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In the above figure, point O is at the center
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14 Feb 2017, 07:33

2

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7

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

64% (02:53) correct 36% (03:19) wrong based on 143 sessions

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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

Re: In the above figure, point O is at the center
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14 Feb 2017, 07:50

3

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

*kudos for all correct solutions

Please follow below fig.

∠BCA = 180-3x (angle of line) so ∠AOB = 2(180-3x) (angle subtended on same chord AB)--------(1) in triangle AOB sum of all angles are 180 ∠ABO + ∠OAB+∠AOB =180 But ∠ABO = ∠OAB =M(let) .... (same radius)---------(2) substituting values from (1) & (2) 2M + 2(180-3x) =180 M=(3x - 90)º

Ans E

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Re: In the above figure, point O is at the center
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14 Feb 2017, 08:00

3

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

*kudos for all correct solutions

Hi,

Few geometric rules are useful here.. 1) angle ACB= 180- angle BCE=180-3x.. 2) important rule:- angle at centre is TWICE the angle on the circumference from CHORD of SAME length.. So angle AOB= 2* angle ACB= 2(180-3x)=360-6x.. 3) Now Triangle AOB is isosceles triangle, so angle ABO = angle BAO=(180-angleAOB)/2=\(\frac{180-(360-6x)}{2}\)=3x-90.. E
_________________

In the above figure, point O is at the center
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14 Feb 2017, 10:57

1

Top Contributor

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

In the above figure, point O is at the center
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07 Jul 2017, 13:43

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 _________________

Kudosity killed the cat but your kudos can save it.

Re: In the above figure, point O is at the center
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20 Sep 2018, 21:16

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