Given one ratio, reconstructing the other five

0:00Here are the six ratios of one acute angle, and here is a question you can ask about them. Suppose somebody tells you just one of the six. The sine of A is one third. That is all you get. Can you work out the other five? It sounds like too little to go on. One number, five answers. But you can, every time, and the reason is worth more than the method.

0:26A ratio is not a description of a triangle. It is a description of a shape, and a shape is enough. Sine of A is one third means the side facing A is one third of the longest side. It does not say the side facing A is one centimetre. It does not say anything at all about how big the triangle is. Here are three right triangles. This one has a facing side of one, this one of three, this one of twelve.

1:00All three satisfy the instruction, because in each one the facing side is a third of the longest. I sorted a hundred and forty four right triangles by the value of the sine at a marked corner. They fell into ninety one groups. And inside every single group, the number of triangles and the number of different sizes were the same number. No two triangles sharing a sine were the same triangle.

1:30The value fixes the shape and never once fixed the size. So here is the move that makes the whole topic work. The instruction is a proportion of one to three. Write the facing side as k, and the longest side as three k. One letter, used twice, holding the proportion and leaving the size open. What is k allowed to be? Any positive number you like. One. Seven. A half.

2:00Each choice picks out a different triangle from the family, and every one of them has the right shape. Notice what has not happened. Nobody has said what k is. Nobody is going to. You will never solve for it, and you will never need to. It is not an unknown. It is a passenger. Two sides are written down. The third comes from Pythagoras. The square on the longest side is the sum of the squares on the other two.

2:32Three k, squared, is nine k squared. The facing side squared is k squared. So the remaining side squared is nine k squared minus k squared, which is eight k squared. Look at what survived the subtraction. Every term carried k squared, and so does the answer. That is not luck. Squaring a proportion gives a proportion, and subtracting two of them leaves a proportion. The multiplier has come through the hardest step of the calculation completely untouched.

3:09Now take the square root, and something needs saying out loud. Eight k squared has two square roots, not one. Two root two k, and minus two root two k. Both of them square to eight k squared. Squaring cannot tell them apart, because squaring destroys the sign. I put seven squared quantities through a routine that returns both roots and then asks which of them is a length. Every time: two candidates, one length.

3:41The negative one is not ignored, and it is not inconvenient. It is rejected, and here is the reason. This is a side of a triangle. A side has a length. A length is not negative. The one case where the routine found no length at all was the square of nothing, where the triangle has run out. Reject the root for the reason, not by habit, and you will reject the right one in problems where it is less obvious.

4:14The triangle is complete. Facing side k, remaining side two root two k, longest side three k. Read off the cosine: the remaining side over the longest. Two root two k, over three k. And there goes the passenger. The k is on the top and on the bottom, and it cancels. Cosine of A is two root two over three. No k anywhere. Tangent of A is k over two root two k, which is root two over four.

4:47Cosecant is three. Secant is three root two over four. Cotangent is two root two. Six values, one given and five found, and not one of them mentions k. If a k ever survives into your answer, you have made a mistake, because a ratio of two sides can never depend on the size. Run the same loop on a different starting ratio. The tangent of A is four thirds.

5:17Tangent compares the facing side with the side along the angle, so write those two as four k and three k. Pythagoras: sixteen k squared plus nine k squared is twenty five k squared, so the longest side is five k. There is the three, four, five, arriving out of a ratio rather than being handed to you. Now all six, and every k cancels. Sine four fifths, cosine three fifths, tangent four thirds as given.

5:51Cosecant five quarters, secant five thirds, cotangent three quarters. Four steps, and the same four steps every time. Read the ratio. Write the two sides with a multiplier. Complete the triangle. Read the rest off. Try a ratio of one. The tangent of A is one. Tangent one means the two shorter sides are equal, so write both of them as k. The longest side squared is k squared plus k squared, which is two k squared, so that side is root two k.

