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Hi there, and welcome to another math lesson with me Dr.
Saada.
In today's lesson we will be looking at the sine and cosine for 30 and 60 degrees.
For today's lesson, you need a pen, a paper, you need a pencil, a ruler and a protractor.
So please pause the video, go grab these, and when you're ready, we can make a start.
For today's lesson, your first task is the following: For each triangle, which side is opposite the marked angle and which side is the hypotenuse? Please pause the video and have a go at the try this task.
It should take you roughly seven minutes to complete.
Resume the video once you're finished.
Welcome back.
How did you get on with this? Really good.
Okay, so the first one has already been done for you, let's do the third one, the green triangle.
Well you actually have two triangles there.
The first part is trying IGH.
So if I take this triangle here, and I started by marking it, I know that this is the opposite side, it's opposite to the marked angle, and now I can label that.
I can say the opposite is GI, I can say that GH is the hypotenuse, and that adjacent side must be HI.
Did you get that right? Good job.
Now for triangle GJH, just this pink one, the biggest triangle, I know that this is the opposite.
It's directly opposite the marked angle.
And this is the hypotenuse, it's opposite the right-angled and it's the longest side of the triangle, and this is the adjacent.
And if I want to label them properly, I can write this down.
So I can say that the opposite is GJ, the hypotenuse is HJ, and the adjacent side is GH.
Okay, for the pink one, again, we have two triangles.
The first one is this small triangle here.
In fact, we are one triangle, and the other one is an enlargement of it.
So for triangle MNL, this here is the marked angle so this one must be the opposite.
Now this is the hypotenuse, it's the longest and it's opposite the right-angled, and this must be the adjacent side.
Did you get that? Good job.
So for the bigger triangle, for triangle MKP, this here is the hypotenuse, it's opposite the right-angled, it's along the side in the triangle.
And this is the opposite, it's opposite the marked angle, and obviously, the adjacent is going to be KM, again.
So I can write them down now, really clearly.
Did you get this? Really good job.
Let's move on to our next task.
So in this demonstration here, I'm going to show you this right-angled triangle.
It has a marked angle of 30 degrees.
And we're going to look at, if I enlarge this triangle, what happens to the ratio between the different sides of the triangle.
So we start by looking at this one here.
It has a hypotenuse of nine, the opposite is 4.
5, and the adjacent side is 7.
79.
What's the ratio between the hypotenuse to the opposite? Well, if I divide the opposite divided by the hypotenuse, 4.
5/9 is 1/2.
So the opposite divide by the hypotenuse is 1/2.
What about if I look at the ratio of the adjacent to the hypotenuse? So if I do the adjacent, divide that by the hypotenuse 7.
79 divided by nine is roughly 0.
87 to two decimal places.
So these are two important values.
We want to check if these values are going to stay the same if we enlarge this triangle.
So I'm going to make this triangle a little bit bigger, so I'm going to go and may stop here.
We'll make the hypotenuse equal to 12.
So now the opposite is six, six divided by 12 is a 1/2.
So we still have that same relationship.
The opposite is still half the hypotenuse.
What about that adjacent and the hypotenuse? What is that adjacent, divided by the hypotenuse? So if you divide 10.
39 by 12, you still end up with 0.
87 to two decimal place.
So this is the second time we're getting the same values.
Let's make it even bigger and see if we maintain that relationship.
So let's go up to here.
Now the hypotenuse is 17.
4, the opposite is 8.
7.
8.
7 divide by the 17.
4 is still 0.
5 it's still a 1/2.
If we divide the adjacent by the hypotenuse, 15.
07, divide that by 17.
4 the answer is still 0.
87 to two decimal places.
So what we have seen here from this demonstration, is that if we have a right-angled triangle and the marked angle is 30, the opposite of that angle divided by the hypotenuse is always going to be 0.
5, the adjacent divided by the hypotenuse is always going to be 0.
87, no matter how big or small that triangle is.
And now we are going to look at a different triangle.
