Monday, January 23, 2012

Problem 718: Intersecting Circles, Midpoint, Angle, Measurement

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 718 details.

Online Geometry Problem 718: Intersecting Circles, Midpoint, Angle, Measurement.

Posted by Antonio Gutierrez at 10:43 AM
Labels: , , , ,

4 comments:

  1. PravinJanuary 23, 2012 at 5:53 PM

    Extend CDO' to cut the circle O' at E.
    Join AE, BE. Let angle AEB = y
    By the result of Problem 717,
    y = 65 deg
    But AEBD is a cyclic quadrilateral.
    So x + y = 180 deg
    Hence x = 115 deg

    ReplyDelete
  2. anway deshpandeFebruary 18, 2012 at 9:01 AM

    By applying the formula that angle bda=90+bca/2 therefore half of 50=25 and by adding 90 we get 90+25=115
    therefore measure of angle bda =115

    ReplyDelete
  3. PravinFebruary 19, 2012 at 4:30 AM

    x
    = angle subtended by major arc AB of circle(O')at D
    = half the ∠ subtended by arc AB of circle (O')at center O'.
    = (1/2)major∠AO'B
    = (1/2)(360°-minor∠AO'B)
    = (1/2)[360°-(180°-∠ACB)] (∵ AO'BC is a cyclic quadrilateral)
    = (1/2)(180°+∠ACB)
    = (1/2)(180°+50°)
    = (1/2)(230°)
    = 115°

    ReplyDelete
  4. MarcoJanuary 12, 2024 at 9:55 PM

    <AO'B=180-50=130
    <ADB=180-<AO'B/2=180-65=115

    ReplyDelete

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