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Thread: 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 $\displaystyle \triangle ABC$ inscribes a circle. D, E, and F are the tangent points. It is given that: $\displaystyle \angle EFD=\alpha$, $\displaystyle \angle FED=\beta$, and AF=a (see picture).
    Using $\displaystyle \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 $\displaystyle \angle BFD=\beta , $ and $\displaystyle \angle AFE=180-\alpha -\beta $, and since the triangle $\displaystyle \triangle AFE$ is an isosceles triangle (tangents drawn from a point outside the circle are equal in length), $\displaystyle \angle AEF=180-\alpha -\beta $ as well. In addition, $\displaystyle \angle D=180-\alpha -\beta $ and $\displaystyle \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.
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  2. #2
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    Hello, loui1410!

    Everything you did was splendid!


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


    We case use the Law of Sines on $\displaystyle \Delta AFE$

    . . $\displaystyle \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:
    $\displaystyle 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 $\displaystyle \triangle ABC$ inscribes a circle. D, E, and F are the tangent points. It is given that: $\displaystyle \angle EFD=\alpha$, $\displaystyle \angle FED=\beta$, and AF=a (see picture).
    Using $\displaystyle \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 $\displaystyle \angle BFD=\beta , $ and $\displaystyle \angle AFE=180-\alpha -\beta $, and since the triangle $\displaystyle \triangle AFE$ is an isosceles triangle (tangents drawn from a point outside the circle are equal in length), $\displaystyle \angle AEF=180-\alpha -\beta $ as well. In addition, $\displaystyle \angle D=180-\alpha -\beta $ and $\displaystyle \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)

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

    b)

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


    The end.
    Last edited by SengNee; Jan 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: .$\displaystyle EF\:=\:-2a\cdot\cos(\alpha + \beta) $ ?
    We need a few identities . . .

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

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

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


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


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

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


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


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

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