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The angle of an isosceles triangle with sides = 10^7 and h = 1 is according to Wolfram $$ 1.000~000~000~000~000~4 \times 10^{-7} \quad \text{radians} $$ which corresponds to $$ 5.727 \times 10^{-6} \quad \text{degrees} $$ but that values gives the sine equal to $$ 9.9955 \times 10^{-8} $$ I imagined that the wolfram value is grossly approximated and spent a long time trying to find how many degrees give that sine but to no avail. The closest I could get was 5.729577951279..... no matter how many 9 I add I can't get that value

can anyone explain this riddle?

EDIT

I'd like to know with max precision the angle of the triangle and the value of its sine. The sine has fourteen 9'sif we use the radians and only 3 if we use the degrees, isn't that a huge difference for a calculator like wolphram's?

user157860
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    So let me get this clear, your real question is, what angle $\theta$ is the solution to $$ \sin{\theta} = 9.9955\times 10^{-8} $$ ? I would approach this with the (quite good!) approximation $$ \sin{\theta} \approx \theta $$ when $\theta$ is small and in radians. Then you just have to convert to degrees and you're done! – Matti P. Dec 17 '19 at 11:12

2 Answers2

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Which riddle? The two numbers are much the same: $1 * 10^{-7}=10 * 10^{-8}$

The two numbers differ for less than $0.1\%$, which is due to approximation.

Alberto Saracco
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  • ciao Alberto, don't you think 11 decimals difference is huge for wolfram? Why the downvote? – user157860 Dec 17 '19 at 17:07
  • What do you mean 11 decimals difference? The difference is less than $0.05%$ – Alberto Saracco Dec 17 '19 at 17:15
  • ii's in the edit : three 9 versus fourteen – user157860 Dec 18 '19 at 07:30
  • That's not how you measure the difference between numbers. It's a very small difference, due to approximation. As stated in other answers and comments, for such a small angle $\theta$, $\theta\simeq\sin\theta\simeq\tan\theta$, i.e. they are all approximately $1*10^{-7}$, which is what Wolfram is telling you. – Alberto Saracco Dec 18 '19 at 07:57
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Consider the expansion $$\sin^{-1}(x)=x+\frac{1}{6}x^3+O\left(x^5\right)$$ or better

$$\sin^{-1}(x)=\frac{x-\frac{17 }{60}x^3}{1-\frac{9 }{20}x^2}$$ Both of them would give (in degrees) the value after multiplying by $\frac {180}\pi$

Claude Leibovici
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