Written by Colin+ in gcse, trigonometry.

My dad tells me that, above the blackboard in his 1960s Scottish high school, was a banner with the letters 'SOH CAH TOA' written out on it. Any questions about the banner were brushed off with a smile and 'you're not old enough to learn about SOH CAH TOA yet."

Which, I have to concede, is a great way to pique kids' interest in the topic. I've often wondered about the idea of telling students they're not old enough to know about maths yet, it's for over-16s only - and then let them get on with finding out the details on the sly.

There are - if you count the SOH CAH TOA way - 11 types of right-angled triangle questions. (Finding the hypotenuse or a leg1 given the other two sides; three versions of finding the angle given the other two sides; and six versions of finding one side given an angle and another side.) If you ask me - and I suggest you do - that's silly.

I say there are only two kinds of right-angled triangle problem: finding an angle if you know all the sides and finding a side if you know all the angles and a side. And all you need to know in order to solve all of these things: Pythagoras and the sine rule.

Now, you know Pythagoras. The square on the hypotenuse is equal to the sum of the squares on the other two sides - or, if you prefer, $opp^2 + adj^2 = hyp^2$. (I prefer this to $a^2 + b^2 = c^2$ because it tells you which side is which.) That means:

- if you know the two short sides, you square them, add them up and square root the answer to get the hypotenuse;
- if you know the hypotenuse and a leg, you square them, take them away (bigger minus smallest, of course) and square root the answer to get the other leg.

That's straightforward. If you're solving triangles the Table of Joy way, it usually makes sense to find the last side, just in case you need it.

Oh! If you have two angles of a triangle, it's easy to find the third, isn't it? Especially if one of them happens to be a right angle. You simply work out $\frac{\pi}{2}$ minus the other angle.

What's that?

Oh, fine. If you MUST use degrees (radians are much better), it's 90 minus the other angle.

At GCSE, you get given the sine rule on the front of the paper. At A-level, nope, you have to remember it. It's not exactly difficult, though:

$$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$$

What you do once you have either all three sides (and the right angle, don't forget) or all three angles (and a side), is label the corners of the triangle like the picture, with each angle (a big letter) opposite its corresponding little letter.

Also, write out the sine rule, and put a circle around all of the information you have, and a square around the thing you don't know. If there's a fraction with one of its numbers unshaped, that's fine - just cross it out. Neatly.

Here comes the clever bit! You can even do this without drawing the whole table - but if you're curious, you can buy Basic Maths For Dummies and/or Numeracy Tests For Dummies to see exactly how the Table of Joy works.

Here's what you do:

- Rewrite your fractions with numbers in the appropriate places and a question mark in the missing space.
- Find the number diagonally opposite the question mark and write it on the bottom of a big fraction.
- Take the other two numbers and write them out on top of the fraction with a times between them.
- Work out the fraction you've just written down.
- If you were looking for a side, you're done. Hooray.
- If you're after an angle, do $\sin^{-1}(Ans)$ and that'll give you the answer.

With this one, we don't really need the bottom (adjacent) side, but let's find it anyway: $16.8^2 - 9.8^2 = 186.2$, so the bottom side is the square root of that - 13.64 units (to 2dp). The Table of Joy would have 9.8 and 16.8 on the top, and $\sin(x)$ next to $\sin(\frac{\pi}{2})$ on the bottom. You'd work out $9.8 \times \sin(\frac{\pi}{2}) \div 16.8 = 0.583$; since we want an angle, we do inverse sine of that *using the answer button* to get 0.623 radians (38.69°, if you must).

This one gives an angle in degrees, tut tut. We can find the other angle by working out 90° - 33° = 57° and then work out the Table of Joy: you've got 8.7 and x on top, and $\sin(57^\circ)$ next to $\sin(90^\circ)$ on the bottom. The sum is $8.7 \times \sin(90^\circ) \div \sin(57^\circ) = 10.37$ units (2dp)

So there you go. A simple, easy way to solve any right-angled triangle without having to mess around with tan and cosine.

