<![CDATA[Math Is Fun Forum / tan 36]]> 2016-04-21T08:57:24Z FluxBB https://www.mathisfunforum.com/viewtopic.php?id=22964 <![CDATA[Re: tan 36]]>

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https://www.mathisfunforum.com/profile.php?id=199410 2016-04-21T08:57:24Z https://www.mathisfunforum.com/viewtopic.php?pid=379634#p379634
<![CDATA[Re: tan 36]]>

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https://www.mathisfunforum.com/profile.php?id=212486 2016-04-21T05:52:42Z https://www.mathisfunforum.com/viewtopic.php?pid=379619#p379619
<![CDATA[tan 36]]> Start with an isosceles triangle ABC with angle C = 36, and A = B = 72.

Bisect A and continue this line to cut BC at D.  E is the 'foot' of the perpendicular from D to AC.

There are two more isosceles triangles in this diagram; ADC is 36,36, 108; and ABD is 36,72,72 again.  I have marked the angles of 36 with a dot to avoid too much labelling on the diagram.

Let AC = BC = x and AB = AD = 1.

As ABC and ABD are similar, the ratios of corresponding sides are equal:

This quadratic has two solutions but as we know x is positive we can discard the negative solution:

So

As x has a square root in its form and the above also I will simplify by working with tan squared.

Now errors can occur if you square and then square root an expression.  But in this case we can safely take the positive root as we know tan(36) will be positive.

If a right angled triangle has an opposite of \sqrt{5 - 2\sqrt{5}} and an adjacent of 1, then the smallest angle will be 36 degrees.

By Pythagoras:

Therefore

and

Bob

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https://www.mathisfunforum.com/profile.php?id=67694 2016-04-01T07:46:47Z https://www.mathisfunforum.com/viewtopic.php?pid=378587#p378587