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A weird trig equation

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Got a question about Student Finance? Ask the experts this week on TSR! 14-09-2014
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    (Original post by ztibor)
    THe square of the area
    \displaystyle A^2=\left (\frac{a^2cos \frac{\theta}{2}}{2}\right )^2
    I think you've made an error here and it should be

    \displaystyle A^2=\left (a^2\cos \frac{\theta}{2}\sin \frac{\theta}{2}\right )^2

    Which, once you substitute for "a", is identical to what you get from Heron's formula, so doesn't lead anywhere.
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    By equating two expressions for b from the diagram I arrived at the equation (3+\sqrt{3})\cos \alpha - (3+\sqrt{3})\sin \alpha = (\sqrt{6}+\sqrt{2})\sin \alpha \cos \alpha+(\sqrt{6}-\sqrt{2})\sin^2 \alpha

    where \alpha = \frac{\theta}{2}.

    Then by substituting using t=\tan(\alpha /2 ) you can arrive at a quartic equation in t...it's fairly ugly but like all the other equations, Wolframalpha can solve it and gives t=\sqrt{2}-1 leading to \theta=\frac{\pi}{2}.

    In case any one is interested..

    http://www.wolframalpha.com/input/?i...1-t%5E2%29%5E2
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    Of course, knowing the answer, we could factorise the quartic to arrive at our solution but that's not very satisfying.
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    I did it with vectors. It was probably even more longwinded than BabyMaths effort so I'm not going to bother to post any working.

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