Hello, finding a point's x and y located in another circle.

[ManyTimes]

Hello there,

I am stumbling at the question of how to find a point in a circle (Do note, picture is not in scale)
http://i50.tinypic.com/2vmj220.jpg

A picture is in the link above, the P is what I am supposed to find.
The information given is:
R = 10
Theta is known by cos(theta) = 4/5 (can someone clear this one out for me? What does this mean? Does it mean that 5 is the hypotenuse and 4 is the adjacent? I mean, it shouldnt mean that, due to the radius (R, which in the link above is the adjacent) is 10)... So what does cos(theta) = 4/5 tell me/you?

What I have done so far is, which also I think is correct due to the fact I drew it on paper afterward, seemed correct:
x = r * cos(theta) = 10 * cos(4/5) = 6,967
y = r * sin(theta) = 10 * sin(4/5) = 7,174
So P(x,y) = (6.967 , 7,174)
So far so good? Or not? Please correct me.

Now what I am trying to do, is to do as the title says, another link is provided, showing a picture of my problem: (picture is NOT in scale)

http://i48.tinypic.com/oqlwzk.jpg

The problem:
The radius from O to S is 65.
The radius from S to P is 10.

A beam from O to S creates an angle Ø with the x-axis in the O-system. This angle is given by cosØ = 5/13. (Ø in this case is "theta")

Find coordinates of P relatively to O.

thanks for suggestions!!!

 

[Denis]

Well, just an observation:
4/5 is same as 8/10 : get it?

Coordinates of point P are obviously (8,6)

 

[ManyTimes]

4/5 same as 8/10, of course, but so what? Does this say that 8 is the x-axis, then draw a circle around origo with radius 10, see where you land at Y-vise, yes it gives 6, it is obviously 8,6... Thanks, but instead of just giving me the answer, I would like to know why/how, does this mean that every time I see "cos(theta)=x1/x2" the X1 always refers to the X-axis or as a percentage of the radius of a circle, you get what Im trying to say? :roll:

>>Coordinates of point P are obviously (8,6)
Whats your calculations behind that statement? Instead of actually draw this everytime? My calculations were wrong:
x = r * cos(theta)
y = r * sin(theta)
What should I rather have done?

But how about the second image, what I have so far is:
cos(Ø) = 5/13 = 25/65
S(x,y) = (25, y)
Angle of cos(Ø) = 25/65 = co^-1(25/65) = about 67 degrees
y = x*tan(a) = 25*tan(67 degrees) = 58.9
S(x,y) = (25, 58.9)

Since I know from the previous calculations how far P was from S, I simply just add it to the coordinates:
25 + 8, 58.9 + 6 = 33, 64.9
Which means; P lays relatively from O with the coordinates (33, 64.9).
Correct?

 

[soroban]

Hello, ManyTimes!

Very confusing . . .
You're mixing \(\displaystyle R\) and \(\displaystyle r\) . . . and \(\displaystyle \theta\) and \(\displaystyle \phi\)

From what I've interpreted so far, this is the diagram:

                     o P
              10  *  |
               * @   |
          S o - - - -o Q
           *:
      65  * :
         *  : 60
        *   :
       *    :
    O * - - *
        25

\(\displaystyle \text{The angle at }O\text{ is fixed (constant): }\;\cos\phi = \frac{5}{13}\)

. . \(\displaystyle \text{Hence, the coordinates of }S\text{ are: }\25,60)\)

\(\displaystyle \text{In right triangle }PQS\text{, we see that: }\:\begin{Bmatrix} SQ \:=\:10\cos\theta \\ PQ \:=\: 10\sin\theta\end{Bmatrix}\)


\(\displaystyle \text{Therefore, the coordinates of }P\text{ are: }\;(25 + 10\cos\theta,\;60 + 10\sin\theta)\)

 

[ManyTimes]

Ehm...
"Coordinates of point P are obviously (8,6)", so this guy who said this was wrong?

Coordinates of P from S is as you say:
10cos(8/10) , 10sin(8/10) (which is 6.97 and 7.17)???

Which one to use, the 8,6 or 10cos(8/10)...??? Guess it is up to me?