Originally posted, 10 Aug 11.

[**EDIT:** I’ve added a section after the post-script, in response to an enquiry about some trigonometry I used to work out the length of a chord, in a minor aside when talking about the definition of a degree. This also led to me correcting the figure I used for the diameter of the circle; also in the aside. I could have made it a new post, but decided that, as it partly involved a correction of this one, it properly belonged here.

*—Daz*]

Had a conversation with a work-mate the other day, about our remembrances of school lessons. It seems we both left with much the same impression; that much of our learning was based more on effect than cause. (We’re both of a similar age, and I have no idea if this is still the case in UK schools.) In history lessons, for instance, we learned a lot of events, much of them almost by rote, but very little about the causes of those events. We were just taught what happened. In science subjects we were taught a lot of laws and facts, but not the underlying causes. For instance we were taught that water expands and becomes less dense as it cools from 4° to 0°, but not *why* it does so. (Turns out it’s to do with the crystalline structure of ice—it has gaps between the molecules. It also turns out that water’s not quite as unique in this as we were led to believe. Gallium, bismuth, antimony and silicon behave in the same way as they freeze.)

Most damningly, though, neither of us were taught, or made to find, proofs of some quite basic mathematical concepts; which really is a pity, as some of them are easy enough to be set as an exercise for quite young students, and encouraging youngsters to look for proofs and evidence would really help them develop a sceptical, ‘scientific’ frame of mind, rather than accept ‘facts’ on authority, which would be of enormous benefit in later science lessons, as well as in ‘real life’.

Anyway, when the conversation turned to maths, my workmate mentioned a few that had puzzled him slightly, and I, having been puzzled enough by the same things as a kid, had worked a couple out on my own (albeit with a hint or two from my maths teacher, when I explained my problem to him), so I managed to explain them to him. The first, involving opposite angles, I managed verbally; the other, involving the angles of a triangle, had to wait for a few hasty back-of-an-envelope diagrams in the canteen at lunch time—which gained a small audience of three or four others who’d apparently had much the same experience of school that we remembered (much to my embarrassment—I began to feel as if I were showing off, somehow!).

All of which got me thinking that maybe there’d be an audience for a blog post or two on the interesting bits of basic maths that we were never taught at school. So here’s a couple I puzzled out for myself as a kid, possibly to be followed by more of the same if enough people find it interesting. I’ve tried to go back to basics as much as possible, so that even those for whom maths lessons were nothing more than a chance to catch forty winks should stand a chance of following it. Those who *did* stay awake may want to skip this next bit. 🙂

*First off, the terminology.*

Imagine a circle with 360 equally spaced marks around the circumference. Now imagine a line is drawn from each mark to the centre. What you should now have in mind is a circle divided into 360 segments. The ‘width’ of each of those segments is 1 degree. (Usually written as 1°.)

What’s important to note is that a degree isn’t a measure of distance. Each segment gets wider, the further from the centre you travel. If the circle has a circumference (the distance around the edge of the circle) of 360 yards (it would, if you want to know, have a diameter of 114.57 yards, give or take a gnat’s tadger), then a person could step one yard along the circumference and have moved one degree. Another person, who’s unlucky enough to be on the edge of a circle of 360-mile circumference (114.57-mile diameter), would have to walk a mile to cover the same fraction of the circle’s edge (assuming she doesn’t cheat by cutting across in a straight line, of course, in which case she’d save a tad under ^{1}⁄_{10} of a mile). If the two circles share a common centre then they will still both have moved one degree, but it’s quite obvious that they won’t have travelled the same distance.

So when an angle is said to be *so-many* degrees, what’s being said is that if the point at which the two lines meet is considered to be the centre of a circle, then the segment (shaded in the diagram) of the circle between them covers *so-many* 360th-sized segments of the circle. A right-angle, for instance (by definition), sweeps out one quarter of a circle, so it’s ^{360}⁄_{4}, or 90°. (If you’re wondering, 360 isn’t used because of some natural law, but merely because it’s both convenient and established by long tradition. We could as easily divide the circle into, say, 500 segments, in which case a right-angle would be ^{500}⁄_{4}, or 125°. There *is* an advantage in using multiples of 60, though. 60 is divisible by 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and itself, whereas 100 is divisible only by 1, 2, 4, 5, 10, 20, 25, 50 and itself. Worth remembering, the next time you run into someone passionate about decimal timekeeping.)

