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In the figure shown, point O is the center of semicircle and [#permalink]
 17 Jul 2009, 15:00
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 In the figure shown, point O is the center of semicircle and points B, C, and 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 60 (2) The degree measure of angle BCO is 40 OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-shown-point-o-is-the-center-of-the-semicircle-88009.html
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Last edited by Bunuel on 30 Jul 2012, 05:37, edited 2 times in total.
Edited the question and added the OA.
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Re: semicircle and triangle [#permalink]
17 Jul 2009, 15:37
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Assume angle OBC=y and angle BAO=z
Using (1)
=========
y = 2z ---(i)(exterior angle is sum of opp interior angles)
z+(180-2y)+60=180 ---(ii) (sum of angle AOB, BOC and COD is 180)
Solving (i) and (ii) we get z=20
Hence sufficient
Using (2)
=========
Angle BCO = y = 40
==> z=40/2 = 20
Hence sufficient
Answer (D).
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Re: semicircle and triangle [#permalink]
17 Jul 2009, 15:38
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asimov wrote: In the figure shown, point O is the center of semicircle and points B, C, and 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 60 (2) The degree measure of angle BCO is 40 I think 2 alone is sufficient Given data : AB = OC which implies = BO. So In triangle ABO, AB = BO , isoceles triangle hence opposite angles BOA =angle BAO similarly triangle We know BO=OC and hence triangle BOC is also isoceles. So if angle BCO = 40, then angle OBC =40 and hence angle ABO = 180-40 = 120 once angle ABO is known, we know angle BOA = BAO, hence 2 angle BAO+ angle ABO = 180 2 BAO = 60 and hence BAO is 30
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Re: semicircle and triangle [#permalink]
18 Jul 2009, 07:32
hi gmanjesh, can you please help us understand why y=2z how do you know BAO and BOA have the same angle? I'm assuming that's the same assumption you used to assume OBC and OCB are equal.
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Re: semicircle and triangle [#permalink]
18 Jul 2009, 08:25
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asimov wrote: hi gmanjesh, can you please help us understand why
y=2z
how do you know BAO and BOA have the same angle?
I'm assuming that's the same assumption you used to assume OBC and OCB are equal. Hi Asimov triangle ABO is isoceles with OB and AB are equal sides, hence the opposite angles BAO and BOA are equal
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Re: semicircle and triangle [#permalink]
18 Jul 2009, 08:42
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alwynjoseph wrote: asimov wrote: hi gmanjesh, can you please help us understand why
y=2z
how do you know BAO and BOA have the same angle?
I'm assuming that's the same assumption you used to assume OBC and OCB are equal. Hi Asimov triangle ABO is isoceles with OB and AB are equal sides, hence the opposite angles BAO and BOA are equal thanks alwynjoseph for explaining it Asimov OB and OC are the radius and hence they are euqal and as per the question AB=OC therefore AB=OC=OB
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Re: semicircle and triangle [#permalink]
18 Jul 2009, 09:17
thanks a lot. I didn't catch on the radius -> isoceles triangle. it makes sense to me now. the OA is D
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Re: semicircle and triangle [#permalink]
20 Jul 2009, 05:05
Hi, guys. If I follow your explanation correctly, I think you're assuming that AC is a straight line and that ACO is a triangle.
If that is what you're doing, is there anything in the question setup that says we can assume that AC is a straight line?
Thanks.
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Re: semicircle and triangle [#permalink]
27 Aug 2009, 13:24
gmanjesh wrote:
y = 2z ---(i)(exterior angle is sum of opp interior angles)
I don't understand this part. Is anyone able to illustrate please?
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Re: semicircle and triangle [#permalink]
21 Apr 2010, 19:09
I understand how statement II is sufficient, but I am having a hard time with statement I. Can someone please help me out?
One thing that seems unexplained with the statement I work done thus far:
I agree that angle BAO = BOA, but in superman's equation he assumed that angle BOC = BOA. How do we know that?
