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

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

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New post 09 Feb 2016, 03:26
Could anyone please be so kind to explain this question with use of diagrams? I have a very hard time understanding the answer to this question. Thank you very much
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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New post 12 Sep 2018, 07:24
we have to connect C to center O and prove that the angle at the center is double the angle CEB.

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

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New post 12 Sep 2018, 08:04
Quick elimination tip here

An triangle formed by the diameter of a circle and any other point on that circle is a right triangle. The diameter is the hypotenuse and the right angle is formed by the two legs extending to the point. So, in this case, we can create a right triangle by drawing a line from, say, E to D. This shows that angle CED is greater than 90, so eliminate C, D and E.
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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New post 12 Nov 2018, 06:52
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

All angles are measured in degrees.

\(? = x = \angle CEB\)

Image

01. Triangle AOC is isosceles with base AC, hence the 66-degrees angle is justified.

02. Angle BAC is inscribed in the circle, hence the 132-degrees red arc (CEB) is justified.

03. Angle x (our FOCUS) is inscribed in the circle and corresponds to the blue arc CADB, hence:

\(? = {{360 - 132} \over 2} = 114\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( A \right)\)


This solution follows the notations and rationale taught in the GMATH method.

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

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New post 12 Nov 2018, 08:17
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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
Image


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


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°
Image


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°
Image

At this point, we have:
Image
So, ∠CEB = 90° + 24° = 114°

Answer: A

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

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New post 12 Nov 2018, 17:14
Angle CBE + Angle ODB = 180 degree as sum of opposite angles of a quadrilateral is 180.
So, CBE =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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New post 13 Nov 2018, 06:08
karishma,

I have a doubt..so from what I understand is that the angle around O should be 360 degree, so shouldnt angle COB be 132 degree? angle COA and angle AOD adds up to 180 degree. angle BOD is 48 degree, making it all add up to 228 degree. Then should angle COB not be 360-228=132 degree?Where am I going wrong, pls help

2nd doubt- is angle COA =48 degree? if not, why? (i assumed angle BOD and COA to be vertically opposite)
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Re: In the figure AB and CD are two diameters of circle. Interse  [#permalink]

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New post 13 Nov 2018, 10:25
1
Quote:
Image

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.
Image
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.
Image
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 &nbs [#permalink] 13 Nov 2018, 10:25

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