point O is the center of the semicircle [#permalink]  03 Aug 2014, 04:38
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Small typo in the copied version of the question :
(2) The degree measure of angle BCD is 40º.
It should actually be angle BCO
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Re: In the figure shown, point O is the center of the semicircle [#permalink]  12 Aug 2014, 01:42
Expert's post
shankyGMAT wrote:
Small typo in the copied version of the question :
(2) The degree measure of angle BCD is 40º.
It should actually be angle BCO

Edited. Thank you.
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Re: In the figure shown, point O is the center of the semicircle [#permalink]  30 Jun 2015, 19:07
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

(1) <BAO +<ACO=<COD=60 degrees (Using exterior angle theorem)
<ACO = <CBO = 2<BAO
So, <BAO + <ACO = 2<BAO + <BOA = 3* (<BAO) = 60 degrees
<BAO = 20 degrees. SUFFICIENT

(2) <BCO=40 degrees --> <BCO=<CBO=40 degrees=2<BAO --> <BAO=20 degrees. 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
gmatprep-2-triangle-semicircle-76801.html
http://gmatclub.com:8080/forum/viewtopic.php?p=607910

For more about the geometry issues check the links below.

What property have you used Bunuel -
<CBO=2<BAO
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Re: In the figure shown, point O is the center of the semicircle [#permalink]  01 Jul 2015, 00:24
Expert's post
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

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

Hope it's clear.
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Re: In the figure shown, point O is the center of the semicircle   [#permalink] 01 Jul 2015, 00:24
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