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

[Brent@GMATPrepNow]

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

Source: GMAT Prep Now
Difficulty level: 700

Answer: E

[regor60]

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

[Brent@GMATPrepNow]

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

Answer: E

[GMATGuruNY]

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.

The correct answer is E.