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Math Help - 3 circle questions

  1. #1
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    3 circle questions

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  2. #2
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    Jelo, dgenerationx2!

    For the first two, we need this theorem:

    The angle between a tangent and a secant (to the point of tangency)
    . . is one-half the intercepted arc.


    1) Given: . \text{arc}(AC) \,=\,50^o,\;\;\text{arc}(AFD) \,=\,190^o,\;\;\text{tangent } BE. . Find: \angle BAC.

    . . (A)\;25^o\qquad (B)\;50^o\qquad(C)\;95^o \qquad (D)\;60^o
    Code:
             D    * * *
              o           *
            *  *            o
           *               F *
                *
          *                   * 190
          *      *            *
        C o                   *
            *     *
           *  *              *
        50 *   *  *        *
              *   *       *
          B - - - * o * - - - - E
                    A

    \text{arc}(AFD) = 190^o,\;\text{arc}(AC) = 50^o \quad\Rightarrow\quad \text{arc}(CD) = 120^o \quad\Rightarrow\quad \angle CAD \,=\,60^o

    \text{arc}(DCA) = 170^o \quad\Rightarrow\quad \angle DAB \,=\,85^o

    Therefore: . \angle BAC \:=\:85^o-60^o \:=\:25^o . . . answer (A)




    Given: .tangent AB and secant BC,\;\;\text{arc}(BDC) = 200^o . Find: \angle ABC
    Code:
                  * * *
              *           *
            *               *  C
           o D                o
                            /
          *                /  *
     200 *               /   *
          *              /    *
                        /
           *           /     *
            *         /     *
              *      /    *
                  * o * - - - - - A
                    B

    We hafve: . \text{arc}(BDC) = 200^o \quad\Rightarrow\quad \text{arc}(BC) = 160^o

    Therefore: . \angle ABC \,=\,80^o . . . answer (A)

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