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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

    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:


    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:


    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:


    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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