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## In the above figure, point O is at the center of the circle

tagged by: Brent@GMATPrepNow

This topic has 2 expert replies and 1 member reply

### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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#### In the above figure, point O is at the center of the circle

Tue Feb 14, 2017 6:33 am
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])

In the above figure, point O is at the center of the circle, and points A, B and C are on the circumference of the circle. If ∠BCE = 3xº, what is the measure of ∠ABO?

A) (x + 10)º
B) (180 - x)º
C) (90 - x)º
D) (2x - 30)º
E) (3x - 90)º

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Difficulty level: 700

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regor60 Master | Next Rank: 500 Posts
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Tue Feb 14, 2017 8:58 am
Since BCE = 3X, its supplementary angle BCA = 180-3X.

Since BCA and AOB subtend the same chord, and one is an inscribed angle and the other, central, AOB = 2 x BCA,

So AOB = 2(180-3X) = 360-6X

Since AO and BO are equal, being radii, OAB and ABO are equal

Since sum of angles = 180, therefore 2 x ABO + 360-6X = 180, and

6x - 180 = 2 x ABO, therefore ABO = 3X-90

Last edited by regor60 on Tue Feb 14, 2017 10:54 am; edited 2 times in total

### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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Tue Feb 14, 2017 10:00 am
Brent@GMATPrepNow wrote:

In the above figure, point O is at the center of the circle, and points A, B and C are on the circumference of the circle. If ∠BCE = 3xº, what is the measure of ∠ABO?

A) (x + 10)º
B) (180 - x)º
C) (90 - x)º
D) (2x - 30)º
E) (3x - 90)º
Given:

Since angles on a line must add to 180 degrees, we know that ∠ACB = (180 - 3x) degrees

Since the inscribed angle ∠ACB and the central angle ∠AOB both contain (hold) the same chord (AB), we know that the central angle is TWICE the inscribed angle.
In other words, ∠AOB = 2(180 - 3x) = 360 - 6x

Finally, since OA and OB are radii, we know that these lengths are EQUAL
If these lengths are EQUAL, then ∆AOB is an ISOSCELES triangle, and ∠OAB = ∠ABO

Let q = ∠OAB = ∠ABO
Since all 3 angles in ∆AOB add to 180 degrees, we can write: (360 - 6x) + q + q = 180
Rearrange: 2q = 6x - 180
Divide both sides by 2 to get: q = 3x - 90
In other words, ∠ABO = 3x - 90

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GMATGuruNY GMAT Instructor
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Tue Feb 14, 2017 1:19 pm
Let x = 55, implying that 3x=155.

Since ∠BCE and ∠ACB must sum to 180, we get:

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 central angle is twice the inscribed angle.
Since inscribed angle ∠ACB and central ∠AOB both intercept points A and B on the circle, ∠AOB must be twice ∠ACB, implying that ∠AOB = 30:

The angles inside of ∆AOB must sum to 180.
Since OA and OB are radii -- and thus are equal -- the angles opposite OA and OB (∠OAB and ∠ABO) must also be equal.
The result is the following figure:

The question stem asks for the value of ∠ABO: 75.
This is our target.
Now plug x=55 into the answers to see which yields our target of 75.
Only E works:
3x - 90 = (3*55) - 90 = 75.

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