For those who are interested, the minute hand of a clock sweeps out a degree every ten seconds. To walk one degree along the Earth’s equator, you’d have to travel a tad over sixty-nine statute miles. (A nautical mile, on the other hand, is defined as ^{1}⁄_{60} of an equatorial degree, so you’d need to row a hand-blistering sixty of those.)

If we’re talking about angles, we need a way of referring to them in such a way that we’re all looking at the right place. What we do, as most will remember from school maths lessons, is give every point of interest, such as the ends of lines and places where lines meet or cross, a letter to identify it. We can then say that we’re referring to line AB, or angle AEC (shaded) and we’ll all know that we’re all on the same page. AEC, for instance, is the angle formed at E by the lines AE and CE. I could even say ‘Now draw a line CB,’ and talk about angle EBC, and you’d all know what angle I meant. Or we could add a point, F, somewhere on the line ED, and talk about the angle AFE or BFD, and we’d still all know what was happening; though if many additions like that were planned, it might help to have put them in the diagram in the first place. Anyway, now you know the basics, here’s the meat of the essay…

*Opposite angles are always equal.*

Well, I was taught that they’re called opposite angles. You may know them as vertically opposite, or as vertical angles. Whatever they’re called, though, what this means is that angles AEC and BED in the diagram will be the same as each other, and that CEB and AED will be the same as each other. We all learned that at school. I don’t remember ever being taught the proof though, or—more importantly—being asked to prove it. Maybe my experience wasn’t typical (though the fact that this essay was inspired by someone else not knowing does show that I’m not alone), but what a pity that is. Going beyond maths, this would have been an easy introduction to sceptical thinking; and it really is easy.

Right, here we go. Angle AEB is, obviously, 180°, being a straight line. Therefore angles AEC and CEB have to add up to 180.

If AEC is, for instance, 20°, then CEB has to be 160°.

CED is also a straight line, and therefore 180°, and as we already know that CEB is 160°, then BED has to be 20°, and following the same reasoning, DEA has to be 160°.

So it’s really just a corollary of the obvious fact that adjacent angles on a line have to equal 180° when added together; which itself is important for our next point, that…

*The internal angles of a triangle always add up to 180°.*

I wasn’t taught the proof of this one, either, but I remember puzzling it out because it troubled me. After all, if I walk around a triangle, starting in the middle of one of the lines, I’ll, obviously, have turned through 360°, not 180°; or else I’d not arrive back at my starting point facing in the same direction I started out in. Hmm, maybe, I thought, it’s the *external* angles I need to look at. Nope, a couple of minutes disproved that one; they always add to 900°. Well, so I thought. A minute or two more got me the answer, and it turns out it *is* the exterior angles, though not by the definition I was using, though I’d eventually figured it out using my own nomenclature anyway. Time for another diagram.

Turned out that what I’d been thinking of as the exterior angle—for instance the PacMan-like grey-shaded angle at ABD—*isn’t* the exterior angle, contrary to what common sense might say. (I don’t know what it *is* called, but it ain’t the exterior angle.) The exterior angle is defined as that which is adjacent to the interior angle, if the line were extended beyond the length of the side, like the blue-shaded angle at EDA, which is adjacent to BDA. And when you stop to think of it, you’ll see that if I was walking along the line, that’s the angle I’d turn through, when making the corner. To turn through the grey-shaded angle, I’d have to have been walking backwards along AB, then have turned (the long way around) to walk forwards along BD. John Cleese might get away with it; the rest of us, I think not. Anyway, I want to avoid using Xs and Ys and such, so lets plug some numbers in.

As you can see, the interior angles do indeed add up to 180°. *(Hurriedly checks arithmetic: Yep!)* What about the exterior angles, though? Remembering that adjacent angles always add up to 180°:

- FAB=110°
- CBD=150°
- EDA=100°

Add ’em together and they do, indeed, make 360°.

All of which proves that the interior angles *can* add up to 180° while the exterior angles add to 360°—which is nice for those of us who don’t want to end up walking backwards round triangles, as well as being the problem I briefly wrestled with in my youth; doubtless while I was supposed to be doing actual homework—but doesn’t prove that they *have to* add up to 180°. For that, we need to briefly consider some parallel lines.

