Earlier today, I was trying to place my storybook into my wardrobe drawer, only to find that I could only place it horizontally as the height of my book exceeds the height of my drawer. Out of curiosity, I tried to rotate the book by holding the spine and observe how much it can rotate, until three of the vertices touched a point of the drawer.
A question came to me as I had reached a point of time when the book could no longer rotate further: By keeping the height of the drawer constant, what is the minimum length of the drawer such that the book can be stored inside it? (A silly question, I know, considering that drawers are always quite long, but aroused my interest nevertheless..)
I proceeded by measuring the dimensions I think I need to solve my problem. I have measured:
The height of the drawer, $17 cm$
The length of the spine of the book, $2.2 cm$
The height of the spine of the book, $17.8 cm$
And I have attempted to represent the problem in the diagram below, solving for $x$ With my mathematics knowledge up to grade 10, initially, I had thought that this will be solved using simultaneous equations, but after pondering for quite some time, I had managed to come up with an approach using R-Formula:
I started by setting $\angle WDA=\theta$, and since the drawer and my book are rectangular, $\angle BAX=\angle DCZ=\theta$
Using these information, I expressed $x$ in trigonometric terms: $x=17.8 \sin\theta + 2.2 \cos\theta$
Using the R-Formula, I get $x=\sqrt(321.68)\sin(\theta+7.045769125^\circ)$.
Next, I attempt to solve for $\theta$ by expressing $WZ=17$ in trigonometric terms as well.
I get: $17=17.8\cos\theta+2.2\sin\theta$
$17=\sqrt(321.68)\cos(\theta-7.045769125^\circ)$ $\theta\approx 25.63217562^\circ$
Finally, I substituted this value of $\theta$ into $x$ and I get: $x=\sqrt(321.68) \sin(32.67794475^\circ)$ $x\approx 9.683637217cm$
My actual measurement of the value of $x$ is $10.5cm$. I did expect inaccuracy as I measured just with a fifteen-centimetre ruler, a pair of wobbly hands and considered parallax error, although I did not expect the discrepancy to be nearly $1cm$.
I have two questions regarding this problem of mine:
Is my approach a valid approach?
Is there an easier approach to solve this problem?