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 7:27 AM


Den söndagen den 21:e juli 2013 kl. 13:21:01 UTC+2 skrev jonas.t...@gmail.com: > 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 UTC5, jonas.t...@gmail.com wrote: > > > > > > > > > > > > > > > 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? > > > > > > > > > > > > > > > > > > > > > > > > > > > > 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 . > > > > > > > > > > > > > > > > > > > > > > > > > > > > For the Hypotenuse of sqrt(10) , pythagoras 1:3 the formula is exact. Mathematics is more precise than the present mathematic. > > > > > > > > > > > > > > > > > > > > > > > > > > > > 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) > > > > > > > > > > > > > > > > > > > > > > > > > > > > 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) > > > > > > > > > > > > > > > > > > > > > > > > > > > > (Base *sqrt(n) value of the hypotenuse* 4 ) > > > > > > > > > > > > > > > > > > > > > > > > > > > > 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. > > > > > > > > > > > > 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. > > > > > > > > > > > > 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.
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