Pls find attached work.

I would really appreciate some help.

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- Jul 10th 2011, 04:36 AMVeronica1999Angle word problem
Pls find attached work.

I would really appreciate some help. - Jul 10th 2011, 05:08 AMkalyanramRe: Angle word problem
Hi Veronica,

There seems to be a contradiction in your results. I cannot make it out because your attempt in the first and second halves are cramped into small space. Here is where I find the contradiction.

You obtained the length of crease $\displaystyle x = 16.\surd 3$ for $\displaystyle \theta = 30$ and $\displaystyle x(\theta)$=$\displaystyle 12.cos\theta . sin(2\theta)$ and substituting $\displaystyle \theta = 30$ we have $\displaystyle x = 12. \frac{\surd 3}{2}. \frac{\surd 3}{2} = 9$ - Jul 10th 2011, 05:24 AMVeronica1999Re: Angle word problem
Then is my crease length of 16 root 3 correct?

- Jul 10th 2011, 05:40 AMkalyanramRe: Angle word problem
Let the length of the crease be $\displaystyle x$ and width of paper be $\displaystyle l$

$\displaystyle \angle CAB = \angle BAD = \theta , \angle EBC = 180 - 2.(90 - \theta) = 2\theta$

Now we have $\displaystyle DB + BE = l \Rightarrow x.sin\theta + x.sin\theta . cos2\theta = l \Rightarrow x = \frac{l}{sin\theta (1 + cos2\theta)}$

$\displaystyle \Rightarrow x = \frac{l}{2.(sin\theta).(cos\theta)^2}$

Now evaluate for minimum of $\displaystyle x$ by differentiating.

Kalyan. - Jul 10th 2011, 06:08 AMkalyanramRe: Angle word problem
Instead of differentiating $\displaystyle x$ w.r.t $\displaystyle \theta$ try maximizing the function $\displaystyle f(\theta) = sin\theta.cos^2 \theta$ this gives $\displaystyle f'(\theta) = cos^3 \theta - 2.sin^2 \theta.cos\theta = 0$ this gives us $\displaystyle \theta = 90 $ or $\displaystyle tan\theta = \frac{1}{\surd 2} $. We take the angle $\displaystyle \theta = tan^{-1} (\frac{1}{\surd 2}) $

Kalyan.