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Hello, I'm Mrs. Lashley and I'm gonna be working with you as we go through the lesson today.

So our lesson today is working with trigonometry to solve problems so we can have a variety of problems where trigonometry can be used in order to help us with them.

This lesson is based around trigonometry and therefore the trigonometric ratios.

So if you want to remind yourself of what they are, they're on the screen now.

You can pause the video and then when you're ready to move on with the lesson, press play.

So our lesson's got two learning cycles and we're gonna break it into two parts.

The first one is about calculating areas and the second is about calculating angles, all using our trigonometry to help.

So let's make a start with the calculating of areas So you'll know from the previous learning that the area of a triangle can be calculated using the base and the perpendicular height.

Isosceles triangle, the base edge could be eight centimetres and the perpendicular height is unknown, so it will need to be calculated.

So if we use this half of the isosceles triangle, and we know it's half because that is the altitude which is dissecting that base edge because it's also the line of symmetry, then we can calculate this perpendicular height, which we've labelled as H.

H is the opposite to the 80 degree angle, and four centimetres is the adjacent, so this is going to be using the trigonometric ratio of tan.

So to find H, it would be four times tan of 80 degrees, and you can do that on your calculator.

This is 22.

685 and there are some further decimals.

So we're going to continue and use that in its most accurate form by keeping that as an answer on your calculator.

So going back to finding the area of this isosceles triangle, we know it's half times base times perpendicular height, that's the formula, and so the base is eight, we just calculated that perpendicular height, so if we substitute that in, the area comes out as 90.

7 square centimetres to one decimal place.

So a check.

First of all, calculate the perpendicular height of this isosceles triangle.

Press pause whilst you're doing that, make sure your calculator is in degree mode and then when you're ready to check your answer, press play.

So the perpendicular height would be 19.

63 to two decimal places, but hopefully you've kept that on your calculator screen because now I'd like you to calculate the area of the triangle.

So pause the video again whilst you're calculating the area of the triangle, and then when you're ready to check that you've got that correct, press play.

So substituting height and using the base edge of 12, then the area is 117.

8 square centimetres to one decimal place.

So this isn't a is isosceles triangle, or we don't know that for certain, so to calculate the area of the triangle, we still need to know the perpendicular height and Pythagoras' theorem and the trigonometric ratios require right angled triangles.

Currently, there are no right angled triangles in that diagram.

So where could a right angled triangle be created? You may wanna pause the video whilst you consider where that could be.

It's a place where we could create a right angled triangle and that's why extending the base edge, that edge of nine centimetres and dropping a perpendicular line to it, and that would therefore meet at a 90 degree angle.

And so you can see below it where the right angled triangle and what we know about that right angled triangle is being put on.

So we have the hypotenuse of 16 centimetres, that was one of the given edges, that angle of 35 degrees, that was within the first triangle, and now we have this edge length which is the opposite to the 35 degree angle we don't have the value of, and the area for the triangle would be half times base times perpendicular height, that perpendicular height would be the length of A.

But where else could you put a right angled triangle? Again, you might wanna pause, perhaps you didn't get the one I got previously and you already know where the next one could be.

So the other place we could do is draw an altitude, and an altitude is always perpendicular to the edge that it meets if it meets an edge within the triangle.

And so again, you can see the right angled triangle that's just been created and we've got a hypotenuse of nine centimetres, a 35 degree angle and a perpendicular height that at this time I've labelled as B.

So the area for this triangle would now be written as half times 16 times B, because the base edge is now not the nine centimetres but instead the 16 centimetres because within the formula really you are multiplying two perpendicular lengths.

We've now got both of those on the screen where we've got our right angled triangles, and both of the areas in terms of A and in terms of B are written on the screen.

So let's actually calculate the area numerically.

Firstly, we need to calculate the perpendicular heights in each of these examples.

So the perpendicular height or A on the left hand version is 16 times sin of 35 degrees.

Because A is the opposite, and we do know that 16 is the hypotenuse, so the ratio between opposite and hypotenuse is sin, substituting that into the area and calculating it would be 41.

