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#1 2016-03-31 19:46:47

Registered: 2010-06-20
Posts: 10,472

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:


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:




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#2 2016-04-20 17:52:42

Registered: 2016-04-16
Posts: 1,086

Re: tan 36

Last edited by thickhead (2016-04-20 18:02:42)

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#3 2016-04-20 20:57:24

Registered: 2013-03-09
Posts: 957

Re: tan 36

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