Lesson video

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

Saada.

In today's lesson, we'll be looking at using the inverse of sine and cosine functions, in order to find the missing angles in right-angled triangles.

For this lesson, you will need a pen, a paper and a calculator.

So, if you do not have these handy, please pause the video, go grab them, when you're ready, resume the video and we can make a start.

To start today's lesson, I would like you to have a go at this.

Write the size of the marked angles in these right-angled triangles.

In order for you to do this, you need to remember what we have done in our previous lessons, where we looked at the relationships between the hypotenuse and the opposite, and the hypotenuse and adjacent sides of a right-angled triangle.

If you're feeling confident, please pause the video now and have a go at this, if not, don't worry, I'll give you a hint in three, two, one.

Okay, so my hint for you is to start by labelling the sides of the triangle, or looking at what information has being given to us.

In this case, the first one we have been given that the hypotenuse is 20 centimetres, and the side that is opposite the marked angle A is 10 centimetres.

The side opposite is half of the hypotenuse, and therefore, that angle must be 30 degrees, excellent.

Now, try and think about those relationships between the two sides that are given to you, think about, is the relationship half, is it 0.

57? What does that mean, what is the angle? Now, with this hint, you should be able to make the start.

The first task should take you about eight minutes.

Please pause the video, and complete it, resume once you're finished.

Welcome back, how did you find the first task? Really good.

Let's go through the answers together so you can mark and correct your work.

So, we've already done the first one and we know that the angle in the first one, a is equal to 30 degrees.

What about the second one? What did you write down? Really good.

So, we look at the second one, adjacent is half the hypotenuse, therefore, the angle must be 60 degrees.

So, m is equal to 60, well done.

All right, in the next one, what did you write down for b? Really good.

The adjacent again is half the hypotenuse, therefore, the angle b is 60 degrees.

And the one after that, what did you write down? Really good.

If we look at this, the relationship is not half this time, the adjacent is actually 0.

87 of the hypotenuse.

Therefore, s must be 30 degrees, and the following one, given, must have been the easiest, wasn't that? It's a right-angled triangle, it's an isosceles triangle.

We know that because the two sides are equal.

We have the adjacent and the opposite sides, are both 80 centimetres, therefore, each angle must be 45 degrees, really good.

And the last one, when I gave you a side for one of the sides, did you find that a little trickier, okay? So, with this one, we have not been given the hypotenuse.

We've not been given the length of that side.

So, I've named it x.

We have been given the adjacent and the opposite sides, so far, we have not covered yet the ratio between the adjacent and opposite sides.

So, we cannot use that.

There is a ratio, but we've not done it yet.

So, we're going to use what we already know.

And the ratios that we know are sine and the cosine, and they involve the hypotenuse.

So, we don't know it now, can we find x? What do we need to use to find x.

Pretty good, I need to use Pythagoras to find x, so, I can say that one squared plus the square root of three, squared, is equal to x squared, or one squared is one, and square root of three squared.

Then the square and the square roots cancel each other out.

It's like multiplying by two and then half, and we're doing the inverse operation.

So, it gives us one plus three equals x squared, really good, which tells us that x squared is equal to four and therefore, x must be two.

So, now I know that this side here, x equal to two, I can start looking at relationship between the hypotenuse and one of the other sides.

In this case, it's really easy to look at the opposite side.

The opposite side is one, therefore, I can say that r is 30, cause the opposite is half of the hypotenuse.

Did you get all of this correct? Good job, well done.

Let's move on to our connect task.

Now, in order for us to be able to use the inverse function of sine and cosine using our calculators, we need to know where these buttons are.

So, I'm just going to go through, how to use the calculator.

So, you need your calculator in your hand, so you can press the same buttons that I'm pressing here.

Do you have it in your hand? You can really get.

So, if you've got your calculator, this is my calculator, I'd like you to start by pressing on this shift button here, even if you have a slightly different calculator, you will still have a shift button there, okay? It would be written shift in yellow, okay.

Once you've pressed that, I want you to go to sine and press down.

And what this means is that you're pressing the inverse function of sine.

You told your calculator to do that because you've pressed the shift, and can you see in yellow, it's written a sine, and what looks like a minus one or negative one next to it.

That's the inverse function of sine.

Have you pressed it? Excellent.

Now, you should see something that looks like this.

So, you were telling your calculator now that you want to work out the inverse function of sine, but for book value, we're going to have a go at this, write down 0.

