topic Re: Some help with trigo in Libraries & objects

Some help with trigo

Anonymous — Tue, 19 Jul 2005 17:16:12 GMT

I would apreciate some help for this trigonometrical problem.
I am searching for H or ang, function of X and ray. I don't find a solution.
Thanks to anyone for the highligth.

Re: Some help with trigo

TomWaltz — Tue, 19 Jul 2005 17:51:42 GMT

What values do you have? Only X? Or do you have "ray" also? From the diagram,

H = X,

then

angle = ATN(H/(x+ray)

Re: Some help with trigo

James Murray — Tue, 19 Jul 2005 18:32:34 GMT

H <> X though. H is equal to the other side of that 45º triangle, less than X.

I don't think it's solvable with the information given. You have one side and no angles, which isn't much.

Hoping I'm wrong,

Re: Some help with trigo

Anonymous — Tue, 19 Jul 2005 18:36:05 GMT

TomWaltz wrote:
What values do you have? Only X? Or do you have "ray" also? From the diagram,
H = X, 
then
angle = ATN(H/(x+ray)

Hi Tom,

Unfortunately, i have only "X" and "ray".
I am searching for "H". Once i will know "H", i can get "ang" automatically.
Vice versa, if i know "ang", of course i get "H" easily.

It looks simple, but i must be tired. Thanks anyway.

Re: Some help with trigo

Anonymous — Tue, 19 Jul 2005 18:38:44 GMT

the triangle 'ray, R and non-included angle (135 degrees)' is solvable

there's a solution here

http://www.teacherschoice.com.au/Maths_Library/Trigonometry/solve_trig_ASS.htm

have fun

bill

Re: Some help with trigo

Anonymous — Tue, 19 Jul 2005 18:53:54 GMT

James wrote:
H <> X though. H is equal to the other side of that 45º triangle, less than X.

I don't think it's solvable with the information given. You have one side and no angles, which isn't much.

Hoping I'm wrong,

Hi James,

Thanks. Yes, H <> X. H is functiun of X (but function of ray too).
The centerpoint of the 45° angle is between ray and X. Hope this is understandable?

I am not sure it is solvable, but this could save me if it was. Why not to try?
Is there some mathematician around? Stuart James?

Re: Some help with trigo

TomWaltz — Tue, 19 Jul 2005 18:54:05 GMT

Sorry, I misread the diagram. I did not see the little extension.

Bill is right.... that is a hard one, and will take a few equations!

Re: Some help with trigo

Anonymous — Tue, 19 Jul 2005 19:08:15 GMT

bill wrote:
the triangle 'ray, R and non-included angle (135 degrees)' is solvable

there's a solution here

http://www.teacherschoice.com.au/Maths_Library/Trigonometry/solve_trig_ASS.htm

have fun

bill

Hi Bill,

Thanks for the link. I take a look. The time to make some translations, that's why i am so slow.

I have tried a solution with addition of angles, and at the end i got... an equality, something like A=A.
Bingo, May be i don't take the problem in the good way.

Re: Some help with trigo

Anonymous — Tue, 19 Jul 2005 21:52:26 GMT

The answer is simple.

You cannot solve it. Why?

Look. You did draw an angle and now you want to measure it knowing only the radius.
To pursuade you. Draw that angle. You know the radius. You can change the angle but keep the radius. Is that right. So?? you cannot from X and r know what ang you have.

You simply cannot do that.

Re: Some help with trigo

Anonymous — Tue, 19 Jul 2005 22:48:24 GMT

"You simply cannot do that"

two sides and one angle of the triangle are knowns

like i said, solvable

the solution, however, gave me a headache, even before i started to scroll thru it

bill

Re: Some help with trigo

James Murray — Tue, 19 Jul 2005 23:11:21 GMT

Law of Sines. I think. Not that bad if I'm right.

a/SIN(A)=b/SIN(B)=c/SIN(C)

The angle we know, A=135. Opposite is the side we know, a=R.

The other side we know, c=ray.

You're looking for B.


R/SIN(135)=ray/SIN(C)

SIN(C)*R=ray*SIN(135)

SIN(C)=(ray*SIN(135))/R

C=ASN((ray*SIN(135))/R)

B=180-A-C.

Something like that.

Re: Some help with trigo

David Collins — Wed, 20 Jul 2005 06:44:54 GMT

Unsolvable, I think.
X minus the endbit (and therefore H) could be be anything, depending on the value of ang.

Re: Some help with trigo

Frank Beister — Wed, 20 Jul 2005 08:39:51 GMT

I think it's solveable, because the triangle over 'X' makes the figure stable.

SIN(ang)=H/R
SIN(ang/2)=s/R
X=d1+d2
d2=H*TAN(45)=H
TAN(ang/2)=d1/H or SIN(ang/2)=d1/(2*s) or COS(ang/2)=H/(2*s)

I have not the time to solve it, maybe later or someone else wants...

Re: Some help with trigo

Frank Beister — Wed, 20 Jul 2005 08:56:17 GMT

Sorry, I haven't seen that James has the solution yet:

ang = 180 - 135 - ASN( ray*SIN(135) /R)

Re: Some help with trigo

Anonymous — Wed, 20 Jul 2005 08:58:37 GMT

James wrote:
Law of Sines. I think. Not that bad if I'm right.

a/SIN(A)=b/SIN(B)=c/SIN(C)

The angle we know, A=135. Opposite is the side we know, a=R.

The other side we know, c=ray.

You're looking for B.
R/SIN(135)=ray/SIN(C)

SIN(C)*R=ray*SIN(135)

SIN(C)=(ray*SIN(135))/R

C=ASN((ray*SIN(135))/R)

B=180-A-C.
Something like that.

Thank you so much, James, you save me. It works perfectly.
The link posted by Bill is helpful, too. Thanks to all for your help.

Re: Some help with trigo

Vitruvius — Wed, 20 Jul 2005 16:01:06 GMT

I think the formulae are:

H = X*SIN(45) --- X being the hypoteneuse of small triangle
Ang = ASN(H/R) --- and then plugged into the large triangle formula

Assuming from your diagram that X, RAY, and 45d are known values.

Cheers

Re: Some help with trigo

Anonymous — Wed, 20 Jul 2005 16:41:23 GMT

Vitruvius wrote:
I think the formulae are:

H = X*SIN(45) --- X being the hypoteneuse of small triangle
Ang = ASN(H/R) --- and then plugged into the large triangle formula

Assuming from your diagram that X, RAY, and 45d are known values.

Cheers

Not exact, H = R*sin(ang)
For the rest, X, ray and 45 are known.

I did i first try with the sum of angles into a triangle, another one with pythagoras (Frank's example), without success.
I can assure you that James's solution works perfectly. The occasion for me to rediscover this "sine rule" (Bill's link).
Thanks anyway.