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In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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Attachment: Circle_Angle.JPG [ 16.89 KiB | Viewed 11211 times ]
In the figure AB and CD are two diameters of circle. Intersecting at angle 48 degree. E is any point on Arc CB. find angle CEB

A. 114
B. 100
C. 80
D. 96
E. 40

Originally posted by PathFinder007 on 03 Aug 2014, 02:28.
Last edited by Bunuel on 04 Dec 2014, 04:32, edited 1 time in total.
Edited the question.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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desaichinmay22 wrote:
smyarga wrote:
Since, CD and AB are two diameters of the circle, so their intersection is the centre. Let's denote this centre O. Angle CEB is inscribed, and angle COB is corresponding central angle (arc CAB). The value of angle COB=180+48=228 degrees. The measure of inscribed angle is always half the measure of the central angle. Angle CEB is equal =228/2=114 degrees.

Hi,

I thought angle COB is 180-42=138 degrees. You have derived angle COB of 228 degrees.

That's why I mentioned arc CAB. You don't really need to understand the notions of major and minor arc. Corresponding central angle means that it stays on the same arc as your inscribed angle.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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Since, CD and AB are two diameters of the circle, so their intersection is the centre. Let's denote this centre O. Angle CEB is inscribed, and angle COB is corresponding central angle (arc CAB). The value of angle COB=180+48=228 degrees. The measure of inscribed angle is always half the measure of the central angle. Angle CEB is equal =228/2=114 degrees.

Tutor Joined: 20 Apr 2012
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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PathFinder007 wrote:
In the figure AB and CD are two diameters of circle. Intersecting at angle 48 degree. E is any point on Arc CB. find angle CEB

A 114
B 100
C 80
D 96
E 40

Kudos from me, because I love problems with central angles:)
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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smyarga wrote:
Since, CD and AB are two diameters of the circle, so their intersection is the centre. Let's denote this centre O. Angle CEB is inscribed, and angle COB is corresponding central angle (arc CAB). The value of angle COB=180+48=228 degrees. The measure of inscribed angle is always half the measure of the central angle. Angle CEB is equal =228/2=114 degrees.

Hi,

I thought angle COB is 180-42=138 degrees. You have derived angle COB of 228 degrees.
Intern  Joined: 31 Jul 2014
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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desaichinmay22 wrote:
smyarga wrote:
Since, CD and AB are two diameters of the circle, so their intersection is the centre. Let's denote this centre O. Angle CEB is inscribed, and angle COB is corresponding central angle (arc CAB). The value of angle COB=180+48=228 degrees. The measure of inscribed angle is always half the measure of the central angle. Angle CEB is equal =228/2=114 degrees.

Hi,

I thought angle COB is 180-42=138 degrees. You have derived angle COB of 228 degrees.

According to me, CB is not the diameter. So central angle doesnt apply in this case. Also, COB is 138 and not 228.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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desaichinmay22 wrote:
smyarga wrote:
Since, CD and AB are two diameters of the circle, so their intersection is the centre. Let's denote this centre O. Angle CEB is inscribed, and angle COB is corresponding central angle (arc CAB). The value of angle COB=180+48=228 degrees. The measure of inscribed angle is always half the measure of the central angle. Angle CEB is equal =228/2=114 degrees.

Hi,

I thought angle COB is 180-42=138 degrees. You have derived angle COB of 228 degrees.

Hi buddy,

COB is indeed 180-48 = 132 degrees . There is a concept of major and minor segment ( please google this ) Now the point O is in the minor segment of the arc and in this case it would be half of the angle subtended in the major segment i.e ( 360-132) = 228 /2 = 114 .
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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smyarga wrote:
desaichinmay22 wrote:
smyarga wrote:
Since, CD and AB are two diameters of the circle, so their intersection is the centre. Let's denote this centre O. Angle CEB is inscribed, and angle COB is corresponding central angle (arc CAB). The value of angle COB=180+48=228 degrees. The measure of inscribed angle is always half the measure of the central angle. Angle CEB is equal =228/2=114 degrees.

Hi,

I thought angle COB is 180-42=138 degrees. You have derived angle COB of 228 degrees.

That's why I mentioned arc CAB. You don't really need to understand the notions of major and minor arc. Corresponding central angle means that it stays on the same arc as your inscribed angle.

Now I got the explanation. Thanks for the help.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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desaichinmay22, specially for you another nice problem with central angle if-the-radius-of-the-circle-below-see-attachment-is-equal-175709.html
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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smyarga wrote:
desaichinmay22 wrote:
smyarga wrote:
Since, CD and AB are two diameters of the circle, so their intersection is the centre. Let's denote this centre O. Angle CEB is inscribed, and angle COB is corresponding central angle (arc CAB). The value of angle COB=180+48=228 degrees. The measure of inscribed angle is always half the measure of the central angle. Angle CEB is equal =228/2=114 degrees.

Hi,

I thought angle COB is 180-42=138 degrees. You have derived angle COB of 228 degrees.

That's why I mentioned arc CAB. You don't really need to understand the notions of major and minor arc. Corresponding central angle means that it stays on the same arc as your inscribed angle.