8 and close the brackets.

So, type in 0.

8, close brackets, and then press enter or equal for me, please.

You should get something that looks like this, a number 53.

130 and so on, okay.

So, what I want to show you from this task is, if you want to press or to find the inverse function of sine, you need to go to shift, sine and write down the value, close the bracket.

If you have to do it for cosine, you need to do exactly the same thing apart from, instead of clicking or pressing the same button, you need to press the cosine button, which is right next to it.

So, then you go shift, cosine, you write down number that you want, you close the bracket, then you click equal.

And we will need to use this in our next task.

And now, the main part of today's lesson as well, we offered to try and find that missing angles in right-angled triangles, using the sine and the cosine ratios that we've been looking at.

So, let's look at the first triangle.

We have a right-angled triangle.

We have an angle marked a, we have two sidelines that have been given to us, one of them is three centimetres and one of them is four centimetres.

And we want to find out what the angle is.

So, step number one is always for us to mark the triangle, to mark the sides.

This is the hypotenuse, it's the longest side.

It's opposite the right angle.

And this side here the three centimetre is the opposite.

Now, I need to ask myself the question, which ratio connect or relate the opposite to the hypotenuse.

Excellent, it's the same ratio.

So, I start by writing that down.

So, sine theta is equal to the opposite divided by the hypotenuse.

Now, out of this equation, what do I know? Do I know theta? Do I know the angle? No, it's what I'm trying to work out.

I know that in the question it's being given the letter a, so I can write sine a equal.

Do we know what the opposite is? Yes, we do, it's three centimetres.

Do we know what the hypotenuse is? Yeah, it's been given to us as four.

So, now I can substitute the values into that equation, so I can write sine of a is equal to three quarters.

Now, I want to find a, so I need to find, I need to use the inverse function for sine.

And this is what I write down, okay? I write down the inverse function of three quarters is equal to a.

I don't want you to get confused or think about this negative one as being anything to do with powers or anything like that.

It's just the way we write down inverse function of sine.

So, I know that sine of a, is three quarters, the inverse of sine, and I want to find the inverse of sine three quarters, so I can get the a.

And at this point you could have your calculator.

So, I want you to grab yours, I'm grabbing mine.

I'm holding mine in my hand, I want you to make sure first it's on.

Then I want you to go shift, sine, and then I want you to write, and you can write 0.

75, or you can write it as a fraction.

I prefer to use fractions with this, when things get complicated, I may not know how to divide it, or round off, if I have a recurring decimal, I don't want this.

So, I'm going to have a fraction here, and then I'm going to write three in the numerator and four in the denominator.

And I'm going to close the brackets carefully and it's going to look like this, okay? I'm going to press enter, equal, did you do that? Okay, so now we should have the same answers.

Mine is showing this, okay? So I'm going to write down, that a is 48.

6 degrees to one decimal place.

That's how I got at the second one.

Again, I have a similar triangle.

Again, I have a three and a four, or what's different this time, where is the four? It's not the longest side anymore, is it? Okay, so we start by doing the same thing as our first step.

What's the first step? Come on, say it to the screen.

Really good, we start by labelling our triangle, the sides of the triangle.

So, if I start by labelling this, this is the opposite.

And I'm given there, adjacent.

At the moment, I don't know what ratio connects the opposite to the adjacent.

There is one, but we're not going to cover it in year nine.

Okay, so what are we going to do? We need to find that longest side.

So, we need to find the hypotenuse and the way for us to find it is using Pythagoras.

I'm going to call that side y for now.

Let's use Pythogras to find it.

So, what does Pythagoras say? The square of the two shorter sides, the sum of the square of the two shorter sides if you add them together, it gives you the square of that longer side.

So, I can say that four squared plus three squared is equal to y squared, that is my first step in the written method.

Now, I want to find y squared, so I need to get my calculator right now, four squared plus the three squared, or you should know that it's 16 plus nine and that's 25.

Now, 25 gives us y squared, so we want y and therefore, y must be the square root of 25, which is five, really good.

So, now I know what the hypotenuse is, I can use any ratio really.

I know the three sides, so I can use sine or cosine, it's entirely up to me.

I'm going to go with sine and I'm going to say that sine of angle m is going to be the opposite out of the hypotenuse.

Now, I know what the opposite is, it's four.

So, I'm going to put four in there, I know what the hypotenuse is, I've just worked it out, it's five, so I'm going to write down, sine m equals four out of five.

