Lesson video

Hello, I'm Mrs. Lashley and I'm gonna be talking you through the lesson today.

I really hope you're willing to try your best.

And even if it gets challenging, remember I'm here to support you.

So our learning outcome is to be able to derive the formula for the cosine rule.

So the cosine rule is a formula used for calculating either an unknown side length or the size of an unknown angle.

And you can see it on the screen.

a is the length of the side opposite angle A, b is the length of the side opposite angle B, and c is the length of the side opposite angle C.

And we are going to be looking at deriving that formula during this lesson.

So in this lesson on the cosine rule, the first learning cycle is going to be deriving the cosine rule.

And then as we get into the second learning cycle, we'll start applying the cosine rule to triangles.

So let's make a start at finding out where the cosine rule comes from.

So when we know all three side length of a triangle but no angles, only one possible triangle can be made.

You can explore this with three strips, as shown, that if you put a split pin in the end of each one, then they will only form one particular triangle.

You could rotate it, you could reflect it, but the triangle itself will be the same.

So they'll form a rigid structure that cannot be changed without altering the lengths.

So you may want to pause the video, make yourself three strips and explore that.

And then press play when you're ready to carry on.

So this also means that given two adjacent sides and the angle between them, we can find the length of the third side with some help from the cosine function.

So if you've got three edge lengths, it will fix the triangle.

So if you know two of the edges and the angle between those two edges, then that third length is fixed and we can calculate it.

So we're gonna have a look at a generalised triangle where our edges are a, b, and c, lowercase a, b, and c, and we've got the vertices A, B, and C.

Here we've dropped the perpendicular, the altitude from the vertex B to the edge AC.

And we are focusing on the highlighted purple right angle triangle.

We can use Pythagoras' theorem to write the relationship between the three edge lengths, and this relationship holds because it's a right angle triangle.

So the edge length DC is the expression b minus r, because if AC is b and A to D is r, then the remaining length will be the difference between the full length and that shorter length of r.

So Pythagoras' theorem would state that p squared plus b minus r all squared is equal to a squared, and that's because of the right angle triangle.

The second line of working that you can see on the screen is where we have the double brackets.

so b minus r all squared can be written as b minus r times b minus r.

And you hopefully you recall how to expand a quadratic.

So then we've got p squared plus b squared minus two br plus r squared equals a squared and that's just Pythagoras' theorem on the right hand purple right angle triangle.

But now if you think about the p squared term, how else we can write that? Well, now we're focusing on the left hand right angle triangle, which is now purple, now highlighted, and once againPythagoras' theorem holds because it's a right angle triangle.

So it's true that c squared, the hypotenuse, is equal to p squared plus r squared, and we can rearrange that to make p squared the subject.

So now p squared can be written as c squared minus r squared.

So previously we used Pythagoras' theorem on the right hand right angle triangle and we had p squared as a term.

We can now substitute this expression for p squared into this.

So we can rewrite it as c squared minus r squared plus b squared minus two br plus r squared equals a squared.

And by simplifying and collecting like terms, this is equivalent to a squared equals b squared plus c squared minus two br.

You may wish to pause the video and just check, go back through those steps that we just did and make sure you are happy with how that third line of working is equivalent to the fourth line.

Press play when you're ready to move on.

So now we want to look at this last term, the product of two b and r.

So we're gonna focus on the r.

R is that length AD, and c is the hypotheses of the right ankle triangle.

So we've used Pythagoras' theorem to get an expression for p squared, but now we're going to look to get an expression for r.

And this time we're gonna make use of the cosine function.

And that's because r is the adjacent to the angle at A.

So cosine of A is equal to the adjacent over the hypotenuse, which is r over c.

So we can rearrange that to say that r, that length AD, is equal to c times cosine of a.

Now we have an expression for r, we can substitute that into that term.

So two br can be written as two bc cosine of A.

And this final equation is called the cosine rule.

so a squared equals b squared plus c squared minus two bc cosine of A.

So look at the general triangle.

Remember our vertices are our capital letters and the opposite edge to that vertex is the lowercase.

So a squared is equal to b squared plus c squared minus two bc cosine of A.

So the subject is a squared and it's cosine of the angle A.

A quick check, which of these statements is correct? Using that general triangle once again, which of those would hold true for this triangle? Pause the video and then when you're ready to check whether you got that one correct, press play.

So a and d are correct.

C squared equals p squared plus r squared is Pythagoras' theorem using the edge length of the left hand triangle, the right angle triangle ABD, and part d, p squared equals a squared minus b minus r all squared is Pythagoras' theorem on the triangle BDC, where a is the hypotenuse.

So after the first task of the lesson and you need to reorder the proof.

So this is the proof of cosine rule, which we saw, and it's been mixed up.

