Teacher: Stationary objects such as triangles, points or circles are not that interesting in their own right. Instead, we will explore motion.

Fix a point O on a piece of paper. Pick any point A. Draw an arrow from O to A. Now begin pushing the arrow OA (keeping A fixed).

What type of motion is this?

Student: This looks like a rotation. The point A will be moving ‘around’ the fixed point O.

Teacher: That is correct. This motion is indeed a rotation. We will be measuring this motion. How do you think we can do that?

Student: We can measure the angle.

Teacher: And what do you understand by an angle?

Student: Suppose after we push OA a little, it has reached OA’. Then the angle is \( \angle AOA’ \).


Teacher: But what IS that? Is it the ‘shape’ of AOA’ that you are referring to?

Student: May be not the shape. Angle is the thing enclosed by OA and OA’. I have read that we can use degrees to measure that.

Teacher: Clearly when you say ‘the thing enclosed’, the idea remains vague. What is this ‘thing’?

Student: Now that I think closely about it, I am not sure what it is.

Teacher: Let us examine it closely. For the purpose of this examination, temporarily forget whatever you have learned about degrees and angles.

Before measuring rotation lets measure another motion: Translation.

Translation is just parallel shift. Draw a segment BC on the paper. Now push it (without fixing any point). BC will slide to B’C’.


How much has BC moved?

Student: We can measure the distance from B to B’.

Teacher: Excellent! The distance between a point and its image is indeed a good way to measure translation.

Student: What do you mean by ‘image‘?

Teacher: Image of a point is the place where it moved to, after the motion. For example, in the above translation motion, B has moved to B’ after BC ‘slide’ to B’C’.

Here, for the point, B, the image (under translation motion) is B’. The point-image distance is the length of the segment BB’.

Point Image distance

Student: I see.

Teacher: Point-image distance is a good way to measure translation because its value does not depend on a particular choice of a point on BC. For example, if you chose C instead of B to measure translation, then the point – image distance would be the length of CC’. This is same as the length of BB’.

Student: But if we change the quantity of translation, the point-image distance will change.

Teacher: Correct! So for every ‘translation’, point-image distance is fixed (invariant), no matter which point you choose to use for your measurement.

Student: I understand.

Teacher: Will the point-image distance be a good way to measure Rotation?

Student: I don’t think so. If I choose a point B closer to the fixed O, for the purpose of measurement, then BB’ will surely be smaller than AA’.

Rotation 1

Teacher: Very good observation. In fact, as point-fixed point distance (OB), decreases, what will happen to point-image distance (BB’)?

Student: It will decrease.

change in point image

Teacher: So the point-image distance changes, even when the quantity of rotation remains same (just by changing our point of observation). Clearly, point-image distance is not a good tool for measuring rotation.

Note that Point-Image distance decreases, as Point-Center distance decreases. On the other hand, Point-Image distance increases, as Point-Center distance increases.

So these two distances increase or decrease simultaneously. Can you guess, what remains constant?

Student: Maybe their ratio?

Teacher: Absolutely! In fact, as you slide A, along the ray OA, \( \Delta OAA’ \) forms a bunch of similar triangles, hence having proportional side lengths.

Let’s record this observation.

For a fixed quantity of rotation, the ratio: \( \displaystyle { \frac {point-image-distance}{ point-center-distance} } \) remains unchanged no matter which point you choose to observe from.

We may use this ratio to measure rotation.

Student: Oh! Is this what ‘angle’ is?

Teacher: Almost. There is a catch though. We won’t be able to add rotations if we use this definition.

Lets explain further. Suppose, pushing OA a little, we get to OA’. Pushing OA’ a bit more, we get to OA”. Also assume the length of OA = OA’ = OA” = R

Rotation 1 = OA to OA’

Rotation 2 = OA’ to OA”

Add Rotations

Define Rotation 3 = Rotation 1 + Rotation 2 = OA to OA”

Measure of Rotation 1 is \( \frac {AA’}{OA} = \frac {AA’}{R} \).

Measure of Rotation 2 is \( \frac{A’A”}{OA’} = \frac{A’A”}{R} \)

Measure of Rotation 3 is \( frac{AA”}{OA} = \frac{AA”}{R} \)

We would like to have Measure of Rotation 1 + Measure of Rotation 2 = Measure of Rotation 3. But that cannot happen because \( \frac{AA’}{R} + \frac{A’A”}{R} \neq \frac{AA”}{R} \)

Can you guess why?

Student: Well, this one is easy. AA’ + A’A” > AA”. This follows from the triangular inequality: sum of two sides in the triangle AA’A” is greater than the third side.

Triangular Inequality

Teacher: Precisely. In fact the shortest path from A to A” is the segment AA”. So A to A’ and then A’ to A” is a longer path from A to A” (this argument is roughly the proof of triangular inequality).

The point is, we cannot add rotations in a natural way, because our formula does not work.

How do you think we can fix this problem?

Student: Now I remember reading something about arcs and radians though I was not sure why people were using arcs.

Teacher: Yes. Using point-image arcs instead of point-image segments will fix the problem. In fact note that arc AA’ + arc A’A” = arc AA”.

Arcs and angle

Hence we will define our measuring tool for rotation to be \( \frac{point-image-arc}{point-center-distance} \). Now everything will work out. Intuitively it is clear that for a fixed quantity of rotation this ratio remains fixed no matter what is your point of observation. Using arcs instead of segments solves the addition problem.

We will call this ratio angle. This ratio (angle) will be our tool for measuring rotation.

To be continued