**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: L**et 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’.

**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’.

**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.

**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”

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.

**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”.

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**