\begin{problem*}{1.62}
In physics, it is important to use mathematical approximations.
Demonstrate that for small angles ($< 20\dg$)
\begin{equation}
 \tan \alpha \approx \sin \alpha \approx \alpha = \pi \alpha ' / 180\dg
\end{equation}
where $\alpha$ is in radians and $\alpha '$ is in degrees.
Use a calculator to find the largest angle for which $\tan \alpha$ may 
be approximated by $\alpha$ with an error less than $10.0$\%.
\end{problem62} % Problem 1.62

\begin{solution}
To kill both birds with one stone,
 a table to show the approximations hold
  and show the \% error of the approximation:\\
\begin{tabular}{|r|r|r|r|r|}
\hline
$\alpha '$&$\alpha$ [rad]&$\sin \alpha$&$\tan \alpha$&\% error\\
\hline
 $0\dg$&0.000&0.000&0.000&$\emptyset$\\
 $5\dg$&0.087&0.087&0.087&$-0.25$\%\\
$10\dg$&0.175&0.174&0.176&$-1.02$\%\\
$15\dg$&0.262&0.259&0.268&$-2.30$\%\\
$20\dg$&0.349&0.342&0.354&$-4.09$\%\\
$31\dg$&0.541&0.515&0.601&$-9.95$\%\\
$32\dg$&0.599&0.530&0.625&$-10.62$\%\\
\hline
\end{tabular}\\
where the \% error is given by
\begin{equation}
 \text{\% error} = \frac{\text{approx.} - \text{actual}}{\text{actual}}
	= \frac{\alpha - \tan \alpha}{\tan \alpha}.
\end{equation}

So $31\dg$ is the largest whole-degree angle with $< 10$\% error.
\end{solution}

