POST BY AVIGYAN

This is going to be a very short blog-post discussing about the notations of angles.

Backstory: This blog-post is written because the notation $\measuredangle (\ell_1, \ell_2)$ confused me so much that I had to ask what it really means in an MSE post. I was still not satisfied with the explanation I got there. As I read a few more handouts, the notations became a bit more clear to me. I hope the idea I am sharing in this blog-post is indeed the correct one.

We start with vanilla angles first. Let's say we have three points $A$ , $B$ and $C$ . Then suppose we want to find the value of the angle $\angle ABC$ . We can extend the lines $AB$ and $BC$ to note that there are four angles that are formed. Which one do we take?

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (31.64835,16.28958); pair B = (40.11708,2.37181); pair C = (44.07794,21.53331); import graph; size(8cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(arc(B,4.60985,78.32093,121.31981)--B--cycle, linewidth(0.6) + blue); draw(arc(B,4.60985,-101.67906,-58.68018)--B--cycle, linewidth(0.6) + blue); draw(arc(B,2.56591,-58.68018,78.32093)--B--cycle, linewidth(0.6) + red); draw(arc(B,2.56591,121.31981,258.32093)--B--cycle, linewidth(0.6) + red); draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw((37.77614,-8.95301)--B, linewidth(0.6) + linetype("4 4")); draw(B--(45.63427,-6.69530), linewidth(0.6) + linetype("4 4")); draw(arc(B,4.60985,78.32093,121.31981), linewidth(0.6) + blue); draw(arc(B,4.27057,78.32093,121.31981), linewidth(0.6) + blue); draw(arc(B,4.60985,-101.67906,-58.68018), linewidth(0.6) + blue); draw(arc(B,4.27057,-101.67906,-58.68018), linewidth(0.6) + blue); dot("$A$", A, NW); dot("$B$", B, dir(180)); dot("$C$", C, N); [/asy]$

So here, we define the notation $\angle ABC$ as the magnitude of the angle with measure $\le 90^{\circ}$ angle of the four angles we found earlier. For this case, it is the blue angle.

But what does the notation $\angle (\ell_1,\ell_2)$ mean? Is it the one which is $\le 90^{\circ}$ or the other one? Does orientation matter?

This is where the doubt first comes up. In vanilla angles however, this question makes no sense as we always define the vanilla angle between two lines as the angle which is $\le 90^{\circ}$ . Vanilla angles do not care about orientation too.

Thus we can now safely conclude that $\angle (\ell_1,\ell_2) = \angle (\ell_2,\ell_1)$ . So far so good.

Before we begin, we should keep in mind that "Anti-clockwise is positive and clockwise is negative".

Let us now move onto directed angles. This is where things start behaving differently because suddenly, reflexive angles start being considered modulo $180^{\circ}$ and orientation also comes into play.

We begin with the same configuration. Let's say we have three points $A$ , $B$ and $C$ (where $A$ , $B$ and $C$ are in anti-clockwise order). Now we want to find the value of the directed angle $\measuredangle ABC$ .

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (31.64835,16.28958); pair B = (40.11708,2.37181); pair C = (44.07794,21.53331); import graph; size(8cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(arc(B,4.60985,78.32093,121.31981)--B--cycle, linewidth(0.6) + blue); draw(arc(B,4.60985,-101.67906,-58.68018)--B--cycle, linewidth(0.6) + blue); draw(arc(B,2.56591,-58.68018,78.32093)--B--cycle, linewidth(0.6) + red); draw(arc(B,2.56591,121.31981,258.32093)--B--cycle, linewidth(0.6) + red); draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw((37.77614,-8.95301)--B, linewidth(0.6) + linetype("4 4")); draw(B--(45.63427,-6.69530), linewidth(0.6) + linetype("4 4")); draw(arc(B,4.60985,78.32093,121.31981), linewidth(0.6) + blue,BeginArcArrow(6)); draw(arc(B,4.60985,-101.67906,-58.68018), linewidth(0.6) + blue); draw(arc(B,4.27057,-101.67906,-58.68018), linewidth(0.6) + blue); draw(arc(B,2.56591,121.31981,251.28453), linewidth(0.6) + red,EndArcArrow(6)); dot("$A$", A, NW); dot("$B$", B, dir(180)); dot("$C$", C, N); [/asy]$

For doing this, we take the line $BA$ and rotate it in an anti-clockwise direction w.r.t. $B$ until it gets super-imposed on the line $BC$ . (If you thought it was the other way around, you were wrong!)

The orientation does not depend on the ordering of the vertices but rather on the direction of rotation. (I had the opposite idea a few weeks back too!) This statement might feel a bit ambiguous right now, but just hold on.

So when we are finding the value of $\measuredangle ABC$ (as shown in the image above), the positive value of the directed angle is $> 90^{\circ}$ (the red angle) for the case in the image. Now if we take this value modulo $180^{\circ}$ to make the magnitude $\le 90^{\circ}$ , we find that the value is actually negative.