3 square centimetres to one decimal place.

You're gonna do all of that onto your calculator and round at the very final stage of working.

So this is still the same triangle.

We're still working out the area of the same triangle, so what should happen is that they should be the same, but this time we're using a different perpendicular height and a different base.

So we need to calculate that perpendicular height.

B this time would be nine times sin 35 because nine is the hypotenuse of that smaller right angled triangle and B is the opposite, and we know that the opposite and the hypotenuse are in a ratio of sin, so substituting that into the area, it calculates as 41.

3 square centimetres.

The two methods result in the same area.

Well, that's what we would expect because it is the same triangle, so the space that it occupies is going to be the same.

So check on this, which of the calculations are correct to find the perpendicular height of this triangle? So the perpendicular height is labelled as H.

Which of those three calculations is the correct calculation to work out that length? Pause the video whilst you decide on that, and when you're ready to check, press play.

Well, it would be B, cannot assume that this is an isosceles triangle.

So unless you know for certain, unless you have two edges that are equal in length and you know that by either the notation on the diagram or by the given lengths or that you have two angles that are equal, you cannot assume it is isosceles just because it may look like it.

So carry it on with areas, we know how to calculate the area of a parallelogram and that's by finding the product of the perpendicular height and its base.

So this parallelogram, which has a base of seven centimetres and a perpendicular height of 4.

5 centimetres, to find its area, we have the product which is 31.

5 square centimetres.

But the second parallelogram on the screen doesn't have the required information.

So can the area be calculated if we do not know the perpendicular height? Well, the perpendicular height is fixed throughout the parallelogram.

It just happens to be on the top one that I've labelled it where I have.

This perpendicular height on the second parallelogram which I have given H as to stand for the perpendicular height is this length here but it's also the same length here or here, we might mark it outside of the parallelogram, or here or here.

And this position creates a right angled triangle with the hypotenuse of 5.

1 centimetres, which happens to be one of the edge lengths and an angle of 62 degrees.

So using this right angled triangle and the sin function, we can calculate the perpendicular height of the parallelogram.

So the perpendicular height H would be 5.

1 times sin of the angle 62 degrees, H is opposite, the focus of 62 degrees, and 5.

1 is the hypotenuse.

And therefore the perpendicular height is 4.

5, not exactly, so I'm going to keep the exact answer in my calculation, and the area would be the base multiplied by the perpendicular height, which is 31.

5 to one decimal place.

Alternatively, the 5.

1 centimetres could be the base of the parallelogram.

Similar to that triangle, we can use an alternative edge as the base, it's like a rotation of the parallelogram.

So for this check, you need to calculate the length of this perpendicular height, again, it's labelled as H.

So pause the video whilst you calculate the length of the perpendicular height marked on that parallelogram.

Press play when you're ready to check your calculation.

So you should have done seven multiplied by sin 62.

We're still using the sin function because H is the opposite and seven is the hypotenuse.

H is 6.

2 centimetres to one decimal place, and if you were to carry on and calculate the area, then you're gonna get the same area because it is the same parallelogram.

We are just using two different perpendicular lengths to find the area.

This time we're moving on to a regular octagon, and a regular octagon doesn't have a formula that we may be able to recall such as the triangle, half times based on perpendicular height, or the parallelogram, based on perpendicular height.

So is it possible to calculate the area of the octagon? It's a regular octagon with an edge length of 10 centimetres.

Andeep doesn't think you can find the area probably because he doesn't have a formula that he can substitute the 10 into.

Alex suggests, well, there must be a way, maybe by sectioning it into other polygons.

Andeep says that might work or perhaps we could surround it, and what they're considering is how we find the area of composite shapes.

For composite shapes, you might section them into other polygons, shapes that you do have formulas for, or potentially surround them and subtract away the extra area that you've created.

So Alex wants to give it a go.

So we've got a regular octagon, edge lengths of 10 centimetres and we're trying to find its area.

Let's first do it by sectioning it into other polygons.

In particular, we want to create polygons that we do have formulae to