Golden Ratio and the Sine of 18Date: 04/17/2001 at 12:42:32 From: David Zaya Subject: Golden Ratio and the sine of 18 I recently found that the 2*sin(18) + 1 is equal to the golden ratio. I was wondering if there was any significance to this. Date: 04/17/2001 at 14:13:24 From: Doctor Jaffee Subject: Re: Golden Ratio and the sine of 18 Hi David, Yes, there is. The connection has to do with what is known as the Golden Triangle. It is an isosceles triangle with vertex angle measuring 36 degrees and base angles measuring 72 degrees. For a picture, see this answer from the Dr. Math archives: Golden Triangle http://mathforum.org/dr.math/problems/byerly.9.19.99.html Like the Golden Rectangle, which has the property that you can cut off a portion of the rectangle that is proportional to the original rectangle, if you bisect a base angle of the Golden Triangle, you form two new triangles, each similar to the original triangle. If we assume that the length of each congruent side of the original triangle is 1 and the length of the base is x, each of the congruent sides of one of the new triangles will be x and its base will have length 1 - x. Setting up the proportion of corresponding sides, we get: x 1 ------- = --- 1 - x x Solving for x we get: -1 + sqrt(5) x = ------------ 2 Now go back to the original triangle and bisect the vertex. You now have two right triangles with acute angles measuring 18 degrees and 72 degrees. Using the definition of sin A = opposite/hypotenuse and the value of x we just determined, the result is -1 + sqrt(5) sin 18 = ------------ 4 I hope this explanation answers your question. Write back if you want to discuss it some more. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/