# Unlabelled triangles

Here’s @nathanday314 with an unusual puzzle:

This is a task that provoked a lot of discussion after a sequence of trigonometry (sine rule, cosine rule etc.) lessons.

— Nathan Day (@nathanday314) June 26, 2021

What approach would you go with? pic.twitter.com/1Pmf8qknuP

### What’s the problem?

I’m currently working on making my problem-solving techniques more explicit, and in that spirit: my first step is to write down exactly what the problem is:

- We have two triangles.
- We’re told (at least implicitly) that the four largest angles in the two triangles are 70º, 75º, 65º and 85º – the 75º angle is marked.
- We’re told that the six sides of the triangles are 4.97, 5.80, 9.00, 9.50, 9.77 and 9.89 units.
- Our job is to label the triangle.

Sounds fun! Have a go – I’ll spoiler it below the line.

### My approach

I could, of course, have used a calculator. However, since The Mathematical Ninja’s recent visit, I’m somewhat fearful of booby-traps, so instead I’ll work with estimates. (You don’t need to… if you’re willing to chance it.)

My plan is to use the sine rule to figure out ratios between the sines of the given angles, and ratios between the sides, to see if anything matches up nicely.

So, what are the sines? They’re all large angles and will be fairly close to 1. I can use the identity $\sin((90-x)º) = \cos(xº)$ and the approximation $\cos(xº) \approx 1 - \frac{3}{20,000}x^2$ to get good estimates.

To divide by 20,000, I would halve and shift by four places; I’m going to leave the numbers as $1-\dots$ for now in case I don’t need to do the sums. I suppose I’m working out something like the vercosine.

- $\sin(85º) = \cos(5º) \approx 1 - \frac{3(25)}{20,000} = 1-0.00375$
- $\sin(75º) = \cos(15º) \approx 1 - \frac{3(225)}{20,000} = 1-0.03375$
- $\sin(70º) = \cos(20º) \approx 1 - \frac{3(400)}{20,000} = 1-0.06$
- $\sin(65º) = \cos(25º) \approx 1 - \frac{3(625)}{20,000} = 1-0.09375$

Now, I can take some shortcuts with the ratios of the angles: $\frac{1-a}{1-b}$ is (for small numbers like we have here) roughly $(1-a)(1+b)$, or $1 + b - a$.

In particular, let’s look at ratios involving the largest angle:

- $\frac{\sin(85º)}{\sin(75º)} \approx 1 + 0.03375 - 0.00375$, or $1.03$
- $\frac{\sin(85º)}{\sin(70º)} \approx 1 + 0.06 - 0.00375$, or $1.05625$
- $\frac{\sin(85º)}{\sin(65º)} \approx 1 + 0.09375 - 0.00375$, or $1.09$

We need to find two sides whose ratios are roughly similar to one of those. Again, I’m going to start with the largest one. I’m also going to rewrite the numbers as differences from 10, which should make the calculations simpler.

- $\frac{9.89}{9.77} = \frac{10-0.11}{10-0.23}$
- $\dots = \frac{1-0.011}{1-0.023} \approx 1 + 0.023 - 0.011 = 1.012$

- $\frac{9.89}{9.50} = \frac{10-0.11}{10-0.5}$
- $\dots = \frac{1-0.011}{1-0.05} \approx 1 + 0.05 - 0.011 = 1.039$

- $\frac{9.89}{9.00} = 1.0888\dots$ by inspection.

The last of those is pretty close to the last of the sine ratios! But we’d better check with the other triangle – is $\frac{\sin(75º)}{\sin(70º)}$ roughly the same as $\frac{9.77}{9.50}$?

Let’s see: $\frac{\sin(75º)}{\sin(70º)} \approx \frac{1-0.03375}{1-0.06} \approx 1.02625$.

$\frac{10-0.23}{10-0.5} = \frac{1-0.023}{1-0.05} \approx 1.027$. I’d say that’s pretty close!

So! Let’s recap: we’ve got a pretty good hunch that one triangle is 85-65-30, with longer sides of 9.89 and 9.00; the other is 75-70-35, with longer sides of 9.77 and 9.50. What about the final side?

Taking the first triangle, I can use sine rule again – $\sin(30º) = 0.5$.

Calling the missing side $x$, I’ve got:

- $\frac{\sin(85º)}{\sin(30º)} = \frac{9.89}{x}$
- $x\sin(85º) = 9.89\sin(30º)$
- $x = \frac{4.945}{\sin(85º)}$

Now, $\sin(85º)$ isn’t quite half a percent less than 1, so $x$ is much more plausibly 4.97 than 5.80.

It’s worth checking the final side of the other triangle ($y$), too – let’s go with 70 and 35, where I have another trick up my sleeve:

- $\frac{\sin(70º)}{\sin(35º)} = \frac{9.5}{y}$
- $2y\cos(35º) = 9.5$, using the double-angle formula
- $y = \frac{4.75}{\cos(35º)}$.

What’s $\cos(35º)$? Certainly less than 0.87 (which corresponds to about 30º), and I know that $\cos(37º) \approx 0.8$ from the 3-4-5 triangle, so somewhere about 0.82 looks good.

That means $y \approx \frac{4.75}{0.82}$. I can’t say I’m a fan. Let’s multiply top and bottom by 10: $y \approx \frac{47.5}{8.2}$. The top is a little less than 48, the bottom is a little more than 8, so $y$ is likely a bit less than 6 – which fits with 5.80.

(In fact, we can be more accurate: the top is a quarter of a percent less than 48; the bottom is two and a half percent more than 8, so we should be about 3% less than 6. That would give 5.82, which isn’t at all bad given all of the approximating up to here.)

I reckon we can conclude that one triangle has angles of 85-65-30 (degrees) and sides of 9.89, 9.00 and 4.97; the other has angles of 75-70-35 and sides of 9.77, 9.50 and 5.80.

### Is there another way?

One of the things I like about this puzzle is that there’s very little by way of scaffolding or instruction: you’re free to tackle it by whatever method makes you happiest. With that in mind, I’m really curious to hear if you took a different approach! Drop me an email!