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

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

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

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

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Mon Nov 12, 2018 6:42 am

BTGmoderatorDC 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
Source: GMAT Prep

All angles are measured in degrees.
$$? = x = \angle CEB$$

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.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Nov 12, 2018 8:12 am

BTGmoderatorDC 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

OA A

Source: GMAT Prep

Let's use some useful circle properties

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

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

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

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 âˆ DEB = 24Â°

At this point, we have:

So, âˆ CEB = 90Â° + 24Â° = 114Â°

Answer: A

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swerve: Legendary Member; Posts: 2535; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

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:

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 = 132Âº, inscribed angle CFB = (1/2)(132) = 66Âº.

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

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

The correct answer is A.

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