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

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by BTGmoderatorDC » Sun Nov 11, 2018 9:35 pm
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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

OA A

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by fskilnik@GMATH » Mon Nov 12, 2018 6:42 am
BTGmoderatorDC wrote: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
Source: GMAT Prep
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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by Brent@GMATPrepNow » Mon Nov 12, 2018 8:12 am
BTGmoderatorDC wrote: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

OA A

Source: GMAT Prep
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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by swerve » Mon Nov 12, 2018 9:41 am
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 the diameter, so angle AEB = 90.
Angle CEB = 90 + 24 = 114.
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angle CEB

by GMATGuruNY » Mon Nov 12, 2018 3:26 pm
BTGmoderatorDC wrote: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.

The correct answer is A.
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