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Math Help - A question based on the sine & cosine formulas

  1. #1
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    A question based on the sine & cosine formulas

    The acute triangle \triangle ABC inscribes a circle. D, E, and F are the tangent points. It is given that: \angle EFD=\alpha, \angle FED=\beta, and AF=a (see picture).
    Using \alpha ,\beta and a, express the following:
    (a) the length of EF.
    (b) the length of DF.

    Well, first thing I thought about is that the angle between a tangent and chord is equal to the subtended angle on the opposite side of the chord, hence \angle BFD=\beta , and  \angle AFE=180-\alpha -\beta , and since the triangle \triangle AFE is an isosceles triangle (tangents drawn from a point outside the circle are equal in length), \angle AEF=180-\alpha -\beta as well. In addition, \angle D=180-\alpha -\beta and \angle A=180-2*(180-\alpha- \beta )= -180+2\alpha +2\beta (is that correct?)

    Does what I did help in solving the question? And what am I supposed to do next? (I must use the sine/cosine formulas)

    Thanks.
    Attached Thumbnails Attached Thumbnails A question based on the sine & cosine formulas-ex6.jpg  
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  2. #2
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    Hello, loui1410!

    Everything you did was splendid!


    Since \Delta AFE isosceles: . \angle AEF \:=\:\angle AFE \:=\:180 - \alpha - \beta


    We case use the Law of Sines on \Delta AFE

    . . \frac{EF}{\sin A} \:=\:\frac{a}{\sin(\angle AEF)}\quad\Rightarrow\quad EF \:=\:\frac{a\cdot\sin(2\alpha + 2\beta - 180)}{\sin(180 - \alpha - \beta)}

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  3. #3
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    Hmm, then how can I reach the textbook's answer:
     EF=-2 * \alpha * \cos ( \alpha + \beta )
    ?
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  4. #4
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    Quote Originally Posted by loui1410 View Post
    The acute triangle \triangle ABC inscribes a circle. D, E, and F are the tangent points. It is given that: \angle EFD=\alpha, \angle FED=\beta, and AF=a (see picture).
    Using \alpha ,\beta and a, express the following:
    (a) the length of EF.
    (b) the length of DF.

    Well, first thing I thought about is that the angle between a tangent and chord is equal to the subtended angle on the opposite side of the chord, hence \angle BFD=\beta , and  \angle AFE=180-\alpha -\beta , and since the triangle \triangle AFE is an isosceles triangle (tangents drawn from a point outside the circle are equal in length), \angle AEF=180-\alpha -\beta as well. In addition, \angle D=180-\alpha -\beta and \angle A=180-2*(180-\alpha- \beta )= -180+2\alpha +2\beta (is that correct?)

    Does what I did help in solving the question? And what am I supposed to do next? (I must use the sine/cosine formulas)

    Thanks.


    a)

    \frac {EF}{sin\angle EAF}=\frac {a}{sin\angle AEF}
    \frac {EF}{sin(2\alpha +2\beta -180^o)}=\frac {a}{sin(180^o -\alpha -\beta)}
    EF=\frac {asin(2\alpha +2\beta -180^o)}{sin(180^o -\alpha -\beta)}
    EF=\frac {a[sin2(\alpha+\beta)cos180^o -sin180^ocos2(\alpha+\beta)]}{sin180^o cos(\alpha +\beta)-sin(\alpha +\beta)cos180^o}
    EF=\frac {a[-sin2(\alpha+\beta)]}{sin(\alpha +\beta)}
    EF=\frac {-a[sin2(\alpha+\beta)]}{sin(\alpha +\beta)}
    EF=\frac {-a[2sin(\alpha+\beta)cos(\alpha+\beta)]}{sin(\alpha +\beta)}
    EF=-2acos(\alpha+\beta)

    b)

    \frac {DF}{sin \beta}=\frac {EF}{sin(2\alpha +2\beta -180^o)}
    DF=\frac {-2acos(\alpha+\beta)sin\beta}{sin(2\alpha +2\beta -180^o)}
    DF=\frac {-2acos(\alpha+\beta)sin\beta}{-2sin(\alpha +\beta)cos(\alpha +\beta)}
    DF=\frac {asin\beta}{sin(\alpha +\beta)}


    The end.
    Last edited by SengNee; January 9th 2008 at 11:41 PM.
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  5. #5
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    Thanks a lot SengNee. The teacher said today that there is still a phase of abstraction that we haven't learnt yet and therefore we don't have to get the same answer as the textbook's, so this is why I didn't know how to proceed.
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  6. #6
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    Hello, loui1410!

    Then how can I reach the textbook's answer: .  EF\:=\:-2a\cdot\cos(\alpha + \beta)  ?
    We need a few identities . . .

    . . \sin(-\theta)\:=\:-\sin\theta

    . . \sin(180^o - \theta) \:=\:\sin\theta

    . . \sin2\theta \:=\:2\sin\theta\cos\theta


    We had: . EF \;=\;\frac{a\cdot\sin(2\alpha + 2\beta - 180)}{\sin(180-\alpha-\beta)} .[1]


    The numerator is: . a\cdot\sin\bigg(-[180 - 2(\alpha + \beta)]\bigg) \;=\;-a\cdot\sin[180 - 2(\alpha + \beta)]

    . . = \;-a\cdot\sin[2(\alpha + \beta)] \;=\;-2a\cdot\sin(\alpha+\beta)\cdot\cos(\alpha+\beta)


    The denominator is: . \sin[180 - (\alpha+\beta)] \;=\;\sin(\alpha+\beta)


    Then [1] becomes: . EF \;=\;\frac{-2a\cdot\sin(\alpha+\beta)\cdot\cos(\alpha+\beta)}{  \sin(\alpha+\beta)} \;=\;-2a\cdot\cos(\alpha+\beta)

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