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Math Help - Triangle related question

  1. #1
    Senior Member pankaj's Avatar
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    Triangle related question

    If D is the mid-point of CA in triangle ABC and \Delta is the area of triangle,then show that
     <br />
\tan(\angle ADB) = \frac{4\Delta}{a^2 - c^2}<br />
    where a=BC,b=CA and c=AB
    Last edited by pankaj; September 14th 2008 at 08:05 AM.
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,
    Quote Originally Posted by pankaj View Post
    If D is the mid-point of CA in triangle ABC and \Delta is the area of triangle,then show that
     <br />
\tan(\angle ADB) = \frac{4\Delta}{a^2 - c^2}<br />
    where a=BC,b=CA and c=AB
    Please have a look at the sketch.

    Triangle related question-aire.jpg

    BE is the altitude of the triangle coming from B.
    Thus \mathcal{A}_{\text{ABC}}=\Delta=\frac{\text{BE} \times b}{2} \implies \text{BE}=\frac{2 \Delta}{b}

    Note that you want \tan(\angle \text{ADB}). Obviously, it is equal to \frac{\text{BE}}{\text{ED}}=\boxed{\frac{2 \Delta}{b \times \text{ED}}} because the triangle BED is a right angle triangle in E.

    Now, you'll have to calculate ED, since you already know the others.

    D is the midpoint of AC. Therefore \text{AD}=\frac{\text{AC}}{2}=\frac b2
    And \text{ED}=\text{AD}-\text{AE}=\frac b2-\text{AE}

    If we multiply by 2, we can write 2\text{ED}=(b-\text{AE})-\text{AE}=\text{CE}-\text{AE}
    Applying Pythagoras theorem in AEB, we have \text{AE}=\sqrt{c^2-\text{BE}^2}
    Applying Pythagoras theorem in BEC, we have \text{CE}=\sqrt{a^2-\text{BE}^2}
    \implies \boxed{\text{ED}=\frac{\sqrt{a^2-\text{BE}^2}-\sqrt{c^2-\text{BE}^2}}{2}}

    --------------------------------------------------------------
    See the two boxed formula ?

    This gives :

    \tan(\angle \text{ADB})=\frac{4 \Delta}{b \times \left(\sqrt{a^2-\text{BE}^2}-\sqrt{c^2-\text{BE}^2}\right)}

    Multiply by \frac{\sqrt{a^2-\text{BE}^2}+\sqrt{c^2-\text{BE}^2}}{\sqrt{a^2-\text{BE}^2}+\sqrt{c^2-\text{BE}^2}}=\frac{\text{AE}+\text{EC}}{\sqrt{a^2-\text{BE}^2}+\sqrt{c^2-\text{BE}^2}}=\frac{b}{\sqrt{a^2-\text{BE}^2}+\sqrt{c^2-\text{BE}^2}}

    And please observe

    (in fact, the important step was to multiply the denominator by its conjugate, and it just popped up that it was the length b)
    ---------------------------------------------------------------------------
    There may be a shorter way, but I don't know... Inspire yourself from it
    Last edited by Moo; September 14th 2008 at 09:25 AM.
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  3. #3
    Senior Member pankaj's Avatar
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    Hello Moo,I am really inspired by your diagram.Here we go:
     <br />
cos \theta=\frac{BD^2 + \frac{b^2}{4} - c^2}{2.BD.\frac{b}{2}}=\frac{\frac{2a^2 + 2c^2 - b^2}{4} + \frac{b^2}{4} - c^2}{b.BD}=\frac{a^2 - c^2}{2.b.BD}<br />

    Now,using sine rule,we have
    \frac{\sin \theta}{c} = \frac{\sin A}{BD} .
    BD=\frac{c\sin A}{\sin \theta}=\frac{ac}{2R\sin \theta}

    Therefore,eliminating BD we get
     <br />
\tan \theta=\frac{abc}{(a^2 - c^2)R}=4\frac{abc}{4R}.\frac{1}{a^2 - c^2}=\frac{4\Delta}{a^2-c^2}<br />
    I would want to hear more about that dyslexic and someone plz solve the other trigonometry question which i posted yesterday
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