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Math Help - circles again

  1. #1
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    Angry circles again

    CD is tangent to the circle and AE is perpendicular to AC. If a radius of the circle measures 12 in., which is the vale of DF + AF?
    Last edited by sanee66; July 12th 2007 at 05:49 PM. Reason: attachment
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  2. #2
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    I can't see the image
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  3. #3
    MHF Contributor red_dog's Avatar
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    Let R=12,CF=x,FA=y\Rightarrow x+y=12.
    In \triangle DCF \ \sin 30=\frac{x}{DF}\Rightarrow DF=2x.
    In \triangle EAF \ \tan 30=\frac{y}{R}\Rightarrow y=4\sqrt{3}\Rightarrow x=12-4\sqrt{3}.
    Then DF+FA=2x+y=24-8\sqrt{3}+4\sqrt{3}=24-4\sqrt{3}
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  4. #4
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    Hello, sanee66!


    In right triangle EAF\!:\;\tan30^o \,=\,\frac{AF}{12}\quad\Rightarrow\quad AF \:=\:12\tan30^o \:=\:4\sqrt{3}

    Then: . CF \:=\:12 - AF \:=\:12 - 4\sqrt{3}\quad\Rightarrow\quad CF\:=\:4(3-\sqrt{3}) .[1]

    In right triangle DCF\!:\;\sin30^o \:=\:\frac{CF}{DF}\quad\Rightarrow\quad DF \:=\:\frac{CF}{\sin30^o} \:=\:2\!\cdot\!CF .[2]

    Substitute [1] into [2]: . DF \:=\:8(3-\sqrt{3})


    Therefore: . DF + AF \;=\;8(3-\sqrt{3}) + 4\sqrt{3} \;=\;24 - 4\sqrt{3}


    [Darn, this is identical to red_dog's solution!]

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  5. #5
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    Using trigo. for this problem is like kill a fly with a Grand Cannon.
    • \measuredangle~AEF=30^\circ~\because~\overline{CD}  \parallel\overline{AE}
    • We have two triangles 30^\circ,\,60^\circ,\,90^\circ
    • Finally \overline{AF}=4\sqrt3\implies\overline{DF}=24-8\sqrt3~\therefore~\overline{DF}+\overline{AF}=24-4\sqrt3
    Properties of triangle 30^\circ,\,60^\circ,\,90^\circ are easily proved applying the Pythagorean Theorem.
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