It’s a property of parallel lines that if they cross a third line, they both cross it at the same angle. (Indeed, if all you have is a protractor and a straight-edge, that’s a perfect way to test whether two lines are parallel; draw a line across them and measure the angles they make with it.) Since we know that opposite angles are equal, and that the parallel lines make the same angles where they meet the bisecting line, we know that angles *a, b, c* and *d* are equal (and all the angles adjacent to those labelled are equal to each other). What does this have to do with triangles, though?

Here’s a triangle drawn with the base parallel to a line which touches the apex, or, to put it another way which should ring bells; at which the two ‘upright’ lines would cross if they were extended. As with the previous example, we can now spot angles that are equal, which I’ve shaded for ease of discussion. The reds are equal to each other, as are the blues. It should be pretty obvious, also (but I’ll point it out anyway, ’cause I’m kind like that), that the three angles at the top—one red, one blue, and the internal angle at the apex (shaded black)—add up to 180°.

Feel free to applaud…

It’s not just triangles though, though I don’t intend to try to prove this. All polygons’ interior angles have a fixed number of degrees which their interior angles add up to:

- Triangle: 180°
- Quadrilateral: 360°
- Pentagon: 540°
- Hexagon: 720°
- Heptagon: 900°
- Octagon: 1080°

The rule for working it out is to count the number of sides (n), take 2 from that figure and multiply by 180°.

In short, that’s; sum of interior angles=(n-2)×180°

If the shape is regular (all sides and angles are equal) and you want to know what each individual angle is, divide the result by the number of sides, so:

interior angle=^{(n-2)×180°}⁄_{n}

*—Daz*

P.S. Came across a feature of Google calculator when I tried to figure out how much distance the person cutting across the circle would save, in my ‘introduction to degrees’. I bisected the triangle and prepared to use trig for the first time since I left school—but, hey, I knew all the angles, so it was just a case of finding a sine or cosine, followed by a bit of multiplication. As a check, I knew that the answer had to be less than 1, that being the long distance around the circumference; the chord *had* to be shorter than that. So imagine my surprise when, [i] the chord came out as seemingly longer than the curve, and [ii], using sin and cos gave different answers. Had I misremembered something as basic as sohcahtoa? Was I just going potty? Nope. Using Windows™ calculator got me a believable answer straight off, as well as providing the comforting sight of sin 0.5 being the same as cos 89.5! Turns out Google calculator assumes you mean radians, not degrees, unless you specify degrees in your query. Worth remembering, that.

**Additional:** This section added, 11 Aug 11.

Had an email asking if I could run through the trigonometry by which I worked out the length of the chord that our cheating walker could have followed, as the enquirer’s remembrance of trigonometry was fuzzy. Since I’d deleted the notepad document I made the calculation on, I had to redo it; which led to the rather embarrassing finding that I’d made a couple of errors, based on a numerical typo. I’ve corrected the figure I gave in the text for the length of the diameter in light of this, but not the distance she saved in her walk, which turns out to be even less than the ^{1}⁄_{10} of a mile that I gave.

I’d like to mention that I’m in no way trying to make myself appear some sort of maths wizard. All this is based on stuff I remember from school, and I barely scraped an O-Level. Where I seem to differ from many people I meet isn’t in my small ability at basic maths, but more in the fact that, through reading popular-science and science fiction, I’ve had the occasional refresher, which has kept what little I do know from fading through disuse.

Anyway, here’s how I worked it out (but—hopefully—correctly this time):

Let’s start with the basics, then: How do we work out the radius of the circle? Well, first we work out the diameter, then halve that to get the radius. Pretty simple:

The length of the circumference equals pi times the diameter. (I’m using pi to the number of decimal places offered by Windows calculator, which is way more accurate than needed, but I’ll be doing the same with all other numbers, then rounding off at the end, so it makes sense to do the same with pi.)

Or, in short, c=πd

To solve for d, we just divide both sides by π, so:

d=^{360}⁄_{π}

d=114.59155902616464175359630962821

Divide that by 2 to get the radius and:

r=57.2957795130823208767981548141.

So that’s the length of the two legs of the triangle in the diagram above (which, though you probably guessed this, is not meant to portray the angles accurately. A triangle sweeping out 1°, at that scale, would hardly look more than a line). What we want to know is what the length of the base (the chord that our intrepid walker would follow) is. This is where the trigonometry comes in.