I want to find m, so I need to do the inverse for sine.

So, again, I'm going to grab my calculator, I'm going to go shift sine, and I should get this, and then I'm going to have a bracket and I'm going to write four fifths as a fraction, and I'm going to close the bracket and we should have something that looks like this in your calculator display.

Do you have that? Okay, I'm going to give you a couple of minutes to make sure you get this, and then I want you to press equals sign and tell me what your answer is.

Okay, what did you get? Really good.

So, I got 53, it's almost equal to 53.

1 to one decimal place, obviously degrees.

If you had this, really good, it means that you're now confident using your calculator for this.

So, I want you to see it now.

What's the difference between the first question and the second question.

The first question, we were given already the hypotenuse, and we were able to use the sine ratio to find the missing angle without any issues.

With the second one, because we couldn't find, we couldn't use the sine, we couldn't use the cosine ratio.

So, we needed to use Pythagoras first.

And that is okay for us to combine different skills.

We've learned about Pythagoras earlier on in the year.

So, we can use that to help us answer questions about right-angled triangle, getting both sine and cosine ratios.

Now, let's look at the further triangle and see what's the same and what's different about it.

Again, we have a three and a four.

We have the four as the longest side, the hypotenuse.

We've got the three as one of the shorter sides, is it the opposite, is it the same as the first one? Look at the first one and look at that last one.

What do you notice? Really good.

So, the first one, the three was the opposite to the marked angle, whereas this one, it's the adjacent to the marked angle.

So, obviously the numbers are going to be maybe slightly different, maybe the same, let's have a look, let's find out.

So, we'll start first by, what's step one? Really good.

We always start by labelling the sides of this triangle.

So, this is the adjacent, and this side here is the hypotenuse.

Now, what ratio connects these two sides? Is it sine, again? No, it's not, excellent.

It's the cosine ratio.

So, I can start by writing that cosine of theta is equal to adjacent divided by hypotenuse.

Now, what is the second step in my written method? What do I need to write next? What would be the next line? Have a little think, you can write it down if you want.

Really good.

So, I can now write down that cosine of theta, but instead of theta, I can write down b if I want to, equal, adjacent is three and hypotenuse is four.

So, I write equals three quarters.

So, cos of b equal to three quarters.

Now, I want to find b, what do I need to do? Have a little think.

Really good.

So, I need to do the inverse of the cosine function, and I do that by pressing what looks like this onto the calculator.

So, go on, grab your calculator and press this.

Let's see if we have the same answer.

How would you press this? What do you need to press first? Good, you'll start with shift then, really good, shift and then the cosine button, and then the three quarters, and then you'd close the bracket.

Really good, you're doing so well.

And therefore, okay.

If you had the same answer as me, then you have done really, really well.

Now, I just want to point something out to you before we move on to the independent task.

If you need to copy down the examples, those three examples that we've gone through, you can pause the video now and write them down, or even make some notes or some hints to help you with the independent task.

Now, it is time for you to have a go at the independent task.

Question number one says, calculate the length of the missing angles in these triangles.

Write your answer, correct to one decimal place.

You will need to refer back to the examples that we've done earlier, and you will need to use your calculator for this.

For question two, it's a little bit more challenging, but it's on the same lines, you need to find some missing sides and some missing angles.

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

If not, I'll be giving a hint in three, two and one.

Okay, so the hint is always remember step one, we have to label the sides of the triangle accurately and correctly.

So, if I look at the first one and we always mark them and we always label them according to the marked angle.

The marked angle is b.

So, I can say, well, if I look at b, this side here is the adjacent, and this is the hypotenuse.

Now, which ratio, and it's identifying that ratio is the most important part here.

Once you have identified the sides, you use that time to decide which ratio you're going to use.

Which ratio connects the adjacent to the hypotenuse or relate to both of them to one another, and that is, it's either the sine or cosine, which one? Really good.

So, if we all have the adjacent from the hypotenuse, we have to use the cosine ratio.

If we have the opposite and the hypotenuse, we have to use the sine ratio.

So, in this case, we're going to use the cosine ratio.

So, you can start by writing it down.

Cosine of theta is equal to adjacent out of hypotenuse because you're using that ratio.

So, always look for that two sides that have been given or you're working with, to help you identify which ratio you're going to use.

Now, with this hint, you should be able to answer question one, please pause the video and have a go at it.

