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In the figure below, AB is the chord of a circle with center

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honeyhani
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In the figure below, AB is the chord of a circle with center [#permalink] New post   21 Oct 2010, 03:02
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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

please help me with answers in details.
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A

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Re: Geomtery, circles [#permalink] New post   21 Oct 2010, 23:08
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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

please help me with answers in details.


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
Answer (A)
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Re: Geomtery, circles [#permalink] New post   21 Oct 2010, 03:22
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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Re: Geomtery, circles [#permalink] New post   21 Oct 2010, 15:35
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

please help me with answers in details.


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

Hence answer is (A)
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Re: In the figure below, AB is the chord of a circle with center [#permalink] New post   20 Aug 2018, 10:25
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

please help me with answers in details.


OA:A
Attachment:
xy.PNG
xy.PNG [ 54.81 KiB | Viewed 8560 times ]


Joining \(OB\),\(\angle BOC =\angle BCO =y\) (As \(OB=BC\))
\(\angle OBA = \angle BOC + \angle BCO\) (An exterior angle of a triangle is equal to the sum of the opposite interior angles.)
\(\angle OBA = 2y\)
\(\angle OAB = \angle OBA =2y\) ( As \(OA=OB=r\))
Considering \(\triangle OAC\),
\(\angle DOA =\angle OAC + \angle OCA\) (An exterior angle of a triangle is equal to the sum of the opposite interior angles.)
\(x =y +2y\)
\(x=3y\) so \(k=3\)
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In the figure below, AB is the chord of a circle with center [#permalink] New post   20 Aug 2018, 11:45
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Quote:
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


OA and OB are radii, so OA=OB.
Since the prompt indicates that OB=BC, we get:
OA=OB=BC.
The following figure is yielded:


Let y=10.
Since the angles inside triangle BCO must sum to 180, and the angles opposite sides OB and BC must be equal, the following figure is yielded:


Since angles ABO and OBC must sum to 180, angle ABO=20.
Since the angles inside triangle ABO must sum to 180, and the angles opposite sides OA and OB must be equal, the following figure is yielded:


Since angles AOD, AOB and BOC must sum to 180, we get:


Since x=30, y=10, and \(k = \frac{x}{y}\), we get:
\(k = \frac{30}{10} = 3\)

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In the figure below, AB is the chord of a circle with center [#permalink]
20 Aug 2018, 11:45
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