mcerd1
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| posted on 3/4/14 at 02:32 PM |
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help - maths fail :(
got a little maths problem here and its been too long since I did this stuff 
I need to solve this for the angle B:
t / cos(B-A) = h / sin(B)
t, A and h are all constants
help !
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carpmart
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| posted on 3/4/14 at 02:46 PM |
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Worryingly, I do not even understand the question, so no hope of providing an answer!
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JAG
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| posted on 3/4/14 at 02:52 PM |
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Re-arranging for B gives;
t / cos(B-A) = h / sin(B)
sin(B) = (h x cos(B-A))/t
therefore;
B = Sin^-1((h x cos(B-A)/t)
I hope 
[Edited on 3/4/14 by JAG]
Justin
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liam.mccaffrey
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| posted on 3/4/14 at 03:08 PM |
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There is a huge chance this is wrong,
B= inverse Tan of ((h/t)-SinA)
first person to correct me wins a 50 pence coin.
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mcerd1
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| posted on 3/4/14 at 03:20 PM |
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quote:
Originally posted by JAG
Re-arranging for B gives;
t / cos(B-A) = h / sin(B)
sin(B) = (h x cos(B-A))/t
therefore;
B = Sin^-1((h x cos(B-A)/t)
I hope
^^ thats what I thought to start with but there is still another B to deal with in the cos(B-A) bit - this is where my math failed (its about 15 years
since i last did any of this stuff)
i dug up the trig identity thats probably the way to solve it - but not sure were to go after this:
cos(X - Y) = cosX * cosY + sinX * SinY
[Edited on 3/4/2014 by mcerd1]
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Slimy38
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| posted on 3/4/14 at 03:25 PM |
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I think (and I'm not 100%) that
cos(b-a) = cos a cos b + sin a sin b
I have no idea whether it helps though!
edit: You beat me to it!
[Edited on 3/4/14 by Slimy38]
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JAG
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| posted on 3/4/14 at 03:55 PM |
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Oh yeah - Bugger 
I didn't spot that second B!
Justin
Who is this super hero? Sarge? ...No.
Rosemary, the telephone operator? ...No.
Penry, the mild-mannered janitor? ...Could be!
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liam.mccaffrey
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| posted on 3/4/14 at 03:57 PM |
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I used that identity to eventually get to my result.
I'm still probably wrong though
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matt_gsxr
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| posted on 3/4/14 at 04:15 PM |
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quote:
Originally posted by liam.mccaffrey
There is a huge chance this is wrong,
B= inverse Tan of ((h/t)-SinA)
first person to correct me wins a 50 pence coin.
Expand cos(B-A) to give cos(B)cos(A) + sin(B)sin(A)
and rearrange
I get
B = inverse tan of (h cos(A))/(t-h Sin(A)), but I haven't checked it.
be a bit wary as inverse tan can hide solutions (i.e. tan(x) = tan(x+pi))
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hughpinder
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| posted on 3/4/14 at 05:14 PM |
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Mr matt_gsxr is correct I believe
Regards
Hugh
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nero1701
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| posted on 3/4/14 at 06:27 PM |
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t/(sin(A) sin(B)+cos(A) cos(B)) = h csc(B)
or if you wanted to be really smart...
(2 t cos(A-B))/(cos(2 (A-B))+1) = -(2 h sin(B))/(cos(2 B)-1)
But this doesn't transpose it...it just makes it bigger....
[Edited on 3/4/14 by nero1701]
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rachaeljf
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| posted on 3/4/14 at 07:06 PM |
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t / cos(B-A) = h / sin(B)
tsinB = hcos(B-A) = h(cosBcosA + sinBsinA) = hcosBcosA + hsinBsinA
tsinB - hsinBsinA = hcosBcosA
sinB(t - hsinA) = hcosBcosA
sinB = hcosBcosA / (t - hsinA)
sinB/cosB = hcosA / (t- hsinA)
tanB = hcosA / (t - hsinA)
B = tan^-1( hcosA / (t - hsinA) )
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james h
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| posted on 3/4/14 at 10:17 PM |
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The website wolfram alpha is your friend for such things As an engineering student I wouldn't be without it
Although not sure about its rearranging skills on this one
here.
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nero1701
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| posted on 3/4/14 at 11:29 PM |
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quote:
Originally posted by james h
The website wolfram alpha is your friend for such things As an engineering student I wouldn't be without it
.
As an engineering lecturer.....neither would I ;-)
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mcerd1
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| posted on 4/4/14 at 07:38 AM |
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quote:
Originally posted by james h
The website wolfram alpha is your friend for such things As an engineering student I wouldn't be without it
Although not sure about its rearranging skills on this one
here.
yeah tried that already, but didn't get anything I could really use
I've actually found a work around that gets me the answer i need
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