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Could someone explain to me why $\tan\theta\approx y/s$? I can follow the other approximations, but here I don't see it. I'm guessing we have similar triangles here, but I wouldn't really know how to show it. It seems to me that I have to make one approximation first before showing the triangles are similar. Also, note: it's only given that we have $\theta_m$ in the triangle $(S_1,S_2,B)$. The other two $\theta_m$ are exactly the angles I try to show.

Sha Vuklia
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1 Answers1

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My suggestion is not to follow the approximations in the textbooks but rather do it in a rigorous way yourself. So we have

$$r_1-r_2=\sqrt{s^2+(y+a/2)^2}-\sqrt{s^2+(y-a/2)^2}$$

For $s>>y,a$, we have

$$r_1-r_2\approx s\left[1+\frac{1}{2}\frac{(y+a/2)^2}{s^2}-1-\frac{1}{2}\frac{(y-a/2)^2}{s^2}\right]$$ $$=\frac{(y+a/2)^2-(y-a/2)^2}{2s}$$ $$=\frac{ay}{s}$$

which is the final result you want.

velut luna
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  • oh, nice, thank you! I didn't realise that $r_1-r_2$ already had all the terms in them that we need (honestly, I didn't even look), but yea, if you have that first expression, then the approximation is simple! thanks! – Sha Vuklia Jun 12 '17 at 12:31