Oh yeah, why is this a better approach than slavishly learning all 11 possible right-angled triangle types? Because this method also works for non-right-angled triangles - although you also need the cosine rule instead of Pythagoras for some of those. It seems daft to me to learn a dozen different ways of doing special cases when you can learn a handful of general cases and be done with it. So there!

(Image by fdecomite used under a Creative Commons by licence)

* Edited 16/12/2013 for formatting.

- a short side' [↩]

## dborkovitz

Great title for your post! Agree that Sohcahtoa is stupid, but I think a bigger issue is that solving triangles isn’t the main point of trigonometry. Here’s my take http://debraborkovitz.com/2011/12/trigonometry-yoga/ .

* Comment edited for formatting.

## Colin

Thanks, Debra! I enjoyed your post 🙂

## lij

what do you do on a non right angled triangle when opposite is hypotenuse

## Colin

Strictly speaking, if it’s not right-angled, you don’t have a hypotenuse, just a longest side. In that case, you’ll need to use either the sine rule or the cosine rule, depending on what information you have. I’ll see if I can get to writing a blog post about that soon!

## Tom

In defence of SOH CAH TOA…

1. It gives an excuse to make up some silly mnemonics which can tell you a lot about a student and how their mind works.

2. It gives a context in which to introduce the sine, cosine and tangent functions, what they do, and what they’re for.

3. I have found that students ‘get’ the SOH CAH TOA rules more intuitively than the sine rule, and after taking advantage of point #2, students seem a bit more open to the sine rule (and the rest of trig) once you’ve had a bit of practise with SOH CAH TOA.

4. I disagree that teaching SOH CAH TOA teaches 11 ways of solving a triangle, at least not explicitly. My kids know that if you’ve got a right-angle and Pythagoras won’t work, then it’s SOH CAH TOA.

I’ve just read dborkovitz’s post, and I really like the Trig Yoga idea. I may well give that a try when introducing sine, cosine and tangent in the future. I feel that allowing students to get a feel for what the $\cos$, $\sin$ and $\tan$ buttons on their calculators actually mean is worth taking a lesson out for before actually pressing them.

* Comment edited for formatting.

## Colin

Hooray, a squabble!

I’ll concede Point 1, happily 😉

I’m curious about Point 3 – is that just because they’re taught SOH CAH TOA first?

My argument is that using the sine rule is simpler because a) it works for all triangles, so you don’t get students trying to use SOH CAH TOA on non-right angled triangles; and b) there’s one method for finding a side and one for finding an angle, end of story.

## Jed Hunsaker

As a programmer, I’m more interested in the performance aspect of the calculations, so if SOH-CAH-TOA takes less logical steps for a right triangle, then that’s what I’m going to use. I can pre-calculate PI/180 to save a fraction of time as well.

## Colin

Now, that

isa good point. I hope you won’t take it the wrong way when I say the article wasn’t addressed to you: it’s more of a clarion call for students who need to solve triangle under exam conditions. Under those circumstances, one method is better than many, but I do take your point.## Cav

I’m intrigued as to why you stop the generalising there? If we’re looking for a wholly general approach why not ditch Pythagoras’s RAT-specific theorem and just use the sine and cosine rule? This gives us tools to solve any triangle problem!

* Comment edited for autocorrect.

## Colin

Well, it is specifically about right-angled triangles – and I’ve softened up my stance a bit since writing this, I think. In principle, though, I agree: better to have one set of methods that work everywhere than hundreds of choices. Pythagoras seems like a more common special case than $\sin$, $\cos$ and $\tan$ to me, though.

## John

A bit late..

Trigonometry is not just about solving triangles. Nor is maths just about methods. SOH CAH TOA are the very base definitions for the trigonometric functions. E.g.: https://www.youtube.com/watch?v=Q55T6LeTvsA (Excuse the daft music). These definitions along are the base for all further understanding in this topic, hence why they were first included for teaching. Unfortunately their use has been narrowed down to solving triangles, neglecting their beauty and varied links to countless other areas of mathematics.

## Colin

Yes, I’m not sure I’d write this article the same way (or at all) today! However, I’d generally define the sine, cosine and tangent functions using a circle rather than defining them with SOH CAH TOA.