Again, we have a right-angled triangle, this time the marked angle is 60 degrees instead of 30 degrees.
If we look at the lengths of the sides that we have, we have the hypotenuse is six, the adjacent is three and the opposite is 5.
2.
What are the ratios between these sides? So what's the adjacent divide by the hypotenuse? Three divide by six is a half, so the adjacent is 1/2, the hypotenuse where we have a 60 degree angle.
What about the opposite and the hypotenuse, what's the relationship there? If I divide 5.
2 divide by six that gives me 0.
87 to two decimal places.
So we are going to see if I enlarge this triangle, are we going to maintain those ratios? What will happen? So I'm going to make.
I'm going to keep that angle 60 degrees, but I'm going to make the triangle a little bit bigger.
Let's look at this one here.
The hypotenuse is nine, the adjacent is 4.
5.
4.
5 divide by nine is 1/2, so we can still see that if we have a 60 degree, the adjacent divided by the hypotenuse is still 0.
5.
What about the opposite divided by the hypotenuse? 7.
79, divide that by nine, that gives me 0.
87 to two decimal places.
So even when I enlarge the triangle, because I still had a marked angle of 60 degrees, I still ended up with the same ratios.
Let's enlarge this triangle a little bit more.
And let's stop here.
Now what do I have? Adjacent divide by the hypotenuse, six divided by 12, again is a 1/2.
Opposite divided by the hypotenuse, so 10.
39 divided by 12 is still 0.
87 to two decimal places.
So we can see that this is not changing even when I am making the triangle bigger.
So let's make it even a bit bigger, and we'll stop here.
Again, adjacent divided by the hypotenuse, 7.
5 divide by two by 15 sorry, is equal to a half, excellent.
And opposite divide by the hypotenuse, so 12.
99 divide by 15, again is 0.
87.
Really good.
So what we need to understand from this demonstration is, whenever I have a right-angled triangle and the market angle is 60 degrees, the adjacent will always be half of the hypotenuse, always will be half of it, and the opposite divided by the hypotenuse, it's always going to be 0.
87 to two decimal places.
And this is really interesting because if we think about the 30 degree triangle that we looked at earlier, we can really see that the relationship is the opposite, isn't it? So in a 30 degree angle, triangle, the opposite divided by the hypotenuse is a half, whereas in a 60, is adjacent divide by the hypotenuse, it's a half.
And in a 30 degree angle, we saw that the adjacent divided by the hypotenuse is 0.
87, whereas in 60 degrees, it's the opposite divide by the hypotenuse is 0.
87.
And this is what we are going to practise in today's lesson.
And now let's use all of this information from the two demos to find the length of the unknown sides in the following triangles.
So let's make a start that first one.
I can label the site as the opposite, and this is the hypotenuse, and always start by labelling your sides.
Now what do I know, I obviously have a 60 degree angle as the marked angle, I know that sine 60 is opposite divided by the hypotenuse, and we know what sine 60 is now, right? It's 0.
87 to two decimal places, is equal to, instead of opposite, I'm going to write a and out of five instead of the hypotenuse.
So I'm going to rearrange the equation so I'm going to write down five multiplied by 0.
87 equal a, and I can do this and find out that a is 4.
35 centimetres.
So what I have done here is to start with label the triangle.
Second thing write down the formula, rearrange it to find the missing side.
Now let's look at the second one.
This is the opposite, and this side is obviously the hypotenuse.
Now I have here a 30 degree angle.
And we already know that if we have a 30 degree as the marked angle, that the opposite is half the hypotenuse.
Therefore, the hypotenuse I can say is 10 centimetres without having to do the written method.
However, I'm going to show you the written method too, just so we know how to write it down formally, okay? So for the written method, I have a marked angle of 30, I'm looking at the relationship between the opposites and the hypotenuse, I'm looking at that sine ratio.
So I can write that sine 30 is equal to opposite over hypotenuse.
Now we know sine 30 is, we know that the ratio must be a half isn't it? So I can write 0.
