# 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 $30^{o}$?"

"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."

Thanks so much.

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But $30\circ 5'$ is a number, in decimals roughly $30.083$. The reason not to replace it by $30$ would be loss of accuracy. In a GPS system it would make you miss your location by several km. But if someone asks you to draw a $30^\circ 5'$ angle, you can use the floor function, replace by $30^\circ$, and the drawing would look the same. So the answer, unfortunately, is "it depends." – André Nicolas Jul 29 '11 at 11:14
As long as the used units are made clear I don't see much harm coming from allowing such a use of the floor function. Granted, there is potential for confusion, so do not allow the use of floor function to physical measurements without specifying the units. Like what's the floor of my height? It could be 6 feet, 75 inches, 1 meter or 0 miles all depending. I would give a silly example like this to drive home the point that specifying the unit is necessary with such use of the floor function. Thus diminishing its usability. Angles are a bit special as they are often treated as bare numbers. – Jyrki Lahtonen Jul 29 '11 at 11:19
I'm not sure I understand your answer. I mean $30^{\circ}\, 5'$ is just a notation for a number indicating its interpretation/intention. I usually read $x^{\circ} = x \cdot \frac{\pi}{180}$, similarly $x^{\circ}\,y'\,z'' = \left(x + y \cdot \frac{1}{60}+ z \cdot \frac{1}{3600}\right)^{\circ}$ by definition. So if you want $[30^{\circ} \,5']$ to mean $30^{\circ}$ all you have to do is to define this as $[30 + 5\cdot\frac{1}{60}]^{\circ}$, but it would probably be better to simply say "rounded down to the next integer degree". – t.b. Jul 29 '11 at 11:21
Well, I don't know: if you are in need to use this rounding off frequently, I'd have absolutely no objection of using this convention, as it is easy to make sense out of it and to explain it. However, I don't think that this is a generally accepted notation (even if self-explanatory). – t.b. Jul 29 '11 at 11:39
Wow, it has been almost 20 years since I last encountered "minutes" (i.e., instead of fractional degrees). I have to wonder about the usefulness of this concept in contemporary society. Admittedly I became a mathematician and not, say, a sextant salesman... – Pete L. Clark Jul 29 '11 at 12:47

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...

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:Thanks for your time. You meant by "...I wouldn't use the floor function and instead consider defining a new one...", that we could define a similar function as [.] for angels defined in degrees? Honestly, my student wanted to impose some conditions on such these measured angels. He concluded, "...if $90^{o}$ mode $a$ = 0 then $b$ is $[90^{o}/a]$..." . – Basil R Aug 2 '11 at 17:11
When I saw his notation, I got his previous wrong mathematical notations. First his using of "mode" which is used for integers numbers and the second was using [.] for angels. The second one led me to ask this question, if "we can use [.] function for angels measured in degrees or not. And as you and other pointed this seemed strange in way I asked, otherwise, we define a new function as desired. – Basil R Aug 2 '11 at 17:17
@ J. M. :plz readmy comments. – Basil R Aug 6 '11 at 18:21
I've read them, @Basil. Now? – J. M. Aug 7 '11 at 4:45

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.

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