The anti-clockwise ordering of the vertices do not matter here, but rather the direction of rotation matters here. In order to get an angle which is $\le 90^{\circ}$ in magnitude, we make the rotation clockwise which results in the value of the angle being negative.

Now in the figure as shown below, $A$ , $B$ and $C$ are ordered in clockwise order. In this case, the value of $\measuredangle ABC$ is actually positive. (The ordering of the vertices does not matter. Only the direction of rotation does.)

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (31.64835,16.28958); pair B = (40.11708,2.37181); pair C = (44.07794,21.53331); import graph; size(8cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(arc(B,4.60985,78.32093,121.31981)--B--cycle, linewidth(0.6) + blue); draw(arc(B,4.60985,-101.67906,-58.68018)--B--cycle, linewidth(0.6) + blue); draw(arc(B,2.56591,-58.68018,78.32093)--B--cycle, linewidth(0.6) + red); draw(arc(B,2.56591,121.31981,258.32093)--B--cycle, linewidth(0.6) + red); draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw((37.77614,-8.95301)--B, linewidth(0.6) + linetype("4 4")); draw(B--(45.63427,-6.69530), linewidth(0.6) + linetype("4 4")); draw(arc(B,4.60985,78.32093,121.31981), linewidth(0.6) + blue,EndArcArrow(6)); draw(arc(B,4.60985,-101.67906,-58.68018), linewidth(0.6) + blue); draw(arc(B,4.27057,-101.67906,-58.68018), linewidth(0.6) + blue); draw(arc(B,2.56591,121.31981,251.28453), linewidth(0.6) + red); dot("$C$", A, NW); dot("$B$", B, dir(180)); dot("$A$", C, N); [/asy]$

To make this even more clear, we take an example from EGMO. This example should make the idea even more vivid.

Now let us get to the original question due to which this blog-post was made. How do we define $\measuredangle (\ell_1, \ell_2)$ ? If you've read through the blog-post till here, you should try this little exercise yourself first before moving on.

For doing this, we define $T=\ell_1\cap \ell_2$ . Now we rotate $\ell_1$ w.r.t. $T$ in an anti-clockwise direction till it gets super-imposed on the line $\ell_2$ . Now if the value of the angle is $\le 90^{\circ}$ , that's basically our desired directed angle and the sign also remains positive (since we rotated in an anti-clockwise direction). However, if the value of the angle turns out to be $\ge 90^{\circ}$ , then we take it modulo $180^{\circ}$ and make its magnitude $\le 90^{\circ}$ . In this case, the sign of the angle is negative. Thus this case is equivalent to taking the line $\ell_1$ and rotating it about $T$ in a clockwise direction till it gets super-imposed on $\ell_2$ .

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ import graph; size(8cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(arc((40.11708,2.37181),5.21970,78.32093,121.31981)--(40.11708,2.37181)--cycle, linewidth(0.6) + blue); draw(arc((62.92721,4.40749),5.21970,78.32093,121.31981)--(62.92721,4.40749)--cycle, linewidth(0.6) + red); draw((54.45847,18.32526)--(62.92721,4.40749), linewidth(0.6)); label("$\ell_1$", (54.45847,18.32526)--(62.92721,4.40749), SW); draw((62.92721,4.40749)--(66.88806,23.56899), linewidth(0.6)); label("$\ell_2$", (62.92721,4.40749)--(66.88806,23.56899), SE); draw((60.58626,-6.91732)--(62.92721,4.40749), linewidth(0.6)); draw((62.92721,4.40749)--(68.44439,-4.65961), linewidth(0.6)); draw((31.64835,16.28958)--(40.11708,2.37181), linewidth(0.6)); label("$\ell_2$", (31.64835,16.28958)--(40.11708,2.37181), SW); draw((40.11708,2.37181)--(44.07794,21.53331), linewidth(0.6)); label("$\ell_1$", (40.11708,2.37181)--(44.07794,21.53331), SE); draw((37.77614,-8.95301)--(40.11708,2.37181), linewidth(0.6)); draw((40.11708,2.37181)--(45.63427,-6.69530), linewidth(0.6)); draw(arc((40.11708,2.37181),5.21970,78.32093,119.21372), linewidth(0.6) + blue,EndArcArrow(6)); draw(arc((62.92721,4.40749),5.21970,78.32093,119.21372), linewidth(0.6) + red,BeginArcArrow(6)); [/asy]$

Thus we can conclude our findings for the directed angles as, $\measuredangle (\ell_1,\ell_2)=-\measuredangle (\ell_2,\ell_1)$ and
$\measuredangle (AB,BC)=\measuredangle ABC=-\measuredangle CBA=-\measuredangle (CB,BA) .$
However I should mention that $\measuredangle (AB,BC)$ is the same as $\measuredangle (AB,CB)$ since both the lines $BC$ and $CB$ are basically equivalent.

I hope I was able to convey my message through this blog-post (plix halp this nub to larn bemter englis :maybe:

).

Thank You!!

amar_04 haw so orsmax :omighty: