# The Wonder and Amusement of Triangles - Part Four : The Cosine Rule

Updated on October 1, 2019 ## Introduction

The Cosine Rule in Mathematics consists of two main equations, A2 = B2 + C2 - 2BC Cos α, which is used to find one unknown length if the other two are known, and Cos α = ( B2 + C2 - A2 ) ÷ 2BC, which is a rearrangement of the previous equation, and is used in order to find a given angle, when all three sides are known. After this, we employ the Sine Rule (qv) to find the other two angles as we did before, so this is a good example of using both formulae in conjunction, where the Sine Rule can do what it can, and the Cosine Rule is used for its own characteristic abilities.

## About the Diagram

From the Triangle above, we are intending to find the length of the Side A, as it is opposite to the angle of 40°, which exist where these two lengths of three and four inches, respectively, join. Let me make it clear that any size of angle can form between two sides of four and three inches, but for this purpose I have decided upon both those lengths subtending that specific angle. When this happens, given these specific values, A is locked in, and can have only one possible value, that is, only if we have a 40° angle between sides of three and four inches, so let’s find out what it is. Note therefore that if we had sides of the same length, but a different angle, then the length of the third side would be predetermined by these three details. But let us look first at this specific example :

## Example of the Cosine Rule, by Chris Lilly on Paintbox™

But, hang on a minute, Wear ‘em out Wilf, this value refers to , not A itself, so we find the square root of this, then we’re right.

√6.615 = 2.572 , which is the correct answer.

It stands also to reason it is the shortest side, since it is opposite the smallest angle.

That done, what we can try next, is to work out what the other two angles of our triangle are, especially as we now know all three sides. Let’s look at our triangle again, this time with special reference to our missing angles, which we’ll find out.

## With all angles included, by Chris Lilly on Paintbox™

But since we do, we can rearrange this, by realising that the square of the side opposite the angle in question is subtracted from the sum of the squares of the other two sides, and all this ( in parentheses in the formula ) is divided by twice the product of the other two lengths. So, following this pattern, we can develop a formula for each of our unknown angles :

## Formulae for the Angles not yet known, by Chris Lilly on Paintbox™

We can find out the Cosines of each of the other two angles, by putting our three length values in their appropriate places, then using the Arcus Cosine Function on the calculator to work out the size of the angles. Let us just be aware, that because Side A is not an exact value, we may get some rounding error, unless we work out a way to avoid that by clever use of the calculator.

## This all leads up to the Proof

And this is all about right. The three angles I got add up to 179.999°, so that isn’t too far off. One way to be super accurate would be to put side A’s value into the calculator’s memory. We also see a Negative Cosine value, which again refers to the wave shapes both Cosine and Sine make on the axes, so that some take positive y values, and others take negative. On that graph, the angle value corresponds to x, while the Trigonometric Function value equals y.

In fact, we shall show some of this in order to prove the Cosine Rule.

The verification of the Cosine Rule comes from a variation in the Pythagorean Theorem.

## Say it is So

But say the angle is not 90º. Now for triangles it should be noted that 0º < θ < 180º, such that none of the single angles within a triangle can reach this 180º, as this number is the sum of all three angles within a Euclidean Triangle, and having even one of 180º will make a flat line, if You can get what I mean ( see illustration below ). When an angle is between 0º and 90º, its Cosine is Positive, and if between and 180º, the Cosine shall be Negative. ( See illustration below ).

## Graph of the Function y = cos(x), by Chris Lilly on Paintbox™

If now we make a triangle like the one before in the illustration entitled Pythagoras, but with the angle less than ninety degrees, causing the side opposite to be shorter than it was before, we shall see.

## What Next

+ = + k, where k is some value. This implies that = + B² - k, where k depends upon the value of θ.

If θ is, as shown above, between 0 and 90 degrees, then k is a positive value, and subtracting it from + will make less than + .

But if our angle θ is such that 90° < θ < 180°, ( noting that in these, 0°, 90° and 180° are never included ), then subtracting the resulting negative value makes > ( + ).

## About the Diagram

From the illustration above, we can see the difference in various values of , depending on what range the angle θ is taking, so this shows us visually what I mean. As the angle increases, if the other two sides remain the same, the side C opposite the angle θ is allowed to widen, thus enlarging the possible square on this side.