So each of those is one line of the proof.

So you need to write each line in order, referencing the triangle on the screen.

So pause the video.

If you get stuck at any point, then go back and re-watch some of the video that goes through the proof.

Press play when you want to move on.

On question two, I'd like you to annotate your proof to show where the following were used.

So in the proof, the Pythagorean theorem was used at least once.

So annotate when we have made use of that.

Part b is substitution, the algebraic skill of substitution.

And finally c, where cosine of theta is equal to the adjacent over the hypotenuse.

So pause the video, go back to your question one solution, and annotate when a, b, and c have been used.

Press play when you're ready to go through the answers to this task.

This is the correct order of the proof for the cosine rule.

So we start by using Pythagoras' theorem on the triangle BDC, we've taken the quadratic out from its factorised form and we've expanded it.

Then we've used Pythagoras' theorem on ABD, substituted our expression for p squared into our previous line of working.

Then we've tidied things up, manipulated the algebra, collected like terms, then we've used the cosine ratio, rearranged that to get an expression for r, and once again substituted it in to finish with the standard form of cosine rule.

It's worth saying that there are some stages of this proof that could have been done in a different order.

It may have been that you've written p squared plus b minus r all squared equals a squared, p squared equals c squared minus r squared.

It would be absolutely fine to substitute the expression for p squared into the top line of the proof and then expand the brackets before simplifying.

Cosine of A equals r over c.

We could have written that relationship down at the very beginning of the proof, but we didn't need it until later on.

So this is the order that I went through it in the explanation, so it's what I would expect you to have also done.

But there are oftentimes, especially with algebraic manipulation, that some of the stages may have been done simultaneously and could be in a slightly different order.

But we do need to finish with that cosine rule at the bottom.

So question two was to annotate the proof.

So using Pythagoras' theorem was the first line of the proof and the third line.

Using the cosine ratio was where we wrote cosine of A equals r over c.

And substitution was when we substituted our expression of p squared and also when we substituted our expression for r.

Okay, so now we know where it comes from and we can derive the cosine formula.

Let's make it use applying the cosine formula or the cosine rule to triangles.

So we can use the cosine rule when we know two adjacent sides and the angle between them.

So if we were looking at this question, we know that this is a fixed triangle because if you've got two sides and the angle between them, that is a proof of congruency, SAS, side-angle-side, and they will fix that third edge.

But what we now can do is also calculate the length of that third edge by using the cosine rule.

So Sam said, "I hope you wrote down the rule." So you will have written that down within the proof in your Task A, but it's a rule that you will need to use regularly and get familiar with the formula.

So it's a squared equals b squared plus c squared minus two bc cosine of A.

And remember that the a and the b and the c don't necessarily have to explicitly be those letters.

So if the triangle was DEF, we would still think about them as A, B, and C in terms of the capital, the vertex label is opposite its lowercase label.

So in this case we're looking to find the length of a, and so our vertex A would be the 55 degree angle, where the 55 degree angle is located.

So we're gonna substitute our values in.

B and c, it doesn't matter which one you think is b and which one you think is c, similar to when we've used Pythagoras' theorem.

The important part is which one was the hypotenuse and the two shorter sides could be either way around, and that's exactly the same here.

Identify where your angle is and the opposite edge will be the corresponding letter.

So here we've substituted in our values.

A is what we're trying to work out.

So there's still the variable of a.

Eight squared, 10 squared, the 160 has come from doing two times eight times 10, because that's what two bc as a term means, the product of two b and c.

55 is our angle at the vertex of A.

Some people will then go on square eight and square 10, so we'd get 64 and 100, which makes 164.

Then they would incorrectly subtract 160 from 164.

Pause the video and think about why we shouldn't subtract 160 from 164.

Press play and I'll explain why that's the case.

So the reason we shouldn't is because the 160 is actually a product with cosine of 55, and 160 cosine 55 is not equal to 160.

So using our order of operations, yes, we would evaluate the squares.

So eight squared is 64, 10 squared is a hundred.

And then we can add, so 64 and a hundred would be 164, but 164 minus 160 cosine of 55 is what we actually need to do.

We'd need to evaluate the 160 cosine 55.

That is a product, it's a multiplication.

So be really mindful, especially if you're doing this without a calculator, not to subtract that constant, the multiple from the sum of the two squares.

So we can on a calculator, the calculator will obviously do this correctly.

If we were not using an calculator, we need to make sure we are doing it in the correct order.

A squared is equal to 72.

227.

I can estimate what a is going to be because I can use the square numbers that are either side.

So eight square is 64 and nine squared is 81.

So I know that my value of a is somewhere between eight and nine.

I'm gonna use the calculator to square root it, and it's 8.