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Re: semicircle and triangle [#permalink]
23 May 2010, 16:25
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ok I found out why y = 2z.
If angle BAO = x, then angle BOA = x (because triangle BAO is Isosceles)
Therefore, angle ABO = 180 - 2x (because the sum of all the angles of the triangle is 180)
Since line segment AC is a straight line that means ABO + CBO = 180.
If you substitute
ABO = 180 - 2x
into
ABO + CBO = 180
CBO = 2x. Then BOC = 180 - 4x. (because triangle BOC is isosceles)
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Re: semicircle and triangle [#permalink]
19 Jun 2010, 10:03
grifter000 wrote: Hi, guys. If I follow your explanation correctly, I think you're assuming that AC is a straight line and that ACO is a triangle.
If that is what you're doing, is there anything in the question setup that says we can assume that AC is a straight line?
Thanks. Good general question. You should almost assume that DS diagrams are not drawn to scale. But there are things about the picture that you can take as fact: (1) If two lines look like they meet, they do. (i.e. no microscopic gaps) (2) If a line looks straight, it is. (i.e. no trick angles of 179.99 degrees)
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Re: semicircle and triangle [#permalink]
22 Jun 2010, 11:33
IMO D. This is simple test of triangle basics. Thanks for sharing this question.
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Re: semicircle and triangle [#permalink]
26 Nov 2010, 08:15
Ok so I understand that BOA = BAO and if BOA = 2z then OBC = z. But I thought that given that OC = AB = OB then BOA = BAO = OBC?
Where is my mistake??
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Re: semicircle and triangle [#permalink]
27 Nov 2010, 08:42
superman wrote: Assume angle OBC=y and angle BAO=z
Using (1)
=========
y = 2z ---(i)(exterior angle is sum of opp interior angles)
z+(180-2y)+60=180 ---(ii) (sum of angle AOB, BOC and COD is 180)
Solving (i) and (ii) we get z=20
Hence sufficient
Using (2)
=========
Angle BCO = y = 40
==> z=40/2 = 20
Hence sufficient
Answer (D). Gud Xplanation Superman!!! Kudos +1
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Re: semicircle and triangle [#permalink]
04 May 2011, 09:25
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asimov wrote: In the figure shown, point O is the center of semicircle and points B, C, and 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 60 (2) The degree measure of angle BCO is 40 Attachment: 
Semicircle_And_Triangle.PNG [ 12.23 KiB | Viewed 2644 times ]
From the stem:
AB=OC=OB=Radius
Thus, ABO and BOC are both isosceles triangles.m\angle{BAO}=m\angle{BOA}=x^{\circ} [Note: OB=AB]m\angle{OBC}=m\angle{OCB}=y^{\circ} [Note: OB=OC]Theorem:
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
m\angle{CBO}=m\angle{BAO}+m\angle{BOA}y=x+x=2x----------------------1Q: What is x?
1.
\angle{COD}=60^{\circ}=tTheorem:Sum of three angles of a triangle is 180^{\circ}m\angle{OCB}+m\angle{OBC}+m\angle{COB}=180y+y+z=180z=180-2y---------------------2Using 1 and 2:z=180-2(2x)z=180-4x-----------------3Theorem:Angles on one side of a straight line will always add to 180^{\circ}m\angle{COD}+m\angle{BOA}+m\angle{COB}=180^{\circ}60+x+z=180z=120-x----------------4Using equations 3 and 4:180-4x=120-x3x=60x=20^{\circ}Sufficient.2.m\angle{OCB}=40^{\circ}=y=m\angle{CBO}y=40^{\circ} ----------------------5Using 1 and 5:2x=40x=20Sufficient.
Ans: "D"
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Re: semicircle and triangle [#permalink]
04 May 2011, 09:46
woooow! fluke, your solution is very helpful thank you so much! 
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Re: semicircle and triangle [#permalink]
06 May 2011, 09:00
each statement alone is sufficient, so... D
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Re: semicircle and triangle
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