From here on in, though I’ll still be showing my working, I’m going to assume that anyone who’s stuck with me will have at least a vague recollection of using sines, cosines and tangents, and is more in need of a memory-jog than a complete explanation of how to find the cosine or whatever of an angle, and such. Oh, and I’m old enough that when I learned trig at school, we looked them up in books of tables like this [pdf], which also contained tables of logarithms. Ah, you youngsters today, you don’t know how easy you have it, etc etc etc…

Ahem. Where was I?

First we work out the angles of the triangle. We already know that BCA is 1°, and the other two are equal, since this is an isosceles, so they have to be ^{189}⁄_{2}=89.5°. Then we bisect the triangle with line CD, forming two right-angle triangles, so that we can apply some trigonometry to ’em.

Hands up, who remembers ‘sohcahtoa’? Our maths teacher told us it was the name of a Red Indian chief, but I’m not too convinced of that. Still, it helped get it embedded in our young minds, so he achieved his purpose. It breaks into three syllables:

Sine = Opposite over Hypotenuse.

Cosine = Adjacent over Hypotenuse.

Tangent = Opposite over Adjacent.

All of which means that, given the length of two sides, we can work out either of the two unknown angles; or, more pertinently for us, given either non-right-angle and a side, we can work out any other side. We have a choice; we can either use the sine of BCD, 0.5°, or the cosine of DBC, 89.5°. Let’s go with the former:

- sin0.5=
^{DB}⁄_{BC} - ∴ DB=sin0.5×BC
- BC=57.2957795130823208767981548141 miles
- sin0.5=0.0087265354983739349648882139735844
- ∴ DB=0.49999365382791892382807286970969 miles
- ∴ AB=0.999987307655837847656145739418 miles.

Let’s round that off to 0.99999 miles, shall we?

Or 1,759 yards, 29.6 inches.

She’d save a whopping great 6.4 inches! If she followed a chord between every degree-mark, she’d shave 64 yards off her 360-mile journey. Or she could book into in a nearby hotel, relax for a week and a half, sneak back to the starting point while no one’s looking, and announce that she’s knackered but proud…

*—Daz*

#### Comments

rustiguzzi

Interesting . . . oddly enough, I’d been sufficiently intrigued recently by an inscription at Ramsgate harbour as to its time difference in advance of Greenwich, to start thinking about degrees of longitude, nautical miles, statute miles and so on, so your little refresher course was worth reading.

I was perhaps more fortunate in my maths teachers at school, in that we did plenty of proofs, though I didn’t appreciate it at the time. I do indeed remember log tables and suchlike, never quite got the hang of them meself, and they weren’t all that accurate compared to the calculators that came in later. I’m sure I still have a slide rule somewhere . . . not particularly accurate, but at least no batteries needed.

Perhaps we shouldn’t be too surprised that taking short cuts by following the chords in that example had so little effect. A circle of 360 miles circumference would surely look, to the hypothetical walker, like a straight line. Anyway, where would we get a pair of compasses and a pencil big enough to try the experiment?

Glad it was of some use to someone!

My O-Level was actually scuppered by log-books, or rather the lack thereof. There were none available in the exam room, and the teacher lent me his calculator, which should’ve been well and good, but I’d never even seen a scientific calculator at that point in my life. I managed to get logs out of it, but couldn’t figure how to get the anti-log at the end of the calculation, and back in them days you still had to use the logs, and show your working; it was just that you were allowed to use a calculator to find em, rather than a log-book. Ah well.

I’m currently trying to remember how to work with trig and scalene triangles. Seem to remember summat about a law of cosines, but I’ll do me best to figure it out for meself before giving up and googling it. More fun that way!

Just thought of another oddity. Even though our walker would feel as if she’d walked a straight line, her left foot—assuming she walked clockwise—would have travelled slightly further than her right, which feels kind of Einsteinian…

Fisherman

all very interseting Daz…….

but where does the fishing come in?????? I was SO disappointed….

Well, the height × half the bass…

on Wednesday, September 7, 2011 at 20:29fishermannice new site daz!!!!

on Wednesday, September 7, 2011 at 22:57DazThanks Fishy.

It’s a pleasure to use, actually. All the really hard work’s done automatically.