MATLAB Visualizing a Segment of a Unit Circle

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The discussion focuses on a MATLAB script that generates a segment of a unit circle from 60° to 120°. The key point of inquiry is the equation "rho1 = rho*sin(a1)./sin(theta)," which describes the transformation of the radius in polar coordinates. The user seeks to understand why this equation produces values ranging from 1 to 0.866 and back to 1, questioning whether this behavior is related to properties of similar triangles or inverse proportions. The conversation also touches on converting between polar and Cartesian coordinates, emphasizing the definitions of rho and rho1 in polar terms. Additionally, a separate request for assistance with a rational function is mentioned, specifically one that has vertical asymptotes at x=2 and x=3 and a horizontal asymptote at y=2.
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a = 60*pi/180;
a1 = (pi - a)/2;
a2 = (pi + a)/2;
theta = a1: a/60: a2;
rho = ones(size(theta));
rho1 = rho*sin(a1)./sin(theta);
polar(theta, rho);
hold on;
polar(theta, rho1)

03.jpg


The above commands will draw a segment of a unit circle which starts from 60^{o} to 120^{o}.

I know everything except the line of "rho1 = rho*sin(a1)./sin(theta);" that I don't know why this equation works from 1 --> 0.866 --> 1 and becomes a straight line. Can anyone tell me the reason? Is it related to a similar triangle or inverse proportion?
 
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try converting from polar to euclidean/cartesian...to see if you can figure out how they got that straight line. Remember what does rho,rho1 stand for in polar.
 
Cartesian to Polar:
r = sqrt(x^2 + y^2)
theta = arctan(y/x)

Polar to Cartesian:
x = r cos theta
y = r sin theta

rho = 1 (for unit circle)

rho1 = rho*sin(a1)/sin(theta) = 1*sin(60)/sin(60) to 1*sin(60)/sin(120) = 1 --> 0.866 --> 1
 
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Write the equation of a rational function that has vertical asymptotes at x=2 and x=3 and a horizontal asymptote at y=2

yeah, i really need help with this problem. Thanks so much
 

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