Using golden triangle to determine the value of \cos 18\degree

A golden triangle is an isosceles triangle with vertex angle of 36\degree and two congruent base angles of 72\degree

Draw a circle with point B as center and radius in the same measure as the length of BC. The circle intersects line AC at point D.

Since \triangle ABC and \triangle BCD are similar , we get

\dfrac{AC}{BC}=\dfrac{BC}{DC}

BC^2 = AC\cdotp DC

And

BC = BD = AD

Let BC =1 , AC =x

1=x(x-1)

x^2-x-1=0

Drop an altitude from point A to BC, which bisects \angle A and intersects BC at point F

Using Pythagorean Theorem

AF = \sqrt{AC^2-(\dfrac{1}{2}BC)^2}

=\sqrt{(\dfrac{1+\sqrt{5} }{2} )^2-\dfrac{1}{4} }

Use the result of (1) and (2) to obtain the final result

\cos 18\degree =\dfrac{AF}{AC} = \dfrac{\dfrac{1}{2}\sqrt{5+2\sqrt{5} } }{\dfrac{1+\sqrt{5} }{2}}

=\dfrac{\sqrt{5+2\sqrt{5} } }{1+\sqrt{5}}

=\dfrac{\sqrt{5+2\sqrt{5} } }{1+\sqrt{5}}\cdotp \dfrac{\sqrt{5}-1 }{\sqrt{5}-1 }

=\dfrac{\sqrt{(5+2\sqrt{5})(\sqrt{5}-1 )^2 }}{4}

=\dfrac{\sqrt{(5+2\sqrt{5})(6-2\sqrt{5} ) }}{4}

=\dfrac{\sqrt{30+12\sqrt{5}-10\sqrt{5}-20 } }{}

=\dfrac{\sqrt{10+2\sqrt{5} } }{4}

Now we have determined the value of \cos 18\degree using the geometric method.