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Math Help - 3 Geometry of Circles, involving parallel lines

  1. #1
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    3 Geometry of Circles, involving parallel lines

    Hi, I have 3 problems which I know the answer to, but I don't know how to reach those answers.

    Note: These diagrams are not to scale at all.

    'O' is always the center of the circle, no matter what it looks like on the diagram

    Lines in red are parallel, no matter how they look in the diagrams

    Sorry for any confusion caused by my hasty drawings

    1) In the figure, EO is parallel to FG, HOG is a diameter and angle OGF = 63 degrees. Find angle HFE.


    2) In the figure, FOH is the diameter. OE is parallel to FG and angle EOF = 48 degrees. Find angle GOH.


    3) In the figure, CB is paralle to OA and angle BOA = 47 degrees. FInd angle OAC.


    Thanks for your explanations.

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  2. #2
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    Lexington, MA (USA)
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    Hello, BG5965!


    You're expected to know these rules:

    [1] If parallel lines are cut by a transversal,
    . . .the corresponding angles are equal.

    [2] If parallel lines are cut by a transveral,
    . . .the alternate-interior angles are equal.

    [3] A central angle is measured by its intercepted arc.

    [4] An inscribed angle is measured by one-half its intercepted arc.



    1) In the figure: . EO \parallel FG,\;\angle OGF = 63^o.\;\;\;\text{Find }\angle HFE


    We are given: . \angle OGF = 63^o

    Then: . \angle HOE = 63^o\;\;{\color{blue}[1]}

    And: . \text{arc}\,\overline{HE} = 63^o\;\;{\color{blue}[3]}

    Therefore: . \angle HFE \:=\:\tfrac{1}{2}(63^o) \:=\:31\tfrac{1}{2}^o\;\;{\color{blue}[4]}




    2) In the figure: . OE \parallel FG ,\;\;\angle EOF = 48^o.\;\;\text{Find }\angle GOH

    We are given: . \angle EOF = 48^o

    Then: . \angle GFH = 48^o\;\;{\color{blue}[2]}

    And: . \text{arc}\,GH = 96^o\;\;{\color{blue}[4]}

    Therefore: . \angle GOH = 96^o\;\;{\color{blue}[3]}




    3) In the figure: . CB \parallel OA,\;\;\angle BOA = 47^o.\;\;\text{Find }\angle OAC

    We are given: . \angle BOA = 47^o

    Then: . \text{arc}\,BA = 47^o\;\;{\color{blue}[3]}

    And: . \angle BCA = \tfrac{1}{2}(47^o) \:=\:23\tfrac{1}{2}^o\;\;{\color{blue}[4]}

    Therefore: . \angle OAC = 23\tfrac{1}{2}^o\;\;{\color{blue}[2]}

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