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help - maths fail :(
mcerd1 - 3/4/14 at 02:32 PM

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 !

carpmart - 3/4/14 at 02:46 PM

Worryingly, I do not even understand the question, so no hope of providing an answer!

JAG - 3/4/14 at 02:52 PM

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]

liam.mccaffrey - 3/4/14 at 03:08 PM

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.

mcerd1 - 3/4/14 at 03:20 PM

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]

Slimy38 - 3/4/14 at 03:25 PM

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]

JAG - 3/4/14 at 03:55 PM

Oh yeah - Bugger

I didn't spot that second B!

liam.mccaffrey - 3/4/14 at 03:57 PM

I used that identity to eventually get to my result.

I'm still probably wrong though

matt_gsxr - 3/4/14 at 04:15 PM

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))

hughpinder - 3/4/14 at 05:14 PM

Mr matt_gsxr is correct I believe
Regards
Hugh

nero1701 - 3/4/14 at 06:27 PM

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]

rachaeljf - 3/4/14 at 07:06 PM

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) )

james h - 3/4/14 at 10:17 PM

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.

nero1701 - 3/4/14 at 11:29 PM

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 ;-)

mcerd1 - 4/4/14 at 07:38 AM

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