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Here's the first part of the proof in Feynman's lectures:

\begin{equation*} dm=2\pi y\mu\,ds=\frac{2\pi y\mu\,dx}{\sin\theta}= \frac{2\pi y\mu\,dx\,a}{y}=2\pi a\mu\,dx, \end{equation*}

I don't get how he replaced ds by dx/sin

Here's the link to the lecture: http://www.feynmanlectures.caltech.edu/I_13.html

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  • $\begingroup$ Because $dx$ and $ds$ form a triangle whose sine of the specified angle $\theta$ is $dx/ds$ $\endgroup$ Aug 23, 2018 at 13:14
  • $\begingroup$ I could guess that. I want to know how the triangle looks like. If dx and ds are two sides, what is the third side? $\endgroup$
    – Osaid
    Aug 23, 2018 at 13:16
  • $\begingroup$ You could call the other side $dy=\sqrt{ds^2-dx^2}$. $\endgroup$ Aug 23, 2018 at 13:17
  • $\begingroup$ Aaargh.how is the angle theta in the triangle? Can you please draw and show? How did the triangle come about exactly? $\endgroup$
    – Osaid
    Aug 23, 2018 at 13:18

1 Answer 1

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Using the picture in the link, you can construct a triangle as such:

Triangle Constuction

As you can see from the gray triangle, $\sin \theta= \frac{dx}{ds}$, or, as questioned: $$ds=\frac{dx}{\sin \theta}$$

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