## Explanation

So in the illustration above, using the Diagrams of the two triangles, which are both in fact the same triangle, but with different features drawn in, along with algebraic manipulation of the Trigonometric identities shown, the Cosine Rule has in one way been proven, and the equation finally found can not only be rearranged for whatever purpose is required, but can be showed to apply to all three angles.

## By Way of Proof, by Chris Lilly on Paintbox™

Actually, looking at the values W and x that I devised to find two expressions for Cos Θ to do with the lengths B and A respectively, we find that :

## Findings, by Chris Lilly on Paintbox™

So the idea is that the three areas on the triangle combine to equal that on the right in the illustration below. This is a rearrangement of + - 2AB CosΘ = + 2AB CosΘ = +, or also , new + AW + Bx = old , and , old - new = AW + Bx.

Then : new = old - ( AW + Bx ) = - AW + - Bx = A( A - W ) + B( B - x ).

## Concerning the Equivalence Relationship

Let us look at the last statement before the previous illustration :

new = old - ( AW + Bx ) - now, where old = + , this means that new = + - ( AW + Bx ) = + - AW - Bx = - AW + - Bx = A( A - W ) + B( B - x ).

That is, new = A( A - W ) + B( B - x ).

This then is saying that the square on the side C of the diagram above will be the sum of the two, as they turn out, rectangular, areas on each of the other two sides, as shown in the diagram below.

## Concerning the Illustration above

As has been noted, new = A( A - W ) + B( B - x ), such that the two rectangular areas, each on sides of the triangle above, combine to equal the square area new on the remaining side.

If we rearrange some equations and alter the diagram above into two related ones, we can see how this is the case.

## In Closing

So there is the introduction to, and proof of, the Cosine Rule, which we can use to either find all the angles when we know all the sides, or some of the sides if we know and angle and the other two sides, and so it is. Use well.

## Disclaimer

As much as some of this Hub contains certain Mathematical knowledge accessible in the public domain, and not subject to any Copyright, other information has been drawn from textbooks which themselves are Copyright, but only in the sense of how they deliver the information which itself is shared and sometimes ancient Mathematical knowledge. Other information can be found on Wikipedia ( Copyright 2014, the Wikimedia Foundation ), which is a good source of information. Part of this is also my own discovery, but may also independently have been found by others. Reference to the comic book Character Wear 'em Out Wilf relates to the British publication Whizzer and Chips, which our family used to enjoy many decades ago, published 18 Oct. 1969 – 27 Oct. 1990 by Fleetway and IPC, and contained 1092 issues. Like me, everything he touched wore out or broke very quickly. That is, he did not wear me out by touching, but rather, if I use something, sometimes it falls to bits. But hopefully these Hubs won't.

The quote for the Title Say it is So comes of course originally from a phrase attributed to Chicago Daily News reporter Charley Owens in reference to baseball player Shoeless Joe Jackson's admission that he cheated in the 1919 World Series, where the reporter said, " Say it ain't so, Joe. " My impression was that those words might have been said by a boy fan at the time, as shown in a movie about the events. It is alleged that these events might have occurred at the behest of gambling king Arnold Rothstein ( 1882 - 1928 ) - some of this text from Wikipedia and online sources through Google.

Some of the illustrations in this particular Hub are my own, and have primarily been done using Microsoft™ Paintbox, edited from illustrations done using Microsoft™ Word. Others are, as noted, from Wikipedia. Any quote or part of this material which seems to belong to any other author should be treated as such, and I claim no ownership of anything I did not myself invent or discover, nor of any obvious copyright, trade mark, or registered trade mark.

The Adventure continues in the next Hub on the Tangent Law .

Also, feel free to check out my non Maths Hubs :

There will also be many More to come on a wide variety of Subjects.

Just take a good look at it, and note how interesting it all is, then see if you can come up with anything else along the same lines. As usual, I would appreciate any comments, feedback and suggestions which would be given due credit, or indeed have a go and publishing Your ideas Yourselves, but firstly, by all means, add Your comments - it's a free Country.

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