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http://geometri-problemleri.blogspot.com/2010/10/problem-94-ve-cozumu.html
Reflect the upper part of the equilateral triangle about he middle parallel line.Then it is easy to see that the tangent of the two angles are in the ratio of (b+a)/(b-a)
Let c be the edge length of the equilateral triangle.* b + a = c*sin(θ)* a = c*sin(θ-α) = c*sin(θ-60°)* b - a = (b+a) - 2*a = c*sin(θ) - 2*c*sin(θ-60°)Therefore,(b+a)/(b-a) = c*sin(θ)/(c*sin(θ)-2*c*sin(θ-60°))= sin(θ)/(sin(θ)-2*sin(θ-60°))= sin(θ)/(sin(θ)-2*(1/2*sin(θ)-sqrt(3)/2*cos(θ))) = sin(θ)/(sqrt(3)*cos(θ))= sin(θ)/(tan(α)*cos(θ))= tan(θ)/tan(α)Please note that sqrt(3) equals to tan(α).
http://geometri-problemleri.blogspot.com/2010/10/problem-94-ve-cozumu.html
ReplyDeleteReflect the upper part of the equilateral triangle about he middle parallel line.
DeleteThen it is easy to see that the tangent of the two angles are in the ratio of (b+a)/(b-a)
Let c be the edge length of the equilateral triangle.
ReplyDelete* b + a = c*sin(θ)
* a = c*sin(θ-α) = c*sin(θ-60°)
* b - a = (b+a) - 2*a
= c*sin(θ) - 2*c*sin(θ-60°)
Therefore,
(b+a)/(b-a) = c*sin(θ)/(c*sin(θ)-2*c*sin(θ-60°))
= sin(θ)/(sin(θ)-2*sin(θ-60°))
= sin(θ)/(sin(θ)-2*(1/2*sin(θ)-sqrt(3)/2*cos(θ)))
= sin(θ)/(sqrt(3)*cos(θ))
= sin(θ)/(tan(α)*cos(θ))
= tan(θ)/tan(α)
Please note that sqrt(3) equals to tan(α).