Geometry Problem

[Ian_Curtis]

I ran into a very practical problem math building a gate this weekend. Basically I had two 2x4s running parallel a certain distance apart, and I wanted to cut a cross piece that for aesthetic reasons ran from one end to the other. This is obvious a simple SOHCAHTOA for a line, but for a beam of finite width it isn’t so simple. I ended up solving it practically (physically do it, and measure), but now I can’t make the math work and it is bothering me.

I end up with an equation, but it has two trig functions that I can’t reduce with a trig identity, so it isn’t directly solvable by an inverse trig function. It seems like it should have an obvious direct solution…

[Cesar_Mugnatto]

θ = inverse sin((b-c) / a)

[Cesar_Mugnatto]

Here is an example of the problem:

https://www.geogebra.org/o/Jh8DADfg

[GeeTwo]

Not the prettiest way, but a single trig identity and the quadratic equation will get you there:

• a = b cotθ + c cscθ
• a - b cotθ = c cscθ
• a² - 2ab cotθ + b² cot²θ = c² csc²θ
• a² - 2ab cotθ + b² cot²θ = c² + c² cot²θ
• (a²-c²) - 2ab cotθ + (b² -c²) cot²θ = 0
• cotθ = (2ab ± √(4a²b² - 4 (b² -c²) (a²-c²) ))/2(b² -c²))
• θ = cot⁻¹((ab ± c√(a²+b²-c²))/(b² -c²))

[Cesar_Mugnatto]

@Cesar Mugnatto

θ = inverse sin((b-c) / a)

Using symbols instead of text…

θ = sin⁻¹((b-c) / a)

[Ian_Curtis]

@Cesar Mugnatto

Here is an example of the problem:

https://www.geogebra.org/o/Jh8DADfg

This tool is super cool, but your answer doesn’t appear to work? sin^-1 ((6-1.2)/10) is 28.6 degrees, not 36.87 degrees? One of the solutions to GeeTwo’s quadratic appears to be the correct answer.

I probably knew this once upon a time, what does the other solution to that quadratic represent?

[Ether]

…

[Cesar_Mugnatto]

Ether’s solution is correct. Mine started out the same way but in the process I dropped / forgot to cross multiply the right side by cos theta.

[GeeTwo]

@Ian Curtis

This tool is super cool, but your answer doesn’t appear to work? sin^-1 ((6-1.2)/10) is 28.6 degrees, not 36.87 degrees? One of the solutions to GeeTwo’s quadratic appears to be the correct answer.

I probably knew this once upon a time, what does the other solution to that quadratic represent?

If you substitute c with -c in either solution, it becomes the other. So the other solution is the answer if you did the problem with c being outside of a, or equivalently if the inside edges of the board (rather than the outside edges) were a apart at span b.

[Ian_Curtis]

@GeeTwo

If you substitute c with -c in either solution, it becomes the other. So the other solution is the answer if you did the problem with c being outside of a, or equivalently if the inside edges of the board (rather than the outside edges) were a apart at span b.

Ah, yes. Thanks! I could get to the top of your post, but I did not remember the trick (square it!) to get to a single trig function you could invert.

I’m still a little surprised the solution is so complicated, I was sure I was missing something obvious and it was going to end up with something like Cesar’s original post.