Lesson video

Hi there and welcome to another maths lesson with me, Dr.

Saada.

In today's lesson, we will be looking at using the sine and the cosine ratio to find the length of missing sides in right angle triangles.

In order for us to do this, you will need a pen, a paper, and you will need a calculator.

So please pause the video, go grab these and when you're ready to make a start, press resume and we can begin.

I would like you to start today's lesson by having a go at this question.

Let's read it together.

Use the cards to create an equation and solve it.

How many solutions can you find? So I'm giving you three cards, number 12, six and an X and I've given you how the equation should look like.

You need to take these cards and place them in different boxes and see how many solutions can you find for X.

You will need about five minutes to complete the task.

So please pause the video and have a go at it.

Resume once you're finished.

Welcome back.

How did you get on with the try this task? Okay, really good.

What were the associations for X? Should we go through them together? Okay, let's do this.

So I started with 12 equal six out of X, and then rearranged the equation where I got X equals six out of 12 and therefore X was a half.

Did you get this as one of your solutions? Really good.

Now, would it have made any difference if I had wrote down six equal 12 out of X? So if I left X in the denominator of the fraction but swapped the numbers.

Really good, it would have, yeah.

And our answer would have been two, really good.

Now the other, so one of the methods is putting X in the denominator.

The other is putting X in the numerator.

So I did 12 equal X out of six and rearranged it and I found that the X equals 72.

Did you get that too? Really good.

And, just to show you the one that I told you about previously, X equal 12 out of six.

If I changed it and then X would have been equal to two.

Okay, would it have been a difference for this one if I swapped one of the numbers whereas if I had six equals X out of 12 select X in the numerator but swapped the numbers down? Really good, it wouldn't have made a difference because if I had six equal X out of 12 and then rearranged it, I would still end up with 12 multiplied by six.

That would give me X and I would still have 72.

Okay, really good job.

Let's move on to our next task.

Okay so as I told you earlier on today, we're going to be looking at using sine and cosine ratios to find the lengths of missing sides in triangles not just when we have 30 degrees or 60 degrees or 45 degrees, but when we have any marked angle.

So let's have a look at this first triangle.

We have a right angle triangle.

The marked angle is 20 degrees and we have been given one of the sides as 10 centimetre and that's the hypotenuse and we have to work out the length of one of the sides and that has been marked for us with that letter a.

So first step is always to label the sides of the triangle according to your marked angle.

So I can start by saying this is the hypotenuse here and this side is the opposite.

Now the second part is I need to ask myself the question, which ratio connects both the hypotenuse and the opposite? Which one relates to both of those two sides? It's the sine ratio, well done.

So I can start by writing sine of theta is equal to the opposite over hypotenuse.

Now, the third step is for us to actually substitute some of the numbers in.

So we know what theta is.

We know what the angle is.

We know it's 20.

We don't know what the opposite is.

We can write it as a because that's what the question tells us, has been marked with the letter a.

We know the length of the hypotenuse and that's 10.

So let's substitute some of the numbers.

So sine 20 is equal to a out of 10.

Now we need to rearrange the formula because we want to find a.

So if I multiply both sides of the equation by 10, that should give me 10 multiplied by sine 20 equal to N.

And now if I grab a calculator and type this into the calculator, I should get an answer.

And that is 3.

4 to one decimal place.

So a is 3.

4 centimetres and I managed to find out the length of one of the missing sides using the sine ratio.

So the first step was? Really good, we mark the sides, we label them.

The second step is? Really good, we identify which ratio we want to use.

Is it sine? Is it cosine? Step three.

We write that ratio down.

So we write sine theta is equal to opposite out of hypotenuse and then, what do we do we do after that? Really good, we substitute the numbers that have been given to us in the equation and then we solve.

Really good, we solve to find that missing lengths.

Let's look at the second one.

Second to try again.

Again, we have the right angle triangle.

One of the angles is 50 degrees, one of the sides is eight centimetres, and one of the sides is given as b.

So first step, we start by labelling.

This is the hypotenuse and this is here, the opposite.

Which ratio connects the opposite to the hypotenuse? Really good, it's sine.

So we write sin theta is equal to opposite over hypotenuse, out of hypotenuse.

It's always like this.

Now, what do we do next? Have a little think.

Really good job, we substitute.

So we say sine of 50 is equal to eight out of b and next what do we do? Have a little think.

Excellent, so we rearrange now.

So we know that b is equal to eight divide by sine 50.

And our last step is to obviously use the calculator, put these numbers in and see what b is.

So b is equal to 10.

4 to one decimal place, centimetres of course.

Okay, now I'm just going to give you a little hint.

If you're using your calculator, be really, really careful with the sine function, okay? So if you are writing eight out of sine 50, you're dividing it.

Either have it as a fraction to make sure that you are dividing, the division is done properly, or make sure that you're including your bracket and closing your brackets properly.

For the first one, that pink one, 10 multiplied by sine 20, when you multiply using your calculator, make sure that you're writing 10 multiplied by and then open brackets to have the sine 20 and then close those brackets.

Do not write sine 20 multiplied by 10 because then you are, if you don't close your brackets properly and use them properly, you will end up with the wrong answer.

Okay, let's have a look at couple more examples.

Okay and our example number three, we have a right angle triangle.

We have one of the side lengths, 19 centimetres, and the other one is a centimetre.

We need to find out the length of that a.

The marked angle is 36 degrees.

Step number one, what do we do? Have a little think.

Say it to the screen.

Excellent job.

We start by labelling the sides of the triangle.

So here I have adjacent and here I have the hypotenuse.

Okay, step two, what do we do? Have a little think.

Excellent.

So we start thinking about what ratio connects those two sides that we have labelled because we didn't label the opposite.

We've not been asked to calculate the opposite and we don't need to use it.

It's not given to us.

So we're not working with the opposite.

So we need to think about the ratio that connects the adjacent to the hypotenuse and that is the cosine ratio, well done.

So we need to write down that.

So we write down cos theta is equal to adjacent over the hypotenuse.

Now, first step, what do we do? Excellent, we need to substitute some of these numbers.

We know what the angle is so we can write now cos of 36 is equal to a out of 19.

And what do we do next? Have a little think.

Say it to the screen.

Really good, we need to multiply both sides of the equation by 19.

So we write 19 multiplied by cos 36 is equal to a and now we use a calculator to find out what a is.

We make sure that we write this properly on the calculator display.

Use brackets if needed.

And therefore, a is equal to 15.

4 centimetres to one decimal place, really good job.

Let's have a look at our last example.

We have another triangle.

Step one, what do we do? Really good, excellent.

We start by labelling the sides.

We've got our adjacent in here and we've got the hypotenuse given to us.

We need to find out the hypotenuse, adjacent is given.

Sorry.

Okay, what do we do next? Let's have a little think.

Which ratio connects the adjacent to the hypotenuse? Really good, it's the cosine ratio so we write cos of theta is equal to adjacent divided by hypotenuse and now what do we do next? Say it to the screen.

Good job, we substitute.

So cos of 20 is equal to 5.

5 out of r and let's rearrange this equation.

What do you get if you rearrange it? Have a little think.

Write it down if you need to.

Really good.

So, r is equal to 5.

5 divided by cos 20.

And now, if we use a calculator and put this into the calculator, do you want to have a go at this yourself? Come on, grab your calculator.

You can do it.

Excellent! If you have r is equal to 5.

9 to one decimal place and that is the length of the hypotenuse and this side.

Again, so you always have to start by marking and labelling the sides of the triangle.

Then think about the ratio.

Which one are you using? Sine or cosine depending if you have the opposite and the hypotenuse, you're using sine.

If you have the adjacent and the hypotenuse, you're using cosine.

Write down that ratio and then substitute and solve.

Make sure you're using your calculator properly Okay, I'm just going to advise you here before we move on to the independent task.

If you feel that you need to pause the video at any point to copy these examples, please do so.

And now it's time for you to have a go at the independent task.

Let's read it together.

Calculate the length of each hypotenuse.

Compare your answers and write down what do you notice.

What happened next? Make a prediction.

This is why I want you to do it.

You have a grid there with nine triangles in there.

The first triangle has an angle marked with 15 degrees.

One of the side lengths is given as five centimetres and you need to find a.

I want you to do that and then it's up to you.

You decide whether you want to go across the row or down the column to find a in the next to triangle.

While you're doing, I want some mathematical thinking about what do I think is going to happen next? Do I think a is going to be bigger? Is it going to be smaller? And why? How is this triangle connected to the previous one? And how is this one connected to the next one? So I want you to be asking yourself all of those questions while you are doing this task on your own, okay? You will need to use a calculator.

It will make this task a lot easier for you.

If you're feeling super confident about this, please pause the video now and have a go at it.

If not, I will be giving you a hint in three, in two, in one.

Okay, if you need a hint, this is my hint.

If you look at the first triangle, we'll always start by labelling the sides, really good.

So this here is hypotenuse and this here is the opposite, really good.

