In the figure below, AB is the chord of a circle with center : GMAT Data Sufficiency (DS)
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# In the figure below, AB is the chord of a circle with center

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In the figure below, AB is the chord of a circle with center [#permalink]

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21 Oct 2010, 02:02
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In the figure below, AB is the chord of a circle with center O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If ACD= y degrees and AOD = x degrees such that x = ky, then the value of k is

A. 3
B. 2
C. 1
D. None of the above

[Reveal] Spoiler: OA

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21 Oct 2010, 02:22
In the figure below, AB is the chord of a circle with center O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If ACD= y degrees and AOD = x degrees such that x = ky, then the value of k is

A. 3
B. 2
C. 1
D. None of the above

ans : OB=BC
thus, in the triangle BOC, angle BOC = angle BCO =y
again, OB=AO (as both are radii of the same circle)
thus, in triangle OAB, angleOAB= angle OBA =y (because OAB and BOC are equal triangle)
thus, we can make the equation like this,
180=y+(180-2y)+x
thus, y=x
thus, ans is
c
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21 Oct 2010, 14:35
honeyhani wrote:
In the figure below, AB is the chord of a circle with center O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If ACD= y degrees and AOD = x degrees such that x = ky, then the value of k is

A. 3
B. 2
C. 1
D. None of the above

OB=BC, So BCO=BOC=x
ABO=2x (exterior angle of triangle BOC, sum of the two opposite interior angles)
OAB=ABO=2x (Isosceles triangle, 2 sides are the radius of the circle)
AOD is an exterior angle of triangle AOC
Hence AOD=OAB+BCO=x+2x=3x=y

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21 Oct 2010, 22:08
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honeyhani wrote:
In the figure below, AB is the chord of a circle with center O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If ACD= y degrees and AOD = x degrees such that x = ky, then the value of k is

A. 3
B. 2
C. 1
D. None of the above

Sol:
OB = BC (given) => Angle BOC = Angle BCO = y => Angle OBC = 180-2y (in triangle OBC)
Hence Angle OBA = 2y
Since AO = OB (Both radius of the circle)
so Angle OBA = Angle BAO = 2y => Angle AOB = 180 -4y
Now we know all the three angles at point O form by a straight line DC
Hence Angle AOD + Angle AOB + Angle BOC = 180
x + 180 - 4y + y = 180
x - 3y = 0
x = 3y
Re: Geomtery, circles   [#permalink] 21 Oct 2010, 22:08
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