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Old 8th April 2003, 12:12   #67
shreyas_potnis
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Join Date: Jan 2003
Location: Mumbai, India
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Conversion of polar co-ords. into rectangular ones and vice-versa:

to convert polar co-ords. into rectangular ones, use:

x=sin(r)*d
y=cos(r)*d

to convert rectangular co-ords. into polar ones, use:

d=sqrt(sqr(x)+sqr(y))
r=atan2(x,y)

I hope I am right, correct me if I am not.

I can explain, how d=sqrt(sqr(x)+sqr(y)) works.
code:

So, here's the deal:

suppose x=.8 and y=.6, then the position of point point would be:


y-axis
^
|
|P Q
|.6- - - - - - -.(.8,.6)
| |
| |
|S |R
x-axis <- - - - - - - - - - - - - - - - >
|(0,0) .8
|
|
|
|
|
|

(this is an approximate figure)
PQRS is the quadrilaeral formed.
Now, we have to find l(SQ) or d(S,Q)
I will solve this as we solve our geomtery problems,
as this will give you a more clear idea.

Given:
1) line x is perpendicaular to line y.
2) d(S,R)=.8 , d(P,S)=.6
3) seg PQ is perpendicular to line y & seg RQ is perpendicular to line x.

to find: d(S,Q) i.e length of seg SQ.
Construction: Draw seg SQ ( I could'nt draw it here ) .
Sol'n:
Statement |Reason
1) In, quad. PQRS, |
m<S=m<P=m<Q=m<R=90 degrees |statement 1 & 3 of Given.
|
2) quad. PQRS is a rectangle. |Statement 1) and def'n of a rectangle.
|
3) seg PS is congreuent to seg QR |statement 2) and opposite sides of a
| rectangle are congruent.
|
4) In triangle SQR, |given statement 2) and statement 3)
seg SR = .8 and seg QR = .6 |
|
5) In triangle SQR, |

SQ^2 = SR^2 + SQ^2 |Pythagorus theorem.
<=> SQ = sqrt ( SR^2 + SQ^2 )
<=> SQ = sqrt ( .8^2 + .6^2 )
<=> SQ = sqrt ( .64 + .36 )
= sqrt ( 1.00 )
= 1
therefore SQ = 1

here, SQ = d
so sqr ( d ) = sqr ( SR ) + sqr ( RQ )
= sqr ( X ) + sqr ( Y ) ( not the lines, but the position of the point )
<=> d = sqrt ( sqr ( X ) + sqr ( Y )

It would be nice if someone explained how r=atan2(x,y) works.



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