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Re: Length of a fold problem

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Originally Posted by

**grelsan** Attachment 30424
I have inserted a picture of the problem here's the text

The lower right hand corner of a long piece of paper 6" wide is folded over to the left-hand edge as shown. The length L of the fold depends on the angle theta. Show that L=3/sin(theta)cos^2(theta)

I honestly can't even figure out where to begin with this problem any help would be much appreciated.

Thanks

Hello,

I've attached a picture too.

You have 2 congruent right triangles whose legs are x and y.

Additionally there is a right triangle with the hypotenuse x and one leg has the length 6'' (in orange).

You'll get:

$\displaystyle \frac6x = \sin(2 \theta) = 2 \cdot \sin(\theta) \cdot \cos(\theta)~\implies~x = \frac3{\sin(\theta) \cdot \cos(\theta)}$

With the grey right triangle you get:

$\displaystyle \frac xL = \cos(\theta)~\implies~L=\frac x{\cos(\theta)}$

Replace x by the first term:

$\displaystyle \frac xL = \cos(\theta)~\implies~L=\frac {\frac3{\sin(\theta) \cdot \cos(\theta)}}{\cos(\theta)}$

Simplify:

$\displaystyle L=\frac3{\sin(\theta) \cdot \cos^2(\theta)}$

Re: Length of a fold problem

Thanks for the assist! I have to say Im slow or thatsbone devious problem even with yoir excellant explantion it still took me a minute to see everything. Thanks!