The independent task should take you about 15 minutes.

Please pause the video and complete it to the best of your ability.

Resume the video once you're finished.

Welcome back.

How did you go on with the independent task? Question one? Okay, let's go through some of the answers.

So, if you look here at the screen, I've already done the first and the third triangle.

With the third triangle, m is equal to 51.

3.

I had to write it, right at the top here, cause I couldn't fit it underneath for you, okay? So, m is 51.

3 degrees, did you get that? Really good job.

So, let's go through the second and the last one together.

If we look at the second one, we have one of the sides being four centimetres and the other is five centimetres.

So, we start by labelling the sides.

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

It's opposite the angle that has been marked as d.

Now, which ratio do we need to use for opposite and hypotenuse? Really good, we need to use sine theta is equal to opposite over hypotenuse.

Now, we know what the opposite and the hypotenuse is so we can substitute, so sine d is equal to four fifths, and now we can have a calculator and we find the inverse function of d.

We press shift on our calculator.

We write four fifths to find the inverse, and therefore, we can get d and d is equal to 53.

1 degrees to one decimal place.

If you had this then really good job.

Let's look at this triangle and start labelling it.

So, this here is the opposite, and here is the adjacent.

Now, we need to find the hypotenuse here and I'm going to call it y.

And the reason I want to find it, because I don't have a ratio at the moment that connects or relates the opposite to the adjacent, okay? The two ratios that we've learned are either hypotenuse and opposite or hypotenuse and adjacent.

So, I would use Pythagoras to start with.

So, y squared is equal to five squared plus six squared, and therefore, y is equal to square root of 61.

Did you get that? Good job.

Now that I know what the hypotenuse is, I can use either the sine or the cosine ratios.

I'm going to use the cosine for a change.

So, I'm going to say cos of theta is equal to adjacent divided by the hypotenuse.

I know what the adjacent is, it's five.

I know what the hypotenuse is, I've just worked it out.

So, I need to substitute these numbers.

So, we're just saying, cos of n is equal to five out of square root of 61.

And I'm keeping my numbers inside, I don't need to change them, okay? It's more accurate like this.

Otherwise, if I write them as decimal rounded off, at an early stage and then wrong again at the second stage or last stage of my calculation, my final answer won't be accurate and will be quite different from what everyone else's answer is going to be.

So, I'll try leave it like this.

Now, to find n, I need to do the inverse of cosine function.

So, I need to type this into the calculator to find n, and therefore, n should be 50.

2 degrees to one decimal place.

Did you get that too? Really good, well done.

Let's move on to question two.

Okay, and question two, let's read it together.

ABC is an isosceles triangle, calculate the following: height of AM, angle CAM, length of CM and the area of ABC.

Now, ABC is an isosceles triangle.

What does that mean? What do you know about an isosceles triangle? Really good.

Plus two equal sides and the base angles are equal.

So, let's just start by finding the side AM.

Now, if we look at the side there, and if I draw a line, I end up creating two smaller triangles, right? So, if I drew this line, I create a smaller triangle.

If I look at one of those triangles, I have a marked angle of 54, that AM is the opposite side to angle 54.

If I look now at this as one triangle here.

So, let me show you.

I look at this side here, as a triangle, this part.

Now, I have a triangle, I have a marked angle of 54, I have this side being the opposite, and this side is the hypotenuse, where 7.

4 is, this is the hypotenuse.

Which means, I can use the same ratio to find the length of the opposite side.

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

Sine 54 is equal to the opposite out of 7.

4.

Now, I can rearrange the formula.

So, 7.

4 multiplied by sine 54, is equal to the opposite side.

And that way, if I use my calculator, I can write that down and I will get the length or the height of AM, which is 5.

986 something.

So, at the moment, I'm not going to do any rounding, I'm going to leave it as this and write it down with centimetres.

You could write down that it is 5.

99 centimetre, and right down to two decimal places, but always have that number as displayed on the calculator.

So, now we found AM.

We need to do the second part, which is finding the angle CAM.

So, labelling that angle here, it's this angle here.

If I want to find this angle, it's not that tricky, is it? What do we know about interior angles in a triangle? Okay, interior angles of a triangle add up to 180 degrees.

I know in this right-triangle or in triangle ACM, I know that I have a 54 degree angle, I know that I have a 90 degree angle, so I can calculate the missing angle.

So, I can say, it's 180 minus 90 plus the 54 is equal to 36 and that is angle CAM.

