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In the figure shown, point O is the center of the semicircle [#permalink]
 19 Dec 2009, 08:28
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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 BCD is 40º. Attachment: 
exteriorAngle.GIF [ 14.09 KiB | Viewed 21180 times ]
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Last edited by Bunuel on 26 Mar 2012, 03:21, edited 2 times in total.
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Re: Exterior Angle in a Semi Circle [#permalink]
19 Dec 2009, 10:25
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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 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 Answer: D. More solutions at: viewtopic.php?p=464547viewtopic.php?p=398461viewtopic.php?p=581082gmatprep-2-triangle-semicircle-76801.htmlviewtopic.php?p=607910For more about the geometry issues check the links below.
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Re: Exterior Angle in a Semi Circle [#permalink]
19 Dec 2009, 12:33
I still miss it. I saw the Circle Chapter from the Maths book as well. Why is <CBO=2<BAO
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Re: Exterior Angle in a Semi Circle [#permalink]
19 Dec 2009, 12:58
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.
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Re: Exterior Angle in a Semi Circle [#permalink]
19 Dec 2009, 13:01
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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 Statement 2 gives BCO = 2x = 40 we can calculate x hence sufficient. Answer is D (Hope this is clear.)
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Re: Exterior Angle in a Semi Circle [#permalink]
19 Dec 2009, 13:06
Thanks a lot guys. Now it is clear. I kinda missed the exterior angle thing. Have to brush up my geometry.
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Re: Exterior Angle in a Semi Circle [#permalink]
15 Feb 2010, 06:27
thanks a lot this single geometry question helped me a lot...
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Re: Exterior Angle in a Semi Circle [#permalink]
28 Oct 2010, 10:16
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. Cheers!
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Re: Exterior Angle in a Semi Circle [#permalink]
02 Nov 2010, 06:09
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. " Hope it helps.
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Hi Bunuel, excuse me for not attaching the related figure for this problem, but I do not know how to draw.
Semicircle with center O and diameter KD.
B and C are two points on the circle, with B closer to point K and C closer to D.
Point A lies outside the semicircle along the diamter KD (to the side of K).
If the length of segment AB is equal to the length of segment OC. what is the measure of angle BAO?
statement 1: the measure of angle COD is 60
statement 2: the measure of angle BCO is 40
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Re: Geometry Problem [#permalink]
26 Mar 2012, 02:31
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.
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Re: Geometry Problem [#permalink]
26 Mar 2012, 03:25
imadkho wrote: Hi Bunuel, excuse me for not attaching the related figure for this problem, but I do not know how to draw.
Semicircle with center O and diameter KD.
B and C are two points on the circle, with B closer to point K and C closer to D.
Point A lies outside the semicircle along the diamter KD (to the side of K).
If the length of segment AB is equal to the length of segment OC. what is the measure of angle BAO?
statement 1: the measure of angle COD is 60
statement 2: the measure of angle BCO is 40 Merging similar topics. Please ask if anything remains unclear. Also, please DO NOT shorten or reword the question you post, ALWAYS type them in EXACTLY as they are stated in the source. As for attaching the picture, it's REALLY easy: make a screenshot and attach.
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Re: In the figure shown, point O is the center of the semicircle [#permalink]
26 Mar 2012, 05:43
Hi,
how did we know that points A,B, and C are collinear, and hence apply the theorem of the exterior angle ?
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Re: In the figure shown, point O is the center of the semicircle [#permalink]
26 Mar 2012, 06:02
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Re: In the figure shown, point O is the center of the semicircle [#permalink]
26 Mar 2012, 07:21
thanks bunuel for the clarification.
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In the figure shown, point O is the center of the semicircle [#permalink]
28 Mar 2012, 12:21
option # 1, angle COD = 60
proves, angle CBO = 30 (angle at center = 2 times angle at any point in circle)
proves angle ABO = 180-30 = 150 (straight line)
proves BAO = (180-150)/2 = 15 (isoceles triangle, equal sides, equal angles)
option # 2,angle BCO = 40
proves angle CBO = 40 (isoceles triangle, equal sides, equal angles)
proves angle ABO = 140 (straight line)
proves angle BAO = (180-140)/2 = 20 (isoceles triangle, equal sides, equal angles)
Answer # D
This would be my approach.
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Re: In the figure shown, point O is the center of the semicircle [#permalink]
27 Jul 2012, 12:21
this is how is understood.
let angle BAO = x
let angle BOA = x
let angle BCO = y
let angle CBO = y
2x + 180 - y = 180 EQN(1)
So, from statement 1, it implies that
2y + 120 - x = 180 EQN(2)
Solve for 2 unknowns done.
Statement 2 is self explainatory. hope it helps someone.
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In the figure shown, point O is the center of the semicircle and [#permalink]
01 Apr 2013, 15:40
Img always makes it pretty clear
Attachments
Note_20130401_173652_01.jpg [ 86.68 KiB | Viewed 998 times ]
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In the figure shown, point O is the center of the semicircle and
[#permalink]
01 Apr 2013, 15:40
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