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In the above figure, point O is at the center

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

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New post 14 Feb 2017, 07:33
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A
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E

Difficulty:

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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

*kudos for all correct solutions

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

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New post 14 Feb 2017, 07:50
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GMATPrepNow wrote:
Image

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  [#permalink]

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New post 14 Feb 2017, 08:00
3
GMATPrepNow wrote:
Image

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

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New post 14 Feb 2017, 10:57
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Top Contributor
GMATPrepNow wrote:
Image

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:
Image


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


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
Image

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
Image

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

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

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Re: In the above figure, point O is at the center &nbs [#permalink] 20 Sep 2018, 21:16
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