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 Quote Link to comment 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. Quote Link to comment 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 Quote Link to comment 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 Quote Link to comment 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 Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.