So, I found that angle.

Now, find out the length of CM.

So, we want to find the length of CM.

There are so many ways of doing it, okay? One of the ways of doing it is looking at the angle CAM.

So, using that angle, if I label, draw a line from here, and I want to find that CM, that CM is the opposite in comparison, the opposite side, if I decide that my marked angle is 36 degrees.

So, if I'm using that CAM as the marked angle, CM is the opposite side.

Now, what do I know? If I have the opposite, I have the hypotenuse, I can use the sine ratio.

So, I can write down sine of theta is equal to opposite divide by the hypotenuse.

I know the angle, I know what theta is.

So, it's 36 because I'm labelling the sides in relation to that angle.

I know that I'm trying to work out the opposite, so I can leave it as opposite, I can give it a letter.

I know what the hypotenuse is, so let's do this, sine 36 is equal to the opposite divided by 7.

4.

Now, rearrange, that will give us 7.

4 multiplied by sine 36 equal the opposite.

And I've just realised here that some of my equals sign is missing, so I will just add it there.

Now, if I put, plug this into the calculator, I should get the length of the opposite, and that is 4.

3 centimetres equal the opposite.

Again, you need to really leave the number as it is in order to make sure that it doesn't affect the last answer.

So, yes, I wrote here 4.

3, but I left the full number on my calculator to use.

Now, the next part is find the area of the triangle ABC.

Now, how do we find the area of a triangle? Excellent.

We find the area by multiplying the base times the height, and we divide by two or half base times height, which is equivalent.

So now, the height is AM, the base is CB.

If we say that we are going to use half base, so half of CB would, where is half of CB? Half of CB is CM.

So, let's write this down.

The area is equal to half base times height, one half of the base is what we already know is 4.

3, multiply that by the height which we calculated earlier as 5.

986.

And that gives us the area of the triangle.

Did you get this correct? Really good, done a really good job.

And for our extra task for today, I want you to start by drawing or sketching on your piece of paper or book, a grid that looks like this, where you have four points by four points, okay? Living one gap between the points.

So, create something that looks like this, but I want you to choose the three points which will join together to make a right-angled triangle.

Calculate all the angles in your triangle, find a different triangle.

How many triangles can you find? Compare the angles inside lengths? What do you notice about that? Change the triangle to one which does not have a right angle.

Can you find the angles for that one, if it's not right angle? So, it's really your opportunity to create your own triangles and really have a go at this, have lots of mathematical thinking when you're creating those triangles and think about, what do you predict? What do you think is going to happen? What actually happens? Now, if you're feeling super confident about this, please pause the video and have a go at this now, if not, I'm going to give you a little hint in three, hint in two, and hint in one.

Okay, so you had, you've got one of the triangles that I've created.

I've used three different points and joined them together to make a triangle.

So, you need to choose any three points in the grid and make a triangle.

To start with, make right-angled triangles.

Then, find out the lengths of the sides and the angles.

So, I can see here, that this is a 90 degree angle, I know that each of these is 45 degrees because I can see that the base and the height have two by two.

So, it's an isosceles triangle, and I can use the Pythagoras to find the length of the hypotenuse and so on.

So, this is one way of starting.

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

I would like you to spend at least 10 minutes on the explore task.

So, please pause the video, resume once you're finished.

How did you go on with this explore task? How many triangles did you draw? Did you spend at least 10 minutes on this task? Really good.

So, one of the classic triangles that usually students draw, for a task like this, is this one, where the triangle has a base of one and a height of the two, and end up with of 63.

4 degrees and 26.

4 degrees for this one.

Some of you may have drawn a triangle that has a base and a height of three units, and that would have given you angles of 45 degrees, which wouldn't have needed any calculation.

You could have calculated the side, the length of the hypotenuse using Pythagoras.

Some of you may have done this, a base of two and a height of three, and that would have given you angles of 56.

3 and 33.

7 degrees.

And hopefully some of you went and managed to draw something, which is not a right.

A triangle which is not a right-angled triangle, and kind of cutting in the middle, use that, use the midpoint to try and find out.

Then let's all the sides and find out even the area from if you want to.

There so many things that you could have done with this task, I would love to see what you have done on it.

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

You've done some fantastic learning, lots of mathematical thinking, you should be super proud of yourself.

Please remember to complete the exit quiz, to show what you know.

Enjoy the rest of your learning for the day, and I will see you next lesson.

Bye.