5 there.
The opposite we know it's five, it's been given to us and it's the hypotenuse that we're trying to find out.
So I can write b instead of this I can write now, 0.
5 equal 5/b.
Rearrange the formula, and b is equal to five divided by 0.
5, and that gives me b is equal to 10 centimetres.
So without the need for all of this written method, I should have been able to find the length of the hypotenuse just by looking at the triangle as well.
Next one, this is the hypotenuse, and I have here the adjacent side.
I have a 60 degree angle, and what do I know if I have a 60 degree angle? What's the relationship between the adjacent and the hypotenuse? Really good.
I know that the adjacent must be half the hypotenuse, so therefore, the hypotenuse must be 10 centimetres.
How do we show that using the written method? I'm looking at the relationship between the hypotenuse and the adjacent.
So I'm looking at which ratio? Really good.
We're looking at cosine ratio so am going to write cos of 60 is equal to the adjacent over the hypotenuse.
In fact, it's cos of any angle, cos of theta, but because we know that the marked angle in this case is 60, we're writing it down.
So we know that cos of 60 is, the ratio is half.
So 0.
5 equal to adjacent is five, and we don't know what the hypotenuse is.
We rearrange the equation.
So c is equal to 10 centimetre, and that's what we worked out.
Let's move on to the next one.
Start again, by marking it.
This is the hypotenuse, and we need to find out the adjacent.
We have a 60 degree angle again, what do we know? Really good.
So we know that that adjacent must be half the hypotenuse and therefore it must be three centimetres, half of six.
I can use the written method too, so I can say cos of 60 is equal to adjacent over hypotenuse.
And now I can substitute, so I can say that 0.
5 is equal to d/6 and therefore d is three centimetres.
And the last one, we start by marking it first.
This is the adjacent, this side here is the hypotenuse.
Now, which ratio relates the adjacent and the hypotenuse together? Really good.
It's cos 30 is equal to adjacent over hypotenuse.
And we know that cos 30 is 0.
87 equal to the adjacent, is 10 divided by e.
e equal to 10 divide by 0.
87, which gives me 11.
5 to one decimal.
It is time now for you to have a good independent task.
Which of these triangles have a 30 degree angle marked? Now remember that in a 30 degree angle triangle, the opposite divided by the hypotenuse is 0.
5.
The opposite is half the hypotenuse, and that adjacent divided by the hypotenuse is 0.
87 to two decimal places.
Please pause the video and have a go at the independent task.
Please use the diagrams that are given to you.
So you can annotate on them.
You can label the sides.
You can highlight things if you need to, use them.
They're not drawn into scale, but you should be able to use them to help you answer the questions.
This task should take you roughly 10 minutes.
Have a go at it and when you're finished, please come back here and resume the video.
Hi there, and welcome back.
How did you get on with this task? Okay, really good.
So the first one, the opposite was half the hypotenuse so we know that the marked angle was 30 degrees.
The second one, the opposite again was half the hypotenuse.
So we know that it was a 30 degree angle.
And the opposite being the side that is opposite the marked angle.
The third one, the opposite was not given to us we had the adjacent.
So the adjacent divided by the hypotenuse was 0.
87 and therefore we know that it must be a 30 degree angle.
Now next one, again, we were not give the opposite.
We were give the adjacent and hypotenuse.
The adjacent divided by the hypotenuse was half.
But if the adjacent is half of the hypotenuse, then the market angle is 60 degrees, not 30 degrees.
Did you get that? Really good, well done.
Next one, again, we were given the opposite, and we were given the hypotenuse.
The opposite was half the hypotenuse, because a is half of two lots of a and therefore the marked angle was 30 degrees.
Now next one, let's go through it together.
So if you look at the triangle that I gave you this time, I didn't give you the hypotenuse, I gave you the adjacent and the opposite.
So I gave you this side is the opposite and this one is the adjacent.
And the hypotenuse, you had no idea what it was, so we can label it as x.
