# Angle word problem

• Jul 10th 2011, 04:36 AM
Veronica1999
Angle word problem
Pls find attached work.
I would really appreciate some help.
• Jul 10th 2011, 05:08 AM
kalyanram
Re: 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 $x = 16.\surd 3$ for $\theta = 30$ and $x(\theta)$= $12.cos\theta . sin(2\theta)$ and substituting $\theta = 30$ we have $x = 12. \frac{\surd 3}{2}. \frac{\surd 3}{2} = 9$
• Jul 10th 2011, 05:24 AM
Veronica1999
Re: Angle word problem
Then is my crease length of 16 root 3 correct?
• Jul 10th 2011, 05:40 AM
kalyanram
Re: Angle word problem
Let the length of the crease be $x$ and width of paper be $l$
$\angle CAB = \angle BAD = \theta , \angle EBC = 180 - 2.(90 - \theta) = 2\theta$
Now we have $DB + BE = l \Rightarrow x.sin\theta + x.sin\theta . cos2\theta = l \Rightarrow x = \frac{l}{sin\theta (1 + cos2\theta)}$
$\Rightarrow x = \frac{l}{2.(sin\theta).(cos\theta)^2}$

Now evaluate for minimum of $x$ by differentiating.

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

Kalyan.