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Topic: How do i go on finding the cordinate of the middle of hypotenuse
of a right angled triangle?

Replies: 14   Last Post: Jul 21, 2013 4:21 PM

 Messages: [ Previous | Next ]
 JT Posts: 1,448 Registered: 4/7/12
Re: How do i go on finding the cordinate of the middle of hypotenuse
of a right angled triangle?

Posted: Jul 21, 2013 7:21 AM

Den söndagen den 21:e juli 2013 kl. 13:10:23 UTC+2 skrev jonas.t...@gmail.com:
> Den söndagen den 21:e juli 2013 kl. 05:52:09 UTC+2 skrev hope...@frontier.com:
>

> > On Friday, July 19, 2013 1:18:40 PM UTC-5, jonas.t...@gmail.com wrote:
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> > > Using a cartesian dot base cordinate system.(computer graphics) I have right sided triangles at differents slopes, and need to find the hypotenuse middle cordinate of them, how do i go on to do that? Can it be done without trigonometry?
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> > A hypotenuse , and its mid value is a relative value , because of the way the horizons are . There is a very precise method for understanding the relative coordinate .
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> > For the Hypotenuse of sqrt(10) , pythagoras 1:3 the formula is exact. Mathematics is more precise than the present mathematic.
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> > 3*(sqrt(10)/2)*4= sqrt(360) that translates the value relative to the horizons. I have proved this that a perfect curve is derived from 1:3 ( angle 360/19). 3 is the base here and 1 perpendicular,The midpoint of the hypotenuse at 3 is exactly sqrt(2.5)
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> > For base 1, Perpendicular 1 it is 1*(sqrt(2)/2)*4=sqrt8 ( 45 degrees, 360/45=8). The value of the mid point is the sqrt(0.5)
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> > (Base *sqrt(n) value of the hypotenuse* 4 )
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> > You cannot relate hypoteuses to hypotenuses! unless you relate each hypotenuse to its relative value. The least score is sqrt(8). Well wish you the best , unless I confused you.
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> In a cartesian cordinate system any line form a rectangle, that is everything needed to realise. As long you have the start and end point cordinate you can find the middle cordinate. I was thinking a bit to to much around triangles and hypotenuse without realising i was working with a line segment.
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> It is solved.

My approach doing curves and geometry without the use trigonmetry and Pi, only using triangles was moot, i should been doing the polygons, triangles and curves using squares and rectangles.

Date Subject Author
7/19/13 William Elliot
7/20/13 JT
7/20/13 JT
7/20/13 JT
7/20/13 Dirk Van de moortel
7/20/13 Brian Q. Hutchings
7/20/13 JT
7/20/13 HOPEINCHRIST
7/21/13 JT
7/21/13 JT
7/21/13 JT
7/21/13 JT
7/21/13 Brian Q. Hutchings
7/21/13 Brian Q. Hutchings