Find angle and diagonal based on width of a board (carpentry kinda) 
Hi, I want to learn how to calculate the measurements of the yellow board. That being the angle at #1 and where the litte cutout in the board will be at #2. Its not important that the cutout vertical height has to be 2cm, but Im assuming it has to be specified to something at least.
Obviously the angle is easy to calculate given the width (1m) and height (0,5m), which is 26,56 degrees. But ofc that changes since the top side of the board is not at the 1m point of the horizontal white board. 
And I also dont know the the width of the board (vertically) without knowing the angle, so Im stuck, I have tried and tried, looked at videos and so on, but they all seem to already have some measurements that I cant figure out. 
If this was a real life thing it would be kind of easy to just hold the board and scribe it with a pen and cut it, but I want to learn how to calculate it if possible.
 A: Let’s label the known measurements: $h = 0.500$ is the height of the frame, $w=1.000$ its width, $b=0.095$ the width of the yellow board and $d=0.020$ the height of the cut. Also, let $\alpha$ be the angle of the board below the horizontal.  
Examining the various similar triangles formed by the figure, we can see that the ratio of the height of the cutout to its width is equal to $\tan\alpha$, and the ratio of the yellow board’s width to the length of its edge abutting the vertical member is equal to $\cos\alpha$. We therefore have $$\tan\alpha = {h-b/\cos\alpha \over w-d/\tan\alpha}.\tag1$$ Using the identity $1+\tan^2\alpha = \sec^2\alpha$, this can be rewritten as $$(d+h)-w\tan\alpha = b\sqrt{1+\tan^2\alpha}.\tag2$$ Squaring both sides produces a quadratic equation in $\tan\alpha$ that can be solved using the standard quadratic formula, which I expect you can remember how to do. This will usually produce two possible solutions, one of which you can reject by substituting back into equation (2). Using the measurements in your question with the above equations results in $\alpha \approx 22.64°$.