Now, the sine and the cosine ratio is all we've been looking at.
Always give us the relationship between the opposite and the hypotenuse or the adjacent and hypotenuse.
So how would we work this out? Do we know anything that can help us find out x in this case? Have a little think.
Good, we can use Pythagoras.
And now if we use Pythagoras we can say that x squared minus 10 square.
Equal, sorry 10 squared plus 17.
3 squared.
And if we work at x it would be 20 centimetres.
In fact on the calculator is 19.
98.
So to which we rounded to the nearest whole number, it's 20 centimetres.
And now if I look at this, I know that the opposite, 10 centimetre, is half the hypotenuse which is 20 and therefore the angle must be 30 degrees.
If we look at the last triangle here, well, we don't have a right-angle there, it's not right-angled triangle so that will not work.
We cannot work out whether that angle is 30 degrees or not, okay? Because the sine and the cosine ratio apply in right-angled triangles.
Did you get that? Really good, you should be super proud of yourself.
Now it's time for you to have a go at this explore task.
Let's read it together.
Find all the right-angled triangles, which have a side length of five centimetres and an angle of 30 degrees.
Put them in order from the smallest area to the largest area.
Again you find a right-angled triangle with 30 degrees in which the perimeter has the same value as the area? Okay, you're feeling super confident at this, please pause the video now and have a go at this.
If not, don't worry, I'll give you some support.
I'll be giving you a hint in three, two, one.
So if we're going to draw a right-angled triangle, we can start by drawing this and saying, well, I want my my angle to be here.
I want the 30 degree angle.
So I have one 90 degree in this corner.
So the 30 and the 60 degrees would have to be here or one there.
Either way, it really doesn't matter.
So I can say we're going to have this side as the five centimetre and then I'm going to draw a 30 degree here.
And you can start by drawing this and measuring the sides or you can even think about sketching it and what can you do to find that the length of the sides? Can you combine the sine ratio, the cosine ratio, and Pythagoras to find the rest of the information? My second hint for you is this; you can draw another triangle by again starting it the same way and we have a right-angle in the same place.
But now instead of saying this side is the five centimetre I can say that, this side is, the vertical side is the five centimetre side.
Okay, now, I still want the 30 degree angle, so I'm going to do it in the same place.
I obviously now have no idea how long that horizontal line is, that side.
I don't know how long it's going to be, how big, how small it is, but I know I need a 30 degree there.
The 30 degree is opposite that five centimetre whereas in the first one, the 30 degree was adjacent to the five centimetre.
Now the remaining angle must be 60 degrees, right? So the last angle over there, would have to be 60 degrees.
And again I have no idea how long this side is going to be.
So you can either try to construct this triangle or you can use, again a combination of the sine, cosine and Pythagoras and try and work out the length of it.
Now, with these two hints, you should be able to have a go at this question.
Please pause the video and spend about 10 minutes exploring this task and when you're done, you can come back here and resume the video.
Off you go.
Welcome back.
How did you get on with this task? Okay, so if we have to draw a right-angled triangle and a 30 degree angle, the third angle must be 60 degrees.
And if we wanted to draw one of the sides to be five centimetres, that side could be either adjacent to the marked angle 30 opposite to it or the hypotenuse of the triangle.
We don't really have any other option, do we? So you may have started with something that looks like this.
A right-angled triangle, one the 30 degree and measured the adjacent side to be five centimetres.
You may have not necessarily measured it and actually done the calculation and worked out using the ratios and using Pythagoras to work out the length of the sides.
Another one, you may have moved the five centimetre.
Now instead of it being there adjacent side of angle 30 it's the opposite side to the angle 30.
Did you do that? Really good.
And the last one, you may have done the five centimetre as the hypotenuse and then said, "Okay, if this is going to be five, what is the opposite side going to be?" Well, it's going to be 2.
5 because it has to be half of it.
And now using Pythagoras worked out what x is, so the adjacent side, which is really five squared minus 2.
5 squared is x squared and find out what x is and in this case is 4.
