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In the above figure, point O is at the center

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Joined: 11 Sep 2015
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GMAT 1: 770 Q49 V46
In the above figure, point O is at the center  [#permalink]

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14 Feb 2017, 06:33
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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

*kudos for all correct solutions

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Joined: 02 Aug 2009
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Re: In the above figure, point O is at the center  [#permalink]

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14 Feb 2017, 07:00
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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
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Re: In the above figure, point O is at the center  [#permalink]

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14 Feb 2017, 06:50
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

∠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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GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4999
GMAT 1: 770 Q49 V46
In the above figure, point O is at the center  [#permalink]

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14 Feb 2017, 09:57
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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

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Joined: 22 Nov 2016
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In the above figure, point O is at the center  [#permalink]

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07 Jul 2017, 12:43
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
$$\angle OBA + \angle OBA = 180 - 2* (180-3x)$$
$$2 \angle OBA = 180 - 360 + 6x$$
$$2 \angle OBA = -180 + 6x$$
$$2 \angle OBA = 6x - 180$$
$$2 \angle OBA = \frac{6x - 180}{2}$$
$$\angle OBA = 3x - 90$$

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Re: In the above figure, point O is at the center  [#permalink]

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08 Jan 2020, 00:34
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Re: In the above figure, point O is at the center   [#permalink] 08 Jan 2020, 00:34