A simple question in trigonometry Yesterday, one of my students asked me:
"Can I use floor($x$)=[$x$] when $x$ is an angle measured in degrees? In fact, is [$30^{o}, 5'$] allowed to be just $30^{o}$?"
I answered him:
"As the floor function is defined for $x$ when $x$ is a real number so, this function can not be used for measured angles when they are in degrees."
Are there any helpful suggestions?
Thanks so much.
 A: As with Theo in the comments, I treat the symbol ${}^\circ$ as convenient shorthand for "multiply by $\pi/180$" (similarly for minutes and seconds) just as I see $\%$ as shorthand for "multiply by 1/100", so using the floor function on quantities involving it would be fine in theory.
However, since $30^\circ5^\prime\approx 0.525053$, the floor of that is $0$. Now, if I wanted "round $30^\circ5^\prime$ to $30^\circ$", I wouldn't use the floor function and instead consider defining a new one. But consider also André's comment...
A: I see it like Theo.
Lets look at the floor function:
$$f: \mathbb{R} \longrightarrow \mathbb{Z}.  x \mapsto \max\{z \in \mathbb{Z}: z \leq x\}$$
Then we have for angles:
$$ \alpha^o=\alpha \cdot \frac{\pi}{180^o}$$
Therefore we could define a function that does exactly what you want:
$$g: \mathbb{R} \longrightarrow \mathbb{R}. x \mapsto f(x \cdot \frac{180^o}{\pi})\cdot \frac{\pi}{180^o}$$
Which would behave like you suggested, whereas the normal floor function $f$ would not produce the desired results when applied to degrees.
