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L=3/sinΘcos²Θ - Proving the equation

by Andrew
(New York)












































The lower right-hand corner of a long piece of rectangular paper 6 in. wide is folded over to the left-hand edge as shown. The length L of the fold
depends on the angle (Theta). Show that: (refer to equation above.)

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Apr 16, 2012
L=3/sinΘcos²Θ - Proving the equation
by: Staff


Question:

by Andrew
(New York)


The lower right-hand corner of a long piece of rectangular paper 6 in. wide is folded over to the left-hand edge as shown. The length L of the fold depends on the angle (Theta). Show that: (refer to equation above.)



Answer:

Refer to the following diagram:

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http://www.solving-math-problems.com/images/triangle-diagram-2012-04-16-01.png




AB ∥DE

AB = 6



△CAD ≅ △EAD

∠CAD = ∠EAD = Θ

∠BAC = 90° - 2Θ



cos(90°-2Θ) = AB/AC = 6/AC

cos(90°-2Θ) = sin(2Θ) = 6/AC



sin(2Θ) = 6/AC

AC = 6/sin(2Θ)



cos(Θ) = AC/L

L = AC/cos(Θ)



L = [6/sin(2Θ)]/cos(Θ)

L = 6/[sin(2Θ)*cos(Θ)]



sin(2Θ) = Sin (Θ + Θ) = Sin(Θ)Cos(Θ) + Cos (Θ)Sin(Θ)= 2 Sin(Θ)Cos(Θ)

sin(2Θ) = 2*sin(Θ)*cos(Θ)



L = 6/[2*sin(Θ)*cos(Θ)*cos(Θ)]

L = 3/[sin(Θ)*cos²(Θ)]




Thanks for writing.

Staff
www.solving-math-problems.com



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