Why enlarging the triangle leaves every ratio unchanged

0:00Last time we wrote down six ratios and gave them names. Sine of A, cosine of A, tangent of A. Look hard at that notation, because it makes a promise it has not yet kept. Writing sine of A says the number depends on the angle and on nothing else. But to work out a ratio you needed a triangle, and the angle did not hand you one. Draw an acute angle on its own. There are infinitely many right triangles sitting inside it: pick any point on the sloping arm and drop a perpendicular.

0:38Each one gives you three sides. Each one gives you a value for the sine. Nothing so far says those values agree. If they disagreed, sine of A would be meaningless, and so would every table of values you will ever use. So this is not a detail. It is the licence. Here is the picture that settles it. One angle at A, with two arms running out from it.

1:06Along the sloping arm, mark three points. From each one, drop a perpendicular straight down to the horizontal arm. Three right triangles, nested inside one another. A small one, a middle one, a large one. Every one of them has its right angle at the foot of its own perpendicular. And every one of them has the same angle at A, because they are all standing on the same two arms.

1:36That is the entire setup. Now the question is whether the sine you read off the small triangle is the sine you read off the large one. Take the small triangle and the middle one, and lift them apart so you can look at them side by side. The angle at A is shared. Not similar, not close — the same angle, because it is the same two arms both times.

2:04And each of them has a right angle where its perpendicular lands. Two angles of one match two angles of the other. That is the AA criterion for similar triangles, and it says these two are similar. Notice what did the work. Not that they look alike. Two angles agreeing is the reason; looking alike is only what the picture happens to show you. I put that to the test on ninety six triangles cut out of twelve different angles, and every single perpendicular cut came out right-angled at its foot with the same angle at A.

2:46Similar is a word. Here is what it actually gives you. In similar triangles, corresponding sides are in one common proportion. Not three separate proportions that happen to be close. One number, shared by all three pairs. Call it t. Then the small triangle's base is t times the middle one's base, its height is t times the middle one's height, and its slanting side is t times the middle one's slanting side.

3:19One t, three sides. That is the whole content of similarity, and it is the only thing the next step needs. I did not take that on trust either. For every pair of those cut triangles I divided each pair of corresponding sides and collected the answers. Six hundred and seventy two pairs, and every time the three answers collapsed to a single number. Now let me show you why that is worth checking rather than assuming.

3:52Go back to the angle at A and cut it again — but this time do not drop perpendiculars. Cut it with slanting lines, each at a different slope. These triangles still share the angle at A. They still stand on the same two arms. Everything the eye uses is still there. But divide their corresponding sides and the three answers do not agree. One hundred and forty four pairs, and all one hundred and forty four gave three different numbers.

4:25Same angle at A, and not similar. So sharing one angle is not enough, and the perpendicular is not decoration — it is the second angle, and it is what makes the criterion apply. Now the argument itself, and it is three lines long. The sine of A in the small triangle is its opposite side over its hypotenuse. Substitute. The opposite side is t times the middle triangle's opposite side. The hypotenuse is t times the middle triangle's hypotenuse.

5:04So the sine of A in the small triangle is t times one length, over t times another length. And there is the whole theorem. A t on the top, a t on the bottom. Cancel them. What is left is the middle triangle's opposite side over the middle triangle's hypotenuse — which is the sine of A read off the middle triangle. The two values are equal, and the only thing that happened was a cancellation.

5:36Nothing in those three lines cared which two sides you picked. Cosine of A is adjacent over hypotenuse. Both are t times the middle triangle's version. The t cancels. Tangent of A is opposite over adjacent. Both scale by t. The t cancels. And the same for the three reciprocals, because turning a fraction over does not bring the t back. Any quotient of two sides survives, because every side carries exactly one factor of t and a quotient has one on top and one underneath.

6:15So far we have only compared the small triangle with the middle one. Here is the rest of it: I compared all six ratios across every pair of those ninety six triangles. Where the angle was the same and the size was different — six hundred and seventy two pairs — all six agreed, every time. Let me state it carefully, because the careless version is wrong. The values of these six ratios do not change when you lengthen or shorten the sides, provided the angle is held fixed.

6:53That last clause is doing real work. What may change is the size. What may not change is the angle. Say it the other way round: the value follows the angle and ignores the size. And now the notation is honest. Sine of A depends on A alone, so writing it with an A and nothing else is legal. That is what this whole argument buys — not a fact about triangles, but permission to use a symbol.

