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Math Help - Kite in circle

  1. #1
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    Kite in circle

    I'm not sure how to start the question. It will need a double angle formula later.
    Attached Thumbnails Attached Thumbnails Kite in circle-scan-1.jpg  
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  2. #2
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    hint ...

    an angle inscribed in a semicircle is a right angle.
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  3. #3
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    Hello, Stuck Man!

    Yes, both double-angle identities are useful.


    12. The diagram shows a kite who leading diagonal is a diameter of the circle.

    Find the exact values of: . (a)\;\sin(\angle XOY) \qquad\qquad (b)\;\cos(\angle XOY)
    Code:
                    Z
                  * * *  _
              *  *  |  *√5*
            * *     |     * *
         X o        |        o Y
            \       |       /
          *  \      |      /  *
          *   \     |     /   *
          *  3 \    |    / 3  *
                \ θ | θ /
           *     \  |  /     *
            *     \ | /     *
              *    \|/    *
                  * * * 
                    O
    As skeeter pointed out: . \angle X =\angle Y =90^o

    Let: \theta =\angle XOZ =\angle YOZ

    Then: . \tan\theta \:=\:\frac{\sqrt{5}}{3} \:=\:\frac{opp}{adj}

    Since opp = \sqrt{5},\;adj = 3\quad\Rightarrow\quad hyp = \sqrt{14}

    Hence: . \sin\theta \,=\,\frac{\sqrt{5}}{\sqrt{14}},\;\;\cos\theta \:=\:\frac{3}{\sqrt{14}}


    (a)\;\;\sin(XOY) \;\;=\;\;\sin(2\theta) \;\;=\;\;2\sin\theta\cos\theta \;\;=\;\;2\left(\frac{\sqrt{5}}{\sqrt{14}}\right)\  left(\frac{3}{\sqrt{14}}\right) . =\;\;\frac{6\sqrt{5}}{14} \;\;=\;\;\frac{3\sqrt{5}}{7}


    (b)\;\;\cos(XOY) \;\;=\;\;\cos(2\theta) \;\;=\;\;\cos^2\!\theta - \sin^2\!\theta \;\;=\;\;\left(\frac{3}{\sqrt{14}}\right)^2 - \left(\frac{\sqrt{5}}{\sqrt{14}}\right)^2 . =\;\;\frac{9}{14} - \frac{5}{14} \;\;=\;\;\frac{4}{14} \;\;=\;\;\frac{2}{7}

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  4. #4
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    Thanks. I didn't know about the right angle. I would probably have done the rest ok.
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