hsc 1987 (1 Viewer)

jxballistic · Oct 23, 2013

Can someone please show me how to do question 7? I couldnt do any of it. Hopefully tomorrows hsc doesnt have a question like this

bangladesh · Oct 23, 2013

lol... 1987... screw that.

nightweaver066 · Oct 23, 2013

possible i made sillies and i know i omitted steps but hopefully it makes sense

a)
Folding the paper across, and constructing the line P'B, you can see that triangle P'BA is congruent with triangle PBA.
So $\angle P'AB = \theta$ and $\angle P'AQ = \pi - 2\theta$ .

b) Looking at $\Delta P'AQ, \; cos(\angle P'AQ) = -cos2\theta = \frac{AQ}{AP'} = \frac{AQ}{w-AQ}$

Looking at $\Delta PAQ, \; cos\theta = \frac{w - AQ}{l}$

$AQ = w-lcos\theta$

$\therefore cos2\theta = \frac{-w + lcos\theta}{lcos\theta}$

$\frac{w}{lcos\theta} = 1 - cos2\theta$

$l = \frac{w}{cos\theta(1 - cos2\theta)}$

c)
$l = w(cos\theta(1 - cos2\theta))^{-1}$

$\frac{dl}{d\theta} = -w(cos\theta(1 - cos2\theta))^{-2}\times (2sin2\theta cos\theta + sin\theta cos2\theta - sin\theta) = 0$

$2sin2\theta cos\theta + sin\theta cos2\theta - sin\theta = 0$

$4sin\theta cos^2\theta + sin\theta(2cos^2\theta - 1) -sin\theta = 0$

$sin\theta(4cos^2\theta + 2cos^2\theta - 1 - 1) = 0$

$sin\theta(6cos^2\theta - 2) = 0$

$0 < \theta < \frac{\pi}{2}$

$cos\theta = \pm \frac{1}{\sqrt{3}}$

$\theta = cos^{-1}\frac{1}{\sqrt{3}}$ ceebs proving it gives minimum
d) plug it in for your answer

kk thats enough procrastination.

jxballistic · Oct 24, 2013

thanks man, don't really understand it tho. Oh well 3u hsc is over

hsc 1987 (1 Viewer)

jxballistic

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