# Math Help - Find distances from angles

1. ## Find distances from angles

Hi,

I have come across a problem I cannot seem to solve by myself, so I was hoping anyone on here could help me with this.
I'm sure it's pretty easy to do what I want, I just can't think of any way...

I'm studying physics at a university (first year) so you may assume that my math level is beyond highschool, but I'll tell you if I don't understand :P

The point P can move along the circle with radius R as shown.
Points 1 and 2 are fixed, seperated by a distance d. They are also centered around the origin so the x-coordinate of point 1 will be -1/2d and the x-coordinate of point 2 will be +1/2d. You get the point.

I'll try to explain in detail what I need, but if you don't need that, I'll ask the question first:
How can I express the lengths r1 and r2 in the angle a0?

Explanation:
I need to find the distances r1 and r2, based on the angle between the line OP and the x-axis (a0).

I need this for a physics project, where points 1 and 2 are two sound sources. I need to find the phase difference between source 1 and 2 at point P. Both sound sources are in phase, but because r2 < r1, there will be a phase difference at point P.
To find the phase difference, I need to know the lengths of r1 and r2.

I know, I could simply apply pythagoras or something, but I need to express these lengths in the angle that r0 makes with the x-axis (a0).

The reason behind this is that the two sound sources form a kind of directional speaker. They are placed at exactly the right distance apart (1/2 * wavelength) so their waves cancel out at the sides and add up right infront. Now I want to create a graph of the soundpressure (which depends on the phase difference) of both speakers combined at point P, which will vary with the angle a0.
So I need to express both lengths r1 and r2 in the angle a0 alone.

Does anyone know any way how to do this?
Thanks!!

2. Originally Posted by Flippy
Hi,

I have come across a problem I cannot seem to solve by myself, so I was hoping anyone on here could help me with this.
I'm sure it's pretty easy to do what I want, I just can't think of any way...

I'm studying physics at a university (first year) so you may assume that my math level is beyond highschool, but I'll tell you if I don't understand :P

The point P can move along the circle with radius R as shown.
Points 1 and 2 are fixed, seperated by a distance d. They are also centered around the origin so the x-coordinate of point 1 will be -1/2d and the x-coordinate of point 2 will be +1/2d. You get the point.

I'll try to explain in detail what I need, but if you don't need that, I'll ask the question first:
How can I express the lengths r1 and r2 in the angle a0?

Explanation:
I need to find the distances r1 and r2, based on the angle between the line OP and the x-axis (a0).

I need this for a physics project, where points 1 and 2 are two sound sources. I need to find the phase difference between source 1 and 2 at point P. Both sound sources are in phase, but because r2 < r1, there will be a phase difference at point P.
To find the phase difference, I need to know the lengths of r1 and r2.

I know, I could simply apply pythagoras or something, but I need to express these lengths in the angle that r0 makes with the x-axis (a0).

The reason behind this is that the two sound sources form a kind of directional speaker. They are placed at exactly the right distance apart (1/2 * wavelength) so their waves cancel out at the sides and add up right infront. Now I want to create a graph of the soundpressure (which depends on the phase difference) of both speakers combined at point P, which will vary with the angle a0.
So I need to express both lengths r1 and r2 in the angle a0 alone.

Does anyone know any way how to do this?
Thanks!!
I can't sketch it, but you can find r2 by the Law of Cosines:
$r_2^2 = r_0^2 + \left ( \frac{d}{2} \right ) ^2 - 2r_0 \left ( \frac{d}{2} \right ) ~ cos(a_0)$

Then you can get the angle between segments 12 and 2P (the supplement of angle a2) by the Law of Sines. (Let $\beta = 180 - a_2$
$\frac{sin(a_0)}{r_2} = \frac{sin(\beta)}{r_0}$
(Solve for $\beta$ and then use that to find a2.)

Use the Law of Cosines again to find r_1:
$r_1^2 = a_2^2 + d^2 - 2a_2d ~ cos(\beta)$

and finally, use the Law of Sines to find a1:
$\frac{sin(a_1)}{r_2} = \frac{sin(\beta)}{r_1}$

-Dan

3. Here is another way.
The coordinates of $P:\;\left( {R\cos \left( {a_o } \right),R\sin \left( {a_o } \right)} \right)\quad 0 \le a_0 \le \pi$ are well known using polar coordinates.
For the other two points we have $P_1 :\;\left( { - \frac{d}{2},0} \right)\,\& \,P_2 :\;\left( {\frac{d}{2},0} \right)$.
Using the common distance formula we get:
$r_1 = \sqrt {\left( {R\cos \left( {a_o } \right) + \frac{d}{2}} \right)^2 +R^2 \sin ^2 \left( {a_o } \right)} \,\& \,r_2 = \sqrt {\left( {R\cos \left( {a_o } \right) - \frac{d}{2}} \right)^2 +R^2 \sin ^2 \left( {a_o } \right)}$.

Now I hope I have understood what you are required to do.

4. Thanks both.

Plato, I have though about using polar coordinates aswell, but never managed to do the last step.

Can you explain the last step to me in more detail? ("Using the common distance formula...")
I understand you're using pythagoras, but how do you get the "cos(a1) + 1/2d" part?

OK nevermind, you have edited it and now I understand :P I was missing a few R's

So yeah you're basically just using pythagoras on the 'x-side' and the 'y-side', where the 'x-side' is (Rcos(a0) + 1/2d) and the 'y-side' is (Rsin(a0)), right?

That seems to work just fine I think!

I'll see where it gets me with my further calculations.

Thanks!!

5. Originally Posted by Flippy
Can you explain the last step to me in more detail? ("Using the common distance formula...")
I understand you're using pythagoras, but how do you get the "cos(a1) + 1/2d" part?
The distance formula follows.
$Ap,q)\,\& \,Br,s)\quad \Rightarrow \quad d(A,B) = \sqrt {\left( {p - r} \right)^2 + \left( {q - s} \right)^2 } " alt="Ap,q)\,\& \,Br,s)\quad \Rightarrow \quad d(A,B) = \sqrt {\left( {p - r} \right)^2 + \left( {q - s} \right)^2 } " />

$\left( {R\cos \left( {a_o } \right) - \left( { - \frac{d}{2}} \right)} \right) = \left( {R\cos \left( {a_o } \right) + \frac{d}{2}} \right)
$