3 to one decimal place.
And then you will have calculated hopefully the areas as 7.
25 centimetre square in the first one, 21.
7 centimetres square to the second and 5.
42 centimetres squared for the third one.
Again, you may have approached the question slightly differently.
And if I wanted to put them in order, I would put that last one as the first or the smallest triangle and then the second one where the adjacent is five centimetre and then the last one where the opposite is five centimetres.
Did you get that? Really good.
The second part of the explore task was for you to try and find a triangle, right-angled triangle that has an angle of 30 degrees where the perimeter and the area are equal.
How did you go on with this? Okay, now there are so many ways to go about for that question.
So I just need to go through my thoughts on this question four with you.
It was not an easy one.
It was quite challenging one it required a lot of mathematical thinking.
So I'm going to go through what I have done with you.
So I started by drawing right-angled triangle and a 30 degree angle.
And then I thought, okay, you know what, let me go for a really basic triangle.
So I'm going to make the opposite equal to one.
So if the opposite is equal to one and I have a 30 degree angle, what's the hypotenuse going to be? It's going to be two, is going to be double.
Now use Pythagoras to find what would the adjacent side be.
And that would be two squared minus one squared square root of three.
So now I have the basic triangle there with dimensions for the three sides.
And I thought, let me just give them names.
So I gave the sides name, D, E, and F, just so I can refer back to them when I'm doing the calculation.
And I said to myself, I want to find the perimeter, it's easier, I have to add the three sides up.
What about if I want to find the area? I need to multiply the base times the height and divide that by two.
Well, I want to try and make those two equals so I'm going to make some generalisation here.
So I said okay, which side is going to be the base and I choice E to be the base.
So I said, if E is x, so if I call it E equal to x instead of one, what would I call D? Well D will be x multiplied by? Root three.
Because it's one times three, it's root three.
And what would I call F? Well, the hypotenuse would be two lots of E.
If E is x, then F will be two x.
So I'm done.
Base equals x.
The height is equal to the square x multiplied by root three and hypotenuse is equal to two x.
I chose the base on the height to be E and D.
Because they meet at a 90 degree angle, okay? After that, I said, okay, well, the perimeter is easy.
It's x plus x, root three plus two x Let me simplify that.
That's three x plus x root three.
And I can even check through I got x common.
So I just took x outside the bracket, I have x bracket, three plus root three.
So this expression tells me the perimeter.
Now what about the area? Well, the area is half times x times x root three, and if I simplify this, it's half x squared root three.
Now I want that expression for the perimeter to be equal to the expression of the area so I can equate them to one another so I can find what x is.
Because once I know what x is, and I know what the base is, I can calculate the height, I can calculate the hypotenuse, I can find out the answer.
So I said, okay, let me equate them to one another.
And let me solve.
So I solved and I found that x is equal to two, root three plus two.
I'm not going to go through the steps for working it out, you should be confident enough about solving the equations.
If not, please go back and watch the videos about that.
Now that I know what x is, I know what the base is.
Remember, we said E was the base, so I know what x is, I can find out the height.
So the base is six plus two root three, that's D.
E is two root three plus two because that's what I found that it's x as it is without having to do anything to it.
And F was double what the x was.
So we found x doubled it and I found the answer.
Now I know what the three sides.
I left everything inside for in my case because I thought it's more accurate rather than having to rounding.
And now if you use a calculator to put these numbers in calculate their perimeter by adding D plus E plus F.
Or find the area by multiplying D times D, D times E, sorry, and divide by two, making sure that you're using brackets in your calculator and then divide by two after you've multiplied you will get the perimeter is equal to the area and is equal to 25.
85 centimetre squared to two decimal places.
Now, this is just my thoughts on it.
You may have done it slightly differently.
I would love to see how you have done it.
This brings us to the end of today's lesson, a huge well done on all the effort that you've put in.
Please remember to do the exit quiz to show what you know.
Take care and I'll see you next lesson.
Bye.
