Help calculating angles for woodworking

Question

I am working on a wood working project and need to cut some 2 x 2's on an angle in order to form an X inscribed inside a rectangle. Visually here is what I am trying to create:

So basically I want to figure out what the angle $\alpha$ and $\beta$ are so that I can cut the wood to form the X inside the rectangle. The rectangle is 30 inches by 16.5 inches and the width of the wood for the X is 2 inches. I tried using simple trigonometry/geometry to figure it out but have been stumped so far as to how to do it. Also, what would the length of each piece of wood making up the X?

Answer 1

If you let $s$ be the length of wood in the bottom left of the $X$ that touches the side, then you get $$\tan(\alpha) = \frac{s}{2}.$$

Additionally, we have that a right triangle with angle $\alpha$, adjacent side length $16.5 - 2 - 2 = 12.5$ and opposite side length $30 - 2 - 2 - s = 26 - s$. This gives another equation of $$\tan(\alpha) = \frac{26 - s}{12.5}.$$ Putting the two equations together gives $$\tan(\alpha) = \frac{52}{29}.$$ This gives $\alpha = \arctan(52/29) \approx 60.85^\circ$. Also, $\beta = 90^\circ - \alpha$.

I'm pretty sure this is right, but would appreciate if someone would double check this (especially since it's actually gonna be built!)

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EDIT: The first equation is wrong, it should be $\cos(\alpha) = \frac{2}{s}$, implying $$\tan(\alpha) = \frac{26 - \frac{2}{\cos(\alpha)}}{12.5} \implies \alpha \approx 60.348^\circ$$.

Answer 2

Given a rectangle origin centered with external dimensions height $h$ and width $w$ and build with $\delta$ thick woods, to construct the needed two diagonals we proceed as follows. First we consider a circle origin centered with radius $\delta/2$. After that a line tangent to this circle and passing through the inner rectangle corner, will give us the desired angle $\alpha$. Calling $\alpha_0 = \arctan\left(\frac{h/2-\delta}{w/2-\delta}\right)$ and $\gamma = \arctan\left(\frac{\delta/2}{\sqrt{(h/2-\delta)^2+(w/2-\delta^2)-(\delta/2)^2}}\right)$ we have $\alpha = \alpha_0-\gamma$ or

$$ \alpha = \arctan\left(\frac{h/2-\delta}{w/2-\delta}\right)-\arctan\left(\frac{\delta/2}{\sqrt{(h/2-\delta)^2+(w/2-\delta^2)-(\delta/2)^2}}\right) $$

now using the trigonometric formula $\tan(a-b) = \frac{\tan a-\tan b}{1+\tan a\tan b}$ we obtain easily

$$ \tan\alpha = \frac{4 \delta^2+h w-\delta \left(\sqrt{7 \delta^2-4 \delta (h+w)+h^2+w^2}+2 h+2 w\right)}{3 \delta^2-4 \delta w+w^2} $$

and for $h=30'',\ h = 16.5'', \ \delta = 2''$ we obtain

$$ \alpha = 1.75659\ \text{rad} = 60.3478^{\circ} $$