Daniel0 Posted September 2, 2008 Share Posted September 2, 2008 Okay, maybe I'm stupid, but in a triangle, if |BC|=2|AB| and |AC|=(5/2)|AB| then how do I figure out what cos© is? According to the cosine relation it's true that cos© = (|AC|^2 + |AB|^2 - |BC|^2) / (2*|BC|*|AC|) Inserting what I know gives me cos© = (((5/2)|AB|)^2 + |AB|^2 - (2|AB|)^2) / (2*2|AB|*(5/2)|AB|) = (((3/2)|AB|)^2) / ((10|AB|)^2) Now what? Can I just say that cos© = 3/2/10 = 3/20? If I call |AB| for x and cos© for y and enter this into my calculator: solve(y=((5/2*x)^2+x^2-(2*x)^2)/(2*(2*x)*(5/2)*x)),y) Then I get 13/40 which is clearly not the same. Am I doing something wrong when deriving cos© without a calculator or am I doing something wrong when I enter things into my calculator. Now, this is for school and there is nothing preventing me from saying "I used a CAS tool and typed X and got Y", but I would still like to do the other and I would like to know how the calculator got to 13/40 instead of 3/20. It does, however, say "Note: Domain of result may be larger" when presenting 13/40, but I'm not sure what it means by that. I wish I could somehow get my calculator (TI-89 Titanium) to tell me the steps between what I entered and what it tells me... I hope there are any math geniuses here. Btw, plaintext math notation sucks Link to comment https://forums.phpfreaks.com/topic/122370-solved-math-geometry/ Share on other sites More sharing options...
Daniel0 Posted September 2, 2008 Author Share Posted September 2, 2008 Duh... so I completely forgot that (ab)^2=a^2*b^2. That helps a lot. I guess stepping away from it for some hours helps. This means that I've got the upper part of the fraction to say: (25/4)|AB|^2+|AB|^2-4|AB|^2 which my calculator says is the same as (13/4)*|AB|^2. How does it get to that? Now that before divided with 10|AB|^2 will give the 13/40 which I got it to say for the entire equation in the above post. I'd still like to know how it gets from (25/4)|AB|^2+|AB|^2-4|AB|^2 to (13/4)*|AB|^2 though, so if anyone could tell me that then it would be great. Link to comment https://forums.phpfreaks.com/topic/122370-solved-math-geometry/#findComment-632114 Share on other sites More sharing options...
The Little Guy Posted September 2, 2008 Share Posted September 2, 2008 I don't know if this helps: SOA CAH TOA SOA = Sin CAH = Cos TOA = Tan SOA = Opposite / Adjacent CAH = Adjacent / Hypotenuse TOA = Opposite / Adjacent Link to comment https://forums.phpfreaks.com/topic/122370-solved-math-geometry/#findComment-632119 Share on other sites More sharing options...
Daniel0 Posted September 2, 2008 Author Share Posted September 2, 2008 Duh... I'm retarded... It obviously just subtracted (25/4)|AB|^2 with 4|AB|^2 and then there is the number fraction left along with |AB|^2. Solved. I don't know if this helps: SOA CAH TOA SOA = Sin CAH = Cos TOA = Tan SOA = Opposite / Adjacent CAH = Adjacent / Hypotenuse TOA = Opposite / Adjacent That's only true for right angled triangles. There is no information about that in my assignment. Thanks for trying though Link to comment https://forums.phpfreaks.com/topic/122370-solved-math-geometry/#findComment-632122 Share on other sites More sharing options...
The Little Guy Posted September 2, 2008 Share Posted September 2, 2008 Oh your right, and also I got SOA wrong it is SOH Link to comment https://forums.phpfreaks.com/topic/122370-solved-math-geometry/#findComment-632125 Share on other sites More sharing options...
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