2
$\begingroup$

Recently I encounter a problem (of trigonometry) where $\sin{x}$ was asked and it was also told that $x$ is acute.

So, I like anyone else, found the general solution, however my general solution provide me with the values of $x$ in all the $4$ quadrants and obviously it also had values exceeding $2\pi$..

[in my lower class I was taught that angles $<90^{\circ}$ are acute and $>90^{\circ}$ are obtuse, but they never went beyond $180^{\circ}$]

My doubt: while writing the answer, would we
$1.$ consider angles $<90^{\circ}$ degrees as acute, or
$2.$ angles in first quadrant as acute?

And what for Obtuse?

Also, my question is not only restricted to trigonometric functions.

AND when the angle is greater than 360 degrees should we take mod360 or not?

$\endgroup$
  • 2
    $\begingroup$ I recall "acute" and "obtuse" being defined for interior angles of triangles. $\endgroup$ – Umberto P. Aug 12 '14 at 15:15
2
$\begingroup$

Acute angles are angles between $0^\circ$ and $90^\circ$.
Obtuse angles are angles between $90^\circ$ and $180^\circ$.
Reflex angles are angles between $180^\circ$ and $360^\circ$.
As far as I am aware, angles of size greater than $360^\circ$ do not have a special name.

I believe these names exist because of reference to the angles inside polygons, which is why negative angles and angles beyond $360^\circ$ do not have names.

The terminology "acute angle" refers specifically to these sizes and is not equivalent to saying "in the first quadrant". Angles in the first quadrant are any angles between $0^\circ$ and $90^\circ$, or between $360^\circ$ and $450^\circ$, or between $720^\circ$ and $810^\circ$, etc.

So in the context of the problem you have, it is most likely specifically asking for angles between $0^\circ$ and $90^\circ$ and not just in the first quadrant.

To answer the title of the question, $361^\circ$ is neither acute nor obtuse.

$\endgroup$
  • 1
    $\begingroup$ It really comes down to whether or not one takes angles modulo $360^\circ$ for the purposes of the definition. For trigonometry that's fine; for arc lengths, not so much! $\endgroup$ – Semiclassical Aug 12 '14 at 15:40
  • 1
    $\begingroup$ I'll have to disagree with you. The names were designed for angles in plane figures and 361$^\circ$ is not such an angle. Also, if you wanted to do it mod 360, then there ought to be a name for angles between 270 and 360 degrees, which there isn't. Finally, my feeling is that the info they were given about acuteness was designed to allow them to locate the angle exactly, which can't be done if you take it as meaning "first quadrant". Of course I would have simply specified what I wanted the angles to be between so it wasn't ambiguous for the students. $\endgroup$ – DavidButlerUofA Aug 12 '14 at 15:47
  • 1
    $\begingroup$ Well, taken modulo $360^\circ$ you'd just have a 'negative' acute angle for angles between $270^\circ$ and $360^\circ$.. But within the context of the problem (inverse trig functions) I'd agree that $361^\circ$ isn't the intended answer. $\endgroup$ – Semiclassical Aug 12 '14 at 15:56
  • $\begingroup$ Yeah although I didn't mention, but the special reference to 360+1 was made so as to ask whether we have to take mod or not.. $\endgroup$ – Harshal Gajjar Aug 12 '14 at 16:14
  • 1
    $\begingroup$ I understood that @HarshalGajjar. :) My answer is that most people would specifically mean by acute that only angles with actual size between 0 and 90 degrees are acute, thus making anything above 360 degrees not acute, even if it was in the first quadrant. $\endgroup$ – DavidButlerUofA Aug 12 '14 at 16:18
2
$\begingroup$

As David mentioned, a $361^{\circ}$ angle is neither acute nor obtuse. If your looking for a special word for this scenario, then you can say that a $361^{\circ}$ angle is coterminal with a $1^{\circ}$ angle.

$\endgroup$
  • $\begingroup$ @HarshalGajjar, your welcome. Glad I could help. $\endgroup$ – k170 Aug 12 '14 at 17:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.