7:25Put numbers on it. A right triangle with legs three and four; three squared is nine, four squared is sixteen, and those add to twenty five, so the hypotenuse is five. Sine of A is four fifths. Cosine of A is three fifths. Tangent of A is four over three. Now double everything. Six, eight and ten. Sine of A is eight tenths — which is four fifths. Cosine is six tenths, which is three fifths. Tangent is eight over six, which is four over three.

8:02Two triangles, one twice the size of the other, and all six numbers identical. Compare that with a triangle stretched one way only: six along the bottom but still four up the side. That is bigger too, and not one of the six survives. Zero out of six. So it was never the size. It was the shape, and the angle is what holds the shape. This licence pays for a habit you will use constantly, and it looks like cheating the first time you see it.

8:36Suppose you are told the sine of an angle is one third, and asked for the cosine. One third is not two lengths. It is a proportion. So write the opposite side as k and the hypotenuse as three k, where k is any positive number you like. Pythagoras: the adjacent side squared is nine k squared minus k squared, which is eight k squared. So the adjacent side is the square root of eight, times k.

9:09Now the cosine. Root eight k over three k — and the k cancels, exactly as this topic promised it would. I ran that with eight different values of k. The squared cosine came out eight ninths every single time. So k is not the length of anything. It is a placeholder that is guaranteed to vanish before you report an answer, and it is guaranteed to vanish because of the invariance.

9:40One caution. The square root has a negative version too. You throw it away, and not because it is awkward — because k is a length, and a length is not negative. Something else falls out of this for free. The hypotenuse is the longest side of a right triangle. Neither leg can reach it. Sine and cosine both have the hypotenuse underneath. So both of them are a smaller number over a bigger one.

10:11Over all ninety six triangles, at both acute corners, the longer of the two legs still did not reach the hypotenuse. Not once, in one hundred and ninety two corners. Now the careful wording. Shrink one leg towards nothing and the sine climbs — a half, then a hundred over a hundred and one, then ten thousand over ten thousand and one. It climbs towards one and never arrives, because the moment a leg reaches zero you no longer have a triangle.

10:47So say it as never exceeding one, rather than as always below one. That way the statement still holds at the two ends, where the figure has run out. And notice this bound came from the hypotenuse being longest, not from the invariance. The invariance says the value does not move. It says nothing at all about where the value sits. Now run the whole thing backwards, because this direction is the one that makes a table of sines usable.

11:22Two right triangles, drawn separately. You are told only that the two marked angles have the same sine. Nothing else. Must the angles themselves be equal? Equal sines means one opposite side over its hypotenuse equals the other opposite side over its hypotenuse. Rearrange, and the two triangles' opposite sides are in the same proportion as their hypotenuses. Call it k. Now Pythagoras recovers the third side of each from the two you already have. Substitute, and a factor of k comes straight out of the square root.

12:04So the third pair is in that same proportion k as well. All three pairs, one proportion — and that is the similarity criterion that works from sides alone. Similar triangles, so the angles are equal. Which is what we wanted. I checked it on two hundred and eight acute angles. Every pair with the same sine had the same angle, and every pair with different sines had different angles. Both mixed cases empty.

12:38That statement carries the word acute, and the word is not decoration. Drop it, and the claim is false. An angle and the angle you get by subtracting it from a straight angle have exactly the same sine. So I added triangles with an obtuse angle in them and ran the same test over all three hundred and twelve angles. This time the mixed cell was not empty: eighty five pairs had the same sine and different angles.

13:09Among acute angles that never happens, because there is only one acute angle with a given sine. That is precisely why you can look a value up in a table and get an angle back. Outside the acute range you cannot, and the failure is not a rounding problem — it is the statement being false. One last thing, because this is the sentence students overgeneralise. The ratios do not depend on the triangle. They depend entirely on the angle.

13:44Change the size, and all six numbers sit still. Change the angle, and all six move at once. I have the figure for that too. Across those ninety six triangles, whenever the angles at A differed, the number of the six that still agreed was zero. Every time, out of more than eight thousand pairs. Not one survivor. So this is an invariance under scaling, and under scaling only. Which is exactly the shape of a good definition. Sine of A is fixed the moment you fix A — no more, and no less.

14:23And that is what makes the next step possible — a table of these values, angle by angle. Without this argument there would be nothing to put in it.