First time poster here and it has me stumped - posting this here for help. I've exhausted my energy and knowledge on this one. AD, BC and alpha are known. I need to derive theta in terms of the knowns. I've tried identical triangle ABD=BCE approaches but just end up chasing my tail. All the help appreciated.
Finding Theta - identical triangles
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Dr.Peterson
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Triangles ABD and BCE are not identical (congruent); they are not even similar (you probably mean CBE). So I'm not sure what you mean.
Can you show us some of your actual work, while I try to work on it myself?
Can you show us some of your actual work, while I try to work on it myself?
Not identical (extremely poor choice of words on my end) however they are similar, ABD and CBE that is. I've tackled it from trying to derive theta in terms of the segments however I always work myself into a loop, proving that AD is indeed AD etc. I haven't done trig in quite a while and even a hint or overall strategy would help.
I think I got it actually now - angle DCA is theta + alpha. From thereon you do a tan(theta), sub in the DC and got yourself a crazy solution via wolframalpha. https://www.wolframalpha.com/input/?i=solve+for+x:+tan(x)+=+a/(b+(a/tan(x+y))+
Dr.Peterson
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I took a similar approach and got a fairly reasonable formula.
Expressing DE in two ways, we get [MATH]\frac{a\sin(\alpha)}{\sin(\alpha+\theta)} = b\sin(\theta)[/MATH]. With some interesting manipulations, I found that
[MATH]\theta = \frac{1}{2}\left(\cos^{-1}\left(\cos(\alpha)-\frac{2a\sin(\alpha)}{b}\right)-\alpha\right)[/MATH].
This agrees with measurements of my figure.
Dr.Peterson
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[MATH]\frac{a\sin(\alpha)}{\sin(\alpha+\theta)} = b\sin(\theta)[/MATH][MATH]a\sin(\alpha) = b\sin(\theta)\sin(\alpha+\theta)[/MATH][MATH]a\sin(\alpha) = b\sin(\theta)(\sin(\alpha)\cos(\theta)+\cos(\alpha)\sin(\theta))[/MATH][MATH]a\sin(\alpha) = b\sin(\alpha)\sin(\theta)\cos(\theta)+b\cos(\alpha)\sin^2(\theta)[/MATH][MATH]a\sin(\alpha) = b\sin(\alpha)\frac{1}{2}\sin(2\theta)+b\cos(\alpha)\frac{1}{2}(1-\cos(2\theta))[/MATH][MATH]2a\sin(\alpha) = b\sin(\alpha)\sin(2\theta)+b\cos(\alpha)-b\cos(\alpha)\cos(2\theta)[/MATH][MATH]b\cos(\alpha)\cos(2\theta)-b\sin(\alpha)\sin(2\theta)=b\cos(\alpha)-2a\sin(\alpha)[/MATH][MATH]b\cos(\alpha+2\theta) = b\cos(\alpha)-2a\sin(\alpha)[/MATH][MATH]\cos(\alpha+2\theta) = \cos(\alpha)-\frac{2a}{b}\sin(\alpha)[/MATH]
Then solve for [MATH]\theta[/MATH]. Everything I hoped might happen, did.
Then solve for [MATH]\theta[/MATH]. Everything I hoped might happen, did.
I should have said - 'alpha' is 0.9deg, 'a' is 2000 and 'b' is 90. With a sketch editor I get 35.77deg for theta. With both the solution I came up with and with the simplification you provided I struggle getting that value. (this is for a calculator for a parametric model)
