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

This topic has 2 expert replies and 1 member reply

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

Post Tue Feb 14, 2017 6:33 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])


    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

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    Post Tue Feb 14, 2017 8:58 am
    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



    Last edited by regor60 on Tue Feb 14, 2017 10:54 am; edited 2 times in total

    Post Tue Feb 14, 2017 10:00 am
    Brent@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

    Answer: E

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course
    Come see all of our free resources

    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!
    Post Tue Feb 14, 2017 1:19 pm
    Let x = 55, implying that 3x=155.

    Since ∠BCE and ∠ACB must sum to 180, we get:
    ttps://postimg.org/image/yesjxs9wj/" target="_blank">

    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:
    ttps://postimg.org/image/d9s8uaztj/" target="_blank">

    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:
    ttps://postimg.org/image/i4nrxvz37/" target="_blank">

    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.

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