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

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

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Updated on: 04 Dec 2014, 04:32
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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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04 Aug 2014, 06:39
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7
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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03 Aug 2014, 02:37
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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.

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

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04 Aug 2014, 06:02
5
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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04 Aug 2014, 13:31
2
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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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06 Aug 2014, 04:25
2
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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12 Nov 2018, 09:17
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Top Contributor
PathFinder007 wrote:
Attachment:
Circle_Angle.JPG
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

Let's use some useful circle properties

First let's add a blue line, to divide ∠CEB into 2 angles

CD is the DIAMETER of the circle
Since ∠CED is an inscribed angle containing (aka "holding") the diameter, we can conclude that ∠CED = 90°

Now recognize that the intersection of AB and CD creates two equal (vertically opposite) angles, we can conclude the angle opposite the 48° is also 48°

Now recognize that we have two angles containing (aka "holding") the arc BD
Property: If a CENTRAL angle and an INSCRIBED angle contain the same arc, then the CENTRAL angle is TWICE the INSCRIBED angle
This means that ∠DEB = 24°

At this point, we have:

So, ∠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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03 Aug 2014, 02:46
1
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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03 Aug 2014, 03:47
1
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.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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04 Dec 2014, 00:07
1
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.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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14 Dec 2014, 21:42
1
Thank you!
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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13 Nov 2018, 11:25
1
1
Quote:

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

An INSCRIBED ANGLE is formed by two chords.
A CENTRAL ANGLE is formed by two radii.
When an inscribed angle and a central angle intercept the same two points on a circle, the inscribed angle is 1/2 the central angle.

Here, inscribed angle CFB and central angle COB both intercept the circle at points C and B.
Since central angle COB = 132º, inscribed angle CFB = (1/2)(132) = 66º.

Rule:
When a quadrilateral is inscribed in a circle, opposite angles sum to 180º.

Since opposite angles inside inscribed quadrilateral CEBF must sum to 180º, angle CEB = 180-66 = 114.

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

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11 May 2019, 00:42
1
ArupRS wrote:

Dear experts,

I thought smaller angle COB is 48 degree. Hence the angle CEB is half of angle COB. Please explain what I am missing here.

Regards,
Arup

No COB is not 48..
Even if you take COB as 48, CEB is not half of COB..
Any point say r in major arc CB that is CADB will make an angle CrB which will be half of COB.
This becomes half of 48 so 24..
Now CEB+CrB will be 180 as sum of opposite angles of a cyclic quadrilateral is 180 and CEBr is cyclic quadrilateral.
Thus CEB = 180-24=156..
But no choice is given as 156.

So you have to take COB as 180-48=132
And CEB = 180-132/2=180-66=114
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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11 May 2019, 01:17
1
ArupRS wrote:

Dear experts,

I thought smaller angle COB is 48 degree. Hence the angle CEB is half of angle COB. Please explain what I am missing here.

Regards,
Arup

Angle COB is the larger angle so it will not be 48 degrees.
Also, inscribed angle is half of the central angle that subtends the same arc.

COB and CEB subtend different arcs.
Angle COB subtends arc CEB.
So angle COB will be twice of angle CAB or CDB.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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03 Aug 2014, 19:52
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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04 Aug 2014, 06:58
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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05 Aug 2014, 06:51
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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05 Aug 2014, 07:07
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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28 Aug 2014, 20:07
Hi there,
Can someone explain me this with the help of proper diagram,unable to get it!
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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15 Dec 2014, 08:11
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] 15 Dec 2014, 08:11

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