Hi Smyarga,
how did you get 180 degrees?
i thought about vertically opposite angles then realised that i was wrong.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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saggii27 wrote:
smyarga wrote:
desaichinmay22 wrote:

Hi,

I thought angle COB is 180-42=138 degrees. You have derived angle COB of 228 degrees.

That's why I mentioned arc CAB. You don't really need to understand the notions of major and minor arc. Corresponding central angle means that it stays on the same arc as your inscribed angle.

Hi Smyarga,
how did you get 180 degrees?
i thought about vertically opposite angles then realised that i was wrong.

Since COD is straight.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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PathFinder007 wrote:
In the figure AB and CD are two diameters of circle. Intersecting at angle 48 degree. E is any point on Arc CB. find angle CEB

A 114
B 100
C 80
D 96
E 40

Join Segment AE in the figure, we know Angle AEC is half of Angle COA(Assuming diameter intersect at point O).

Angle AEC = 24
Ab is diameter, so Angle AEB = 90
Angle CEB = 90+24 = 114
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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Hi there,
Can someone explain me this with the help of proper diagram,unable to get it!
Manager  B
Joined: 06 Aug 2013
Posts: 65
Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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smyarga wrote:
desaichinmay22 wrote:
smyarga wrote:
Since, CD and AB are two diameters of the circle, so their intersection is the centre. Let's denote this centre O. Angle CEB is inscribed, and angle COB is corresponding central angle (arc CAB). The value of angle COB=180+48=228 degrees. The measure of inscribed angle is always half the measure of the central angle. Angle CEB is equal =228/2=114 degrees.

Hi,

I thought angle COB is 180-42=138 degrees. You have derived angle COB of 228 degrees.

That's why I mentioned arc CAB. You don't really need to understand the notions of major and minor arc. Corresponding central angle means that it stays on the same arc as your inscribed angle.

another approach,

join C and A. since AB is the diameter, A, C, E and B form a cyclic quadrilateral. now opposit angles of cyclic quadrilaterals sum up to 180 degrees. considering the point of intersection of the 2 diameters as O.

angle AOC = 48 degrees. ........... given
OA = OC, radius of the circle, hence trianlge OAC is isoceles.
thus angle ACO = angle OAC = 66 degrees.

now,
angle OAC + angle CEB = 180 degrees......... since opposit angles of cyclic quadrilaterals sum up to 180 degrees.
therefore, angle CEB = 180 - 66 = 114 degrees.

kudos if you like. GMAT Tutor G
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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Thank you!
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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nktdotgupta wrote:
Thank you!

Moved to PS forum. Thank you for noticing this.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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Hi,

Can any1 explain me how angle COB=180 degrees. Why is the sum 180+48?
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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Shree9975 wrote:
Hi,

Can any1 explain me how angle COB=180 degrees. Why is the sum 180+48?

Angle COB is not 180.

Note that the angle around O will be 360 degrees. There is angle COB, the smaller angle (the angle that arc CEB subtends at the center) and angle COB, the larger angle (shown by red arrow around center O in the second diagram). The major arc CADB subtends larger angle COB at the center and inscribed angle CEB at the circle. The central angle will be twice the inscribed angle.

The larger angle COB = angle COA + angle AOD + angle DOB = 180 + 48 = 228 = Central angle

Inscribed angle = 228/2 = 114

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Veritas Prep GMAT Instructor

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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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VeritasPrepKarishma wrote:
Shree9975 wrote:
Hi,

Can any1 explain me how angle COB=180 degrees. Why is the sum 180+48?

Angle COB is not 180.

Note that the angle around O will be 360 degrees. There is angle COB, the smaller angle (the angle that arc CEB subtends at the center) and angle COB, the larger angle (shown by red arrow around center O in the second diagram). The major arc CADB subtends larger angle COB at the center and inscribed angle CEB at the circle. The central angle will be twice the inscribed angle.

The larger angle COB = angle COA + angle AOD + angle DOB = 180 + 48 = 228 = Central angle

Inscribed angle = 228/2 = 114

Hi Karishma

can't we take this way that smaller COB is 132 and then CEB = 180 - (132/2) ? using cyclic quadrilateral property.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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VeritasPrepKarishma wrote:
Shree9975 wrote:
Hi,

Can any1 explain me how angle COB=180 degrees. Why is the sum 180+48?

Angle COB is not 180.

Note that the angle around O will be 360 degrees. There is angle COB, the smaller angle (the angle that arc CEB subtends at the center) and angle COB, the larger angle (shown by red arrow around center O in the second diagram). The major arc CADB subtends larger angle COB at the center and inscribed angle CEB at the circle. The central angle will be twice the inscribed angle.

The larger angle COB = angle COA + angle AOD + angle DOB = 180 + 48 = 228 = Central angle

Inscribed angle = 228/2 = 114

Hi Karishma

can't we take this way that smaller COB is 132 and then CEB = 180 - (132/2) ? using cyclic quadrilateral property.

All vertices of a cyclic quadrilateral must lie on the circle. Angle COB does not lie on the circle. As given, the figure has no cyclic quadrilateral.
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Veritas Prep GMAT Instructor Re: In the figure AB and CD are two diameters of circle. Interse   [#permalink] 26 Oct 2015, 02:14

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