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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

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AB is a chord shared by two triangles: AOB and ACB.
Hence the \(\angle AOB = 2* \angle ACB\)..................(1)

From the given information \(\angle ACB = (180-3x)^ {\circ}\)...........(2)

Using (1) and (2):
\(\angle AOB = 2* (180-3x)^ {\circ}\)...........(3)

Triangle AOB is an isosceles triangle with OA = OB hence \(\angle OAB = \angle OBA\)
Since the three angles make up 180, we can write the equation
\(\angle OBA + \angle OBA = 180 - 2* (180-3x)\)
\(2 \angle OBA = 180 - 360 + 6x\)
\(2 \angle OBA = -180 + 6x\)
\(2 \angle OBA = 6x - 180\)
\(2 \angle OBA = \frac{6x - 180}{2}\)
\(\angle OBA = 3x - 90\)

Answer is E
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