When are angles negative?

In earlier grades we learnt that in for instance an equilateral triangle all angles are 60 degrees. Now in high school I am taught that measuring angles anti-clockwise and clockwise make a difference.

So does that mean in an equilateral triangle one angle is 60 which is measures anti-clockwise and one is -60 degrees which is clockwise and.. I don't know, I am pretty confused.

How does it add to 180 then?

I was taught about this negative angle concept in trigonometry in unit circles which I understood very well but why is it if not applicable to any geometric figure?

It looks like I am lacking some common sense.

Solution 1:

The following line segment on the number line has length four.

    ---------
0 1 2 3 4 5 6 7

The following arrow drawn on the number line represents a displacement of $-4$.

    <--------
0 1 2 3 4 5 6 7

The following arrow drawn on the number line represents a displacement of $4$.

    -------->
0 1 2 3 4 5 6 7

You can do the same thing with angles; sometimes you only care about the magnitude, but other times you pick an orientation: you decide which leg of the angle you start at and which leg you end at. When we orient angles, we typically pick counterclockwise to be the positive direction, and clockwise to be the negative direction.

Even more interesting is an idea of angular position versus angular displacement. If we consider an angle whose vertex is the origin and one leg (the starting leg) is the positive $x$ axis, then putting the other leg at $270^\circ$ would be the negative $y$ axis, but $-90^\circ$ would also be the negative $y$ axis.

The idea that $270^\circ$ and $-90^\circ$ represent the same angle is the idea of angular position.

However, there is a difference between actually going counterclockwise $270^\circ$ around the plane from the positive $x$ axis to the negative $y$ axis, and going $90^\circ$ clockwise (i.e. $-90^\circ$ counterclockwise) from the positive $x$ axis to the negative $y$ axis. When this difference matters, we call it "angular displacement". We could even consider going completely around the origin to return to the $x$ axis then continuing another three quarters the way around: this would be a displacement of $630^\circ$!

As an addendum, if we orient the two legs of the angle so that they point in opposite directions in relation to the vertex, then we usually consider the leg pointing away from the vertex to be the starting leg, and the leg pointing towards the vertex to be the ending leg.

This way, if we orient the three sides of a triangle so that the sides all point in a counter-clockwise direction around the triangle, then all of the angles are oriented counter-clockwise too.

Solution 2:

In plane geometry, ‘negative angles’ are used when defining angles of a pair of vectors; it's a measurement for the rotation which takes the first vector to the second – so that $ \mathrm{angle}(\vec u,\vec v)=-\mathrm{angle}(\vec v,\vec u)$. As an example in usual life, screwing is not the same as unscrewing.

The difference between geometric angles and ‘algebraic’ angles is similar to the difference betwen a segment and a vector.

Solution 3:

Negative angle measurement don't exist. Sorry.

In a geometric figure like a triangle all angles have positive measurement.

So in a triangle our angles could have measures of $30^o, 60^o, 90^o$ or $\pi/6, \pi/3, \pi/2$ if you like. These are measurements of physical angles which are never negative.

However the measurements are done in degrees or radians which is a numerical system and we can think about a number as $-60^o$ or $-\pi/6$. These are numbers plain and simple.

You probably saw "negative angles" in trignometry class when you were confusing them with negative numbers of angle measurements. Functions like $\sin \theta, \cos \theta, \tan \theta$, etc. are easy to define for $\theta$ between $0^o$ and $360^o$ but we can just as easily define them for any other measurement including numbers like $-60^o$ by realizing that on the unit circle going $-60^o$ (or $60^o$ clockwise) is the same as going $300^o$ counterclockwise.

Hope this helps!

Solution 4:

Angles are not negative, however in trigonometry we easily talk about them anyway. The thing we use to make sense of them is modular arithmetic.

Modular arithmetic arranges the numbers in a clock or cycle, for example modular 3 we count $0,1,2,0,1,2,0,1,\dots$. We simply only have $3$ integers, and if we reach $3$ we simply go back to $0$. We therefore say that $3$ modulu $3$ is $0$, which is written as $3\equiv0\pmod{3}$

When using modular numbers, we don't have negative numbers either, and if we go below zero, we end up at the top again, so $-1\equiv2\pmod{3}$

We have a rule here and that is, we can add or subtract the number we're doing the modulus to as many times we want, and it'll remain the same.

When using degrees, a full turn in a circle is $360^\circ$, and when we have performed a full turn in a circle, we're just back where we started. This is just like the modulus operation I talked about above, and we can use the modulus concept to explain how angles operate when going below $0^\circ$ or above $360^\circ$.

When working with degrees, we go back to zero at $360^\circ$, therefore in degrees it makes sense to say that degrees work using modulus $360^\circ$. If we understand angles this way, we can explain what a negative angle really is. Let's for example take $-90^\circ$, we can apply modular arithmetic, which allows us to add $360$ to any angle without changing it: $-90^\circ\equiv270^\circ\pmod{360^\circ}$ and if you draw those two angles on the unit circle, notice how the angles point at the same place.

All this also explains why going above $360^\circ$ allows you to go below again, namely $400^\circ\equiv40^\circ\pmod{360^\circ}$, and therefore the angle $40^\circ$ and $400^\circ$ is just the same thing.

When talking about radians, it's much the same, except a full turn is $2\pi$ instead of $360$, and therefore we have to work modulus $2\pi$ with radians.