50 to two decimal places.

So having got the two sides and the angle between them, we can use the cosine rule to work out the length of the third side.

So here's the check for you.

What is the value of a on this triangle to two decimal places? Pause the video and then when you're ready to check, press play.

So the answer is 9.

25 to two decimal places.

So you should have done nine squared plus 10 squared minus two times nine times 10 times cosine of 58 degrees and then square rooted it to finish with our answer of 9.

25.

So, the cosine rule can be used to find the third edge length, but also it can be used to find the angle.

So if we have all three edges of the triangle, we can calculate any of the angles.

In this particular one, we are looking to work out the angle of theta, which is opposite the 13.

And this is where it's really important to make sure that you are familiar with the idea of the corresponding angle and edge.

so a squared equals b square plus c squared minus two bc cosine of A.

Well, 13 is therefore our lowercase a because our theta is opposite it.

So our angle A in using the labelling of the cosine rule is our theta.

So we substitute in the parts we have, the values we have, which is the a, the b, and the c.

So 13 squared.

Once again we know that a, lowercase a is 13 because it is opposite the angle we're trying to calculate.

And then b and c can be either way around, so b is nine in this case, c is 10.

Two times nine times 10 is 180.

So 13 squared equals nine squared plus 10 squared minus 180 cosine of theta.

We can then rearrange this to make cosine of theta the subject.

And so just pause the video and think about how the second line where we've substituted the values has been rearranged to become the third line.

Press play when you're happy with that rearrangement.

Okay, so how would we have done this? Well, we could have added 180 cosine theta to both sides and then we can subtract the 13 squared, and then divide by the 180.

Cosine theta is now the subject.

So now I need you to think back to how do we find an angle in trigonometry? What do we use when we want to go from cosine theta to only theta? Pause the video.

Think about that.

You've met it before with right angle trigonometry.

How do we get from cosine theta to theta? Press play when you think you've got the answer.

So we'd use the inverse of cosine, arccos or arccosine.

And we've recognised that notation hopefully.

So the answer for this angle is 86.

18 degrees to two decimal places.

So the numerator, nine squared plus 10 squared minus 13 squared and over 180 simplified down to 1/15, so arccosine of 1/15.

So I'm gonna go for another one where we're finding the angle and then there'll be one for you to try by yourself.

So here I've got a triangle, I've got all three edges, and I'm looking to find the angle opposite the 23, which means that if I'm using cosine rule, which is a squared equals b square plus c squared minus two bc cosine of A, then a is 23.

So I can substitute 23 in that position of a, 13 and 19 into the positions of b and c.

And then finally, angle A is gonna be theta.

So I can rearrange it and evaluate it using arccosine to get 89.

88 degrees to two decimal places.

So again, that stage from the first line of work to the second line of work is a rearrangement.

So I've added 494 cosine theta.

Remember, that is a term, it's a product that has to stay together at this point.

Subtract the 23 squared, then divide by the 494.

So here's one for you.

So pause the video, you're gonna use your calculator and round your answer to two decimal places.

Press play when you're ready to check.

So your a was 22, your b and c either way round, it doesn't matter, rearrange it and you would've got cosine of theta is equal to 1/17.

So to get the value of theta, we're gonna use arccosine on that fraction, and it's 86.

63 degrees to two decimal places.

So here is the second task of the lesson, and there's two questions, they're both on the screen.

So question one is finding the angle A, which is opposite side a to two decimal places for the following triangle measures.

And question two is find the side b opposite angle B to one decimal place for the following triangle measures.

And you can see the general triangle on the screen.

So there's eight questions effectively for you to do.

So pause the video, work through question one and question two.

And when you press play, we'll go through the answers.

So question one was all about finding the angle.

So you were going to need to rearrange and use arccosine to get that.

So question one, part a was 28.

96 degrees, b was 53.

13 degrees, c was 41.

41 degrees, and d was 38.

94 degrees.

Question two, you are working out on edge length.

This time it was for the side b.

So really, you're gonna think of the cosine rule as b squared equals a squared plus c squared minus two ac cosine B, because you needed the angle B, and therefore the subject would be b squared, or you could just relabel it to make them a and A, if that makes your life easier.

So the edge length b would be 5.

3 centimetres on triangle a, 5.

9 centimetres on triangle b, 2.

5 centimetres on triangle c, and 6.

1 centimetres on triangle d.

So this lesson today was all about the cosine rule and there are multiple ways to prove the cosine rule, and we looked at a particular way.

When deriving or proving it, it's sometimes useful to derive other information.

And if you remember, we used Pythagoras' theorem and the cosine ratio and a lot of algebraic manipulation by substitution and rearrangement.

Really well done today and I look forward to working with you again in the future.