[MATH] distance in X between two parallel lines

[MATH] distance in X between two parallel lines

Ariesnl

Member #2,902

November 2002

Hey everyone, I ran into a math problem, I hope someone can help me out

When the distance of two parallel lines is given and the angle of those lines with the Y axis is also known I want to know the distance in x

Anyone ? thnx

Perhaps one day we will find that the human factor is more complicated than space and time (Jean luc Picard)
Current project: [Star Trek Project ] Join if you want ;-)

Niunio

Member #1,975

March 2002

What about pitagoras? I mean, trace a line perpendicular to both paralels. Then trace a line with the given angle from one of the new vertex. This should give to you a triangle and you know all angles (the rect one, the one given and its supplementary) and one of the cathetus, so you should be able to calculate the hypotenuse, can't you?

I'm not able to doodle now but I hope you can picture it.

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Current projects: Allegro.pas | MinGRo

Edgar Reynaldo

Major Reynaldo

May 2007

http://2000clicks.com/MathHelp/GeometryPointsAndLines3D.aspx

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King Piccolo will make it so Then I could just flip it with some sweet, sweet matrix math. In either case, I’ll ride this wagon until the wheels fall off. Thanks to those that keep it running

Peter Hull

Member #1,136

March 2001

$<math>x=d/{cos \theta}</math>$

Polybios

Member #12,293

October 2010

Distracting... I think it's x = d / cos(phi).

Edit: beaten.

Edit2: We need a sketch:
{"name":"610966","src":"\/\/djungxnpq2nug.cloudfront.net\/image\/cache\/e\/c\/ec67707f7a1439cd8c62ba6517fbbef9.png","w":437,"h":357,"tn":"\/\/djungxnpq2nug.cloudfront.net\/image\/cache\/e\/c\/ec67707f7a1439cd8c62ba6517fbbef9"}

So:
$<math>\varphi' = 90^\circ - \varphi</math>$
and
$<math>\frac{d}{x} = \sin(\varphi') = \sin(90^\circ - \varphi) = \cos(\varphi) \Rightarrow x = \frac{d}{\cos(\varphi)}</math>$

Ariesnl

Member #2,902

November 2002

Thanks a lot !

Perhaps one day we will find that the human factor is more complicated than space and time (Jean luc Picard)
Current project: [Star Trek Project ] Join if you want ;-)

Niunio

Member #1,975

March 2002

Polybios said:

Edit2: We need a sketch:

Just that. Thank you.

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Current projects: Allegro.pas | MinGRo