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Thread: Iscosceles triangle

  1. #1
    mms
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    Iscosceles triangle

    AB=AC
    find x

    thank you
    Attached Thumbnails Attached Thumbnails Iscosceles triangle-dibujo1.jpg  
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  2. #2
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    Quote Originally Posted by mms View Post
    AB=AC
    find x

    thank you
    $\displaystyle 180^0-50^o-60^o=70^o$
    the angle opposite of it is also $\displaystyle 70^o$ and the other two angles are $\displaystyle 50^o$ and $\displaystyle 60^o$
    the two triangles are similar, and i can't exactly remember which rule, but x would be $\displaystyle 60^o$
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  3. #3
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    Hello mms
    Quote Originally Posted by mms View Post
    AB=AC
    find x

    thank you
    This is surprisingly tricky. The answer is $\displaystyle x = 110^o$, which I can prove using trigonometry, but I'm not sure whether there's a simpler explanation. My solution is this.

    Let the points on the lines $\displaystyle AB$ and $\displaystyle AC$ be $\displaystyle D$ and $\displaystyle E$ respectively, and suppose that $\displaystyle AB = AC = y$.

    Then $\displaystyle \angle ABC = \angle ACB = 80^o$ (isosceles $\displaystyle \triangle ABC$)

    $\displaystyle \Rightarrow \angle ABE = 20^o$ and $\displaystyle \angle ACD = 30^o$

    So using Sine Rule on $\displaystyle \triangle$ís $\displaystyle ABE$ and $\displaystyle ACD$:

    $\displaystyle AE = \frac{y\sin20}{\sin140} = 0.5321y, AD = \frac{y\sin 30}{\sin130}= 0.6527y$

    Now use Cosine Rule on $\displaystyle \triangle ADE$:

    $\displaystyle DE^2 = 0.5321^2y^2 + 0.6527^2y^2 - 2\times 0.5321\times 0.6527 y^2\cos20 = 0.0564y^2$

    $\displaystyle \Rightarrow DE = 0.2376y$

    $\displaystyle \Rightarrow \sin\angle EDA = \frac{AE\sin20}{DE}=\frac{0.5321\sin20}{0.2376}=0. 9395$

    $\displaystyle \Rightarrow \angle EDA = 50^o$

    $\displaystyle \Rightarrow x = 110^o$

    Perhaps someone can come up with a more direct method?

    Grandad

    PS Correction!

    Sorry, I calculated the wrong angle. It's $\displaystyle \angle DEA$ that's $\displaystyle 110^o$. bjhopper is quite right. $\displaystyle x = 140 - 110 = 30^o$
    Last edited by Grandad; Jul 10th 2009 at 05:33 AM. Reason: Add PS
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    isosceles triangle originally posted by mms

    your diagram is a bit distorted..the line forming the x angle should be oblique.this is not an easy problem.in fact it is more of a puzzle.it comes uo on the internet as the hardest / easiest problem in geometry.

    the answer is 30.proof requires construction lines sorry i cannot put drawings in this reply.other responders migth help


    bjh
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