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In the figure below, AB is the chord of a circle with center [#permalink]
21 Oct 2010, 02:02
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Question Stats:
43% (02:15) correct
57% (01:13) wrong based on 7 sessions
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
Re: Geomtery, circles [#permalink]
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
Re: Geomtery, circles [#permalink]
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
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
Re: Geomtery, circles [#permalink]
21 Oct 2010, 22:08
1
This post received KUDOS
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)
gmatclubot
Re: Geomtery, circles
[#permalink]
21 Oct 2010, 22:08
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