SAS, and why the angle has to be the included one

0:00There is another thing you could measure instead of the sides. The angles. So try it. Three angles: thirty degrees, seventy, and eighty. They add to a hundred and eighty, so a triangle with those angles does exist. Here is one. Thirty at the bottom left, seventy at the bottom right, eighty at the top. Now here is another triangle with exactly the same three angles. And a third. The same three angles again.

0:31Each one is bigger than the last, and there is no end to that list. Three angles, and the triangle is not pinned down at all. So what did the angles settle, and what did they leave open? They settled the shape. All three of those triangles are the same shape. Multiply every side of the first one by one point six and you get the second, exactly. By two point four and you get the third. Every side by the same factor.

1:04That is a real relationship, and it has a name. Those triangles are similar. But similar is not congruent. Congruent means the same shape and the same size. The angles handed us the shape for free and said nothing whatever about the size. So we need at least one length. The question is how much more than that. Try this instead. Two sides, and the angle at the corner where those two sides meet.

1:35The side A B is six centimetres. The side A C is five. And the angle at A, the corner where they meet, is thirty degrees. Look carefully at where that angle is. A B starts at A. A C also starts at A. The angle sits between them. It is the included angle, and that word does all the work here. So: is that enough? Draw it and find out.

2:01Put the six down, swing the five off the same corner at thirty degrees, and join up. And notice that there was never a moment in that where you had a choice. Everybody who draws that gets the same triangle. Here is why. The third side is not something you choose. It is something you can calculate. Two sides with the angle between them fix the distance across the gap.

2:30With six, five, and thirty degrees, that third side comes out at about three point zero one centimetres. Exactly, its square is sixty-one minus thirty root three. And that is one number, not two. There is no second answer sitting there to be picked. So now all three sides are known, and three known sides already forced congruence. Turn it, flip it, draw it upside down. Everyone's triangle stacks onto everyone else's.

3:03That is the second condition, and it has a name. Two sides, and the angle included between them. Side, angle, side. S A S. And the middle letter really is in the middle. The order of those three letters is telling you where the angle has to sit. Three measurements again, exactly as before, and again they are enough. But be careful here, because there is a version of this that fails.

3:32And it fails while looking almost exactly the same when it is written down. So change one thing, and watch what happens. Keep the six centimetre side. Change the five to a four. That is not the important change. This is the important change. Move the angle. Instead of thirty degrees at A, put thirty degrees at B. The two sides we are naming are still A B and A C.

4:01But B is not where those two meet. B is one corner along from it. So the angle is no longer included. It is sitting beside the pair instead of between them. Two sides, and an angle that is not the one in the middle. Draw that, and something goes wrong. Sketch it roughly first, the way you would plan any construction. Three corners. Call the one carrying the angle P, and the far end of the six Q.

4:32P Q is the six. The thirty degrees is at P. And the four runs from Q to the third corner. Now build it properly, in three steps. Step one. Draw P Q, six centimetres. Step two. From P, draw a ray at thirty degrees to it. The third corner has to be somewhere along that ray. Step three. The third corner is also four centimetres from Q, so swing an arc of four with Q as its centre.

5:04Wherever that arc meets the ray, that is where the corner goes. So swing it, and watch the ray. The arc crosses it here. And then it crosses it again, out here. Two crossings. Call the near one R and the far one S. How far along? The perpendicular from Q down to the ray is exactly three centimetres. That is because thirty degrees is the angle whose sine is exactly one half, and half of six is three.

5:37Our arc has a radius of four, which is longer than three, so it reaches the ray and cuts right through it. R lands about two and a half centimetres along. S lands about seven point eight. And both of them are in front of P, on the same ray, on the same side. So join them up. Both of them. Triangle P Q R is squat and wide. Triangle P Q S is tall and leaning over.

6:08Now read off their measurements. Both have a base of six. Both have a side of four running out to the third corner. Both have thirty degrees at P. Six, four, thirty. Six, four, thirty. Identical. And they are plainly not the same triangle. Their third sides are two and a half against seven point eight. Neither drawing is wrong. Both obey every instruction given. There simply were not enough instructions.

6:41Now hold on a moment. We have met two crossings before, and it was fine. Three sides gave two crossings as well, and those two triangles turned out to be congruent. So why is this one fatal? Look at where the crossings sit. In that earlier construction the two arcs met above the base, and below it. One point on each side. So the base was a fold line, and one triangle folded straight onto the other.

7:11Here, both crossings sit on the same ray, going the same way, on the same side. There is no line to fold along. Nothing at all carries R onto S. Two crossings was never the problem. Where they sit is the problem. How often does this go wrong? I swept it. Every whole-number base up to twelve, every whole-number swung side up to twelve, every angle in five degree steps.

7:42Five thousand and forty sets of measurements in all. Three hundred and eighty of them produced two different triangles. Two thousand five hundred and twenty produced exactly one, and the rest produced none at all, the arc never reaching the ray. And there is a pattern in the failures. Every single ambiguous case had an acute angle. Not one had a right angle. Not one had an obtuse angle either.

8:11And whenever the swung side was longer than the base, there was exactly one answer. Every time. So here is the honest summary of all that. Two sides with the angle between them. That is S A S, and it settles the triangle completely. Two sides with the angle beside them. That is S S A, and it does not. But notice what that does not say. It does not say S S A is a rule that gives wrong answers.

8:44Nothing is being calculated incorrectly. There is simply not enough information to go on. And it does not say a non-included angle is useless, either. There are situations where one settles everything, and those are worth coming back to. What fails in general does not fail always. The word to hold on to is included.