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

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

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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
[Reveal] Spoiler: OA

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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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

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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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18 Jul 2009, 07:32
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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18 Jul 2009, 08:25
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asimov wrote:
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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18 Jul 2009, 08:42
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alwynjoseph wrote:
asimov wrote:
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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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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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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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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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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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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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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22 Jun 2010, 11:33
IMO D.

This is simple test of triangle basics. Thanks for sharing this question.
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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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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

Gud Xplanation Superman!!! Kudos +1

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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
D

Sol:

Attachment:

Semicircle_And_Triangle.PNG [ 12.23 KiB | Viewed 20921 times ]

From the stem:
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.

$$m\angle{CBO}=m\angle{BAO}+m\angle{BOA}$$

$$y=x+x=2x$$----------------------1

Q: What is x?

1.
$$\angle{COD}=60^{\circ}=t$$

Theorem:
Sum of three angles of a triangle is $$180^{\circ}$$

$$m\angle{OCB}+m\angle{OBC}+m\angle{COB}=180$$

$$y+y+z=180$$

$$z=180-2y$$---------------------2

Using 1 and 2:

$$z=180-2(2x)$$

$$z=180-4x$$-----------------3

Theorem:
Angles on one side of a straight line will always add to $$180^{\circ}$$

$$m\angle{COD}+m\angle{BOA}+m\angle{COB}=180^{\circ}$$

$$60+x+z=180$$

$$z=120-x$$----------------4

Using equations 3 and 4:

$$180-4x=120-x$$

$$3x=60$$

$$x=20^{\circ}$$

Sufficient.

2.

$$m\angle{OCB}=40^{\circ}=y=m\angle{CBO}$$

$$y=40^{\circ}$$ ----------------------5

Using 1 and 5:

$$2x=40$$

$$x=20$$

Sufficient.

Ans: "D"

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04 May 2011, 09:46
woooow!
thank you so much!
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06 May 2011, 09:00
each statement alone is sufficient, so... D
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Re: semicircle and triangle   [#permalink] 06 May 2011, 09:00
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