6:28The sine is k over root two k. The cosine is the same. Both come to root two over two. Now multiply them together and double the result. Root two over two, times root two over two, is one half. Twice that is one. Exactly one. I checked that in surds rather than in decimals, so it is exact and not nearly. That angle, by the way, is forty five degrees, and this little result is the first glimpse of something much bigger about doubling angles.

7:04Keep it in mind. It comes back. Here is a mistake worth having on purpose. You are given a number and you start writing sides down. Which two sides? The number alone does not say. The name does. So I ran the reconstruction with every name offered against every value. When the name matched the number, all one thousand seven hundred and twenty eight attempts recovered all six ratios. When it did not, three thousand five hundred and fifty two of them refused to build a triangle at all.

7:42Four thousand nine hundred and forty four built one, and it got none of the six right. Not one out of six. None. And a hundred and forty four of them agreed anyway. Every single one of those was a triangle whose two shorter sides are equal, where the sine really does equal the cosine. That is a fact about those triangles, not a hole in the method. The name is not a label on the number. It is half the data.

8:15Sometimes you are handed lengths instead of a ratio. A right triangle with the longest side twenty nine units and one of the others twenty one. No multiplier needed here, because the size is already fixed. The third side squared is twenty nine squared minus twenty one squared. Which is eight times fifty, which is four hundred. So that side is twenty. The sine of the marked angle is twenty over twenty nine, and the cosine is twenty one over twenty nine.

8:49Now add their squares. Four hundred plus four hundred and forty one is eight hundred and forty one, over eight hundred and forty one. One. Exactly one. That is not a coincidence of this triangle. I added those two squares at every acute corner of all hundred and forty four triangles. The answer was one, every time, with no exceptions. Subtract them instead and you get ninety one different answers, which is what makes the first result worth noticing.

9:26One more kind of data, and it needs a different tool. A right triangle where one side is seven centimetres, and the longest side is one centimetre more than the remaining one. Nothing here is a proportion, so a multiplier would be the wrong instrument. Call the remaining side x. Then the longest is x plus one. Pythagoras: x plus one, all squared, equals forty nine plus x squared. Expand the bracket and the x squared appears on both sides and goes.

10:01What is left is linear. One plus two x equals forty nine, so x is twenty four, and the longest side is twenty five. Real lengths, in centimetres, because the data fixed a size. I searched every whole number up to two hundred for a triangle fitting that description. Exactly one. Compare that with the ratio case, where the same search found fifteen triangles in fifteen different sizes. Recognising which kind of data you have been handed is half the skill here.

10:39Last question. Is every number a possible value? The reconstruction answers this without any extra thought, because it refuses when the third side would not come out positive. I offered fifty five candidate values to each of the six names. For the sine, twenty seven were possible and twenty eight were refused. Same for the cosine. For the tangent, all fifty five were possible. Same for the cotangent. For the secant and the cosecant, twenty seven possible and twenty eight refused, but the other way round.

11:18So a sine of four thirds is impossible, because the side facing the angle cannot beat the longest side. A secant of twelve fifths is perfectly possible, because a secant is the longest side over a shorter one and is always above one. And a tangent above one is ordinary, not exotic. One routine, one refusal, and every one of those questions answered the same way. So here is the whole thing, in four boxes.

11:52Read the ratio, and note which two sides it names. Write those two sides as multiples of one positive number. Complete the triangle with Pythagoras, keeping the positive root because a length is positive. Read off whichever of the six you were asked for, and watch the multiplier cancel. The shapes that come out of this are not tricks to memorise. Three, four, five turned up. So did five, twelve, thirteen, and eight, fifteen, seventeen, and twenty, twenty one, twenty nine, and seven, twenty four, twenty five.

12:31Five different whole-number shapes, and there are eleven of them with both short sides under sixty. The shapes are scenery. The loop is the content. One ratio, and the other five follow, because the ratio was never about the triangle in front of you. It was about the shape all of them share.