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In the figure shown, point O is the center of the semicircle [#permalink]
19 Dec 2009, 07:28

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Difficulty:

65% (medium)

Question Stats:

45% (02:21) correct
55% (01:21) wrong based on 233 sessions

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 BCO is 40º.

Attachment:

exteriorAngle.GIF [ 14.09 KiB | Viewed 29126 times ]

Re: Exterior Angle in a Semi Circle [#permalink]
19 Dec 2009, 09:25

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Expert's post

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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 ?

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

Re: Exterior Angle in a Semi Circle [#permalink]
19 Dec 2009, 11:58

5

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Expert's post

msunny wrote:

I still miss it. I saw the Circle Chapter from the Maths book as well. Why is

<CBO=2<BAO

First of all note that ABO is isosceles triangle. Why? Given that AB=OC, OC is radius, but OB is also radius, hence AB=OC=OB=r --> two sides in triangle ABO namely AB and BO are equal. Which means that angles BAO and BOA are also equal.

So we have <BAO=<BOA.

Next step: angle <CBO is exterior angle for triangle BAO. According to the exterior angle theorem: An exterior angle of a triangle is equal to the sum of the opposite interior angles. --> <CBO=<BAO+<BOA, as <BAO=<BOA --> <CBO=<BAO+<BAO=2<BAO.

Hope it's clear.

For more about the triangles check the link abot triangles below.

Re: Exterior Angle in a Semi Circle [#permalink]
19 Dec 2009, 12:01

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Let angle BAO=x since AB=BO we have angle BOA=x SInce Angle CBO is an exterior angle to BAo and BOA it is equal to the sum of their individual angles Angle CBO = x+x=2x

BO and CO are the two radii hence they subtend equal angles thus BCO = 2x and BOC = 180-4x We need x Statement 1 gives COD Since COD+BOC+AOB = 180 60+180-4x+x=180 We cans olve for x - sufficient

Re: Exterior Angle in a Semi Circle [#permalink]
28 Oct 2010, 09:16

Hi,

What have I missed? I too marked this as C. GMAT Prep says its D, so fine, agreed. But here is my reasoning. The reason is that it is not mentioned in the question that ABC is one single line. (points A,B,C all lie on the same line).

Why do we need to consider it as one straight line? If we do not consider them on the straight line, we cannot apply Exterior angle sum rule. If we cannot apply, we will not be able to solve this without both stmt1 and stmt2.

Re: Exterior Angle in a Semi Circle [#permalink]
02 Nov 2010, 05:09

Expert's post

gmatretake wrote:

Hi,

What have I missed? I too marked this as C. GMAT Prep says its D, so fine, agreed. But here is my reasoning. The reason is that it is not mentioned in the question that ABC is one single line. (points A,B,C all lie on the same line).

Why do we need to consider it as one straight line? If we do not consider them on the straight line, we cannot apply Exterior angle sum rule. If we cannot apply, we will not be able to solve this without both stmt1 and stmt2.

Cheers!

Ian Stewart:

"In general, you should not trust the scale of GMAT diagrams, either in Problem Solving or Data Sufficiency. It used to be true that Problem Solving diagrams were drawn to scale unless mentioned otherwise, but I've seen recent questions where that is clearly not the case. So I'd only trust a diagram I'd drawn myself. ...

Here I'm referring only to the scale of diagrams; the relative lengths of line segments in a triangle, for example. ... You can accept the relative ordering of points and their relative locations as given (if the vertices of a pentagon are labeled ABCDE clockwise around the shape, then you can take it as given that AB, BC, CD, DE and EA are the edges of the pentagon; if a line is labeled with four points in A, B, C, D in sequence, you can take it as given that AC is longer than both AB and BC; if a point C is drawn inside a circle, unless the question tells you otherwise, you can assume that C is actually within the circle;if what appears to be a straight line is labeled with three points A, B, C, you can assume the line is actually straight, and that B is a point on the line-- the GMAT would never include as a trick the possibility that ABC actually form a 179 degree angle that is imperceptible to the eye, to give a few examples).

So don't trust the lengths of lines, but do trust the sequence of points on a line, or the location of points within or outside figures in a drawing. "

Re: Geometry Problem [#permalink]
26 Mar 2012, 01:31

You can draw using MSPaint, if you want to draw it indeed. Else, take a screenshot and attach the pic file with your post. Please do this ASAP, and also (because it's a GMATPrep question), search in GMATClub to see if this has been posted earlier.

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

So, if what appears to be a straight line is labeled with three points A, B, C, you can assume the line is actually straight, and that these points are actually on the line in the order given on the diagram.

Re: In the figure shown, point O is the center of the semicircle [#permalink]
30 Jul 2013, 03:42

Expert's post

Qoofi wrote:

Where does it say that Points A, B and C lie on the same line? Assumption?

OG13, page 272: A figure accompanying a data sufficiency problem will 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 can be assumed to be straight and lines that appear jagged can also be assumed to be straight. You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees. All figures lie in a plane unless otherwise indicated.