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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

850.0. "trig question/proof?" by GORP::MARCOTTE (George Marcotte SWS Santa Clara) Thu Mar 31 1988 22:20

    
    
    I was reading a book on plotting for sailing. One of the methods
    use to maintain a save distance from rocks was to use the angle between
    to points:
                            points a and c are land marks and are on the 
         __|__              chart. If you draw a circle 
    boat \___/              through points a b and c and move point b along
           *b               the arc of the drawn circle will the angle <abc
                            remain constant?  The book on plotting says it 
                            will.  The Idea is to create a circle on the 
                                   chart that is outside the rocks then measure
                       x                    the angle on the chart between 
                     x  rocks     /------\  a b c. Then use a sextant to measure
        ------\                  /   c*   \     actual points. If the angle is
       /  *a   \----------------/          \    larger than the one calculated 
                                            \   from the chart your in trouble.

    Can anyone prove this?
    That is: prove that the vertex angel formed by three points on a circle 
    will remain constant as the vertex point is allowed to move along the
    circle and the other two points are fixed.

T.R	Title	User	Personal Name	Date	Lines
850.1	the easy case	ZFC::DERAMO	Take my advice, I'm not using it	`Fri Apr 01 1988 00:11`	8
	Well, if point a and point c are opposite ends of a diameter, then as b moves along the circle angle abc is always a right angle. If points a and c are not opposite ends of a diameter, well, this is just too much geometry for two days! Dan
850.2	here's a sketch of proof	PULSAR::WALLY	Wally Neilsen-Steinhardt	`Mon Apr 11 1988 16:24`	20
	Yes, the theorem in .0 is true and provable. I don't think I could give the full proof without a diagram, but here is a sketch: Given points ABCD on a circle with center O. Then there are 5 isosceles triangles OAD, OAB, OAC, OCD, OBC. This leads to 5 equalities among angles. And there are two equalities based on the fact that the sum of angles in any two triangles, and therefore ABC and ADC, is constant. There are two more equalities based on angles contained within angles (here's where the diagram is necessary). Combine all these equalities in the right way and you get <ABC = <ADC
850.3	almost got it	GORP::MARCOTTE	George Marcotte SWS Santa Clara	`Mon Apr 11 1988 21:49`	29
	> Given points ABCD on a circle with center O. > Then there are 5 isosceles triangles OAD, OAB, OAC, OCD, OBC. > This leads to 5 equalities among angles. well <oab = <oba <obc = <ocb <OCD = <ODC <OAD = <ODA That is 4 whats the 5th one? > And there are two equalities based on the fact that the sum > of angles in any two triangles, and therefore ABC and ADC, is > constant. I can see that: <ABC + <ADC + <bad + <dcb = 360 how is <ABC and <ADC constant? > There are two more equalities based on angles contained within > angles (here's where the diagram is necessary). name an example Its been tooooooo long... George :-)
850.4	more hints	PULSAR::WALLY	Wally Neilsen-Steinhardt	`Tue Apr 12 1988 17:05`	24
	re: < Note 850.3 by GORP::MARCOTTE "George Marcotte SWS Santa Clara" > -< almost got it >- > That is 4 whats the 5th one? <OAC = <OCA > how is <ABC and <ADC constant? What is constant is the sum of angles in a triangle: <ABC + <ACB + <BAC = <ADC + <ACD + <DAC >> There are two more equalities based on angles contained within >> angles (here's where the diagram is necessary). > name an example <ABC = <ABO - <CBO Note that the sign here depends on the way I drew the diagram, since I follow the Euclidean convention that all angles are positive. If you draw a different diagram, your sign may be different, but the proof will still work out.