Now, which ratio connects the opposite to the hypotenuse? It's, the sine ratio so I know I'm going to use sine.

So sine 15 is equal to the opposite which is five out of a.

Now let's rearrange.

Use a calculator and you'll be able to find what a is.

Now with this hint, you should be able to make a start on your own.

Please pause the video and complete the independent task.

This should take you about 15 minutes to complete.

Resume the video once you're finished.

Off you go.

Welcome back.

How did you get on with this task? Did you manage to find the hypotenuse for the nine triangles that I gave you? Really good.

What did you notice about the nine triangles? Excellent.

So in all the nine triangles, the hypotenuse was missing.

All of them required you to use the sine ratio because in each triangle here, I gave you the angle, the marked angle and I gave you the opposite side and I asked you to find the hypotenuse so that means we have to use the sine ratio.

Anything else? Really good.

If you've noticed that for the second column, we didn't even need to use a calculator.

We should have been able to look at it and say, okay I have a 90, I have a right angle triangle, I have a 30 degree angle.

Therefore the hypotenuse must be double the opposite.

So for the first one in the middle column, we knew that the a is 10 and the one underneath it, a is 12, double the six.

And the last one, a is 14 centimetres, double the seven.

Really good.

What did you notice in each column? What was happening as I went down each column? Okay, really good.

In each column, my answers were getting bigger.

The reason for this is that the opposite was getting bigger.

I was keeping the angle the same but the opposite side was getting bigger.

And if the opposite side gets bigger, then the hypotenuse would obviously get bigger to maintain that ratio between the opposite and the hypotenuse, really good.

What about in each role? What was happening? Excellent job.

So in each role, the hypotenuse was getting shorter.

And why do you think this was happening? Really good.

So, in each row the angle was getting bigger and if the angle was getting bigger then that ratio is getting obviously bigger.

So the opposite divided by the hypotenuse is bigger.

Now to get a bigger opposite divided by the hypotenuse, we have to divide by a smaller number then.

So we end up dividing by a smaller hypotenuse.

So the hypotenuse gets a bit smaller.

If you managed to find the hypotenuse in all those sides and if you managed to write some sentences about the observations across the rows and down the columns, then you should be super proud of yourself, well done.

And for the explore task for today's lesson, you have five statements.

I want you to read each statement carefully.

I want you to decide if these statements are sometimes true, never true or always true.

You would find it really, really helpful to refer back to the table from the independent task with the nine triangles because some of the statements here describe some patterns that you had in the table.

So please do refer back to it.

If you find it helpful, get a calculator, do some calculation, put in some numbers and test each statement to see whether it always works or it doesn't.

It's time for you now to have a go at the explore task.

Please pause the video and complete it.

Resume once you're finished.

Welcome back.

How did he get on with this task? Did you find it helpful to go back to that table? Really good, let's mark this together.

The first statement, the hypotenuse is longer than the opposite.

What did you write down? Really good, it's always true.

The hypotenuse is the longest side in a right angle triangle so it's always true.

Second one, the opposite divided by the hypotenuse is equal to one.

What did you write down? Was a tricky one, wasn't it? Yeah, for this one we're going to write down sometimes because if the marked angle is 90, so if we are thinking about 90 degree, then the opposite divided by the hypotenuse can be one.

And the reason for that is that sine 90 is equal to one.

You can check that by grabbing your calculator and type sine 90 equal.

You'll get one.

If it's one, it means that the opposite divided by the hypotenuse must be one.

If you would like to find out more information about this particular statement, you can research sine of 90 and sine of zero and there will be lots of interesting facts and videos that you can find on the internet.

Next one, increasing the angle decreases the length of the hypotenuse.

We've just done that in the table and we know that that's always true.

Next one, increasing the length of the opposite increases the length of the hypotenuse.

Again, this was one of the patterns that we observed in our table when we increased the length of the opposite side what was happening and the hypotenuse was increasing to maintain that ratio.

And next one, the opposite divided by the hypotenuse is less than one.

Okay, and this is also always true.

The opposite divided by the hypotenuse is always less than one.

If we have the hypotenuse is always bigger than the opposite.

So if we imagine that the opposite is X, then the hypotenuse is going to be X plus something even if that thing is very, very small number, okay? And that would give us less than one.

Again, unless we are talking about sine 90.

This brings us to the end of today's lesson.

You have done some fantastic learning and you should be super proud of yourself.

Please do not forget to complete the exit quiz.

This is it for me for today.

Enjoy the rest of your learning and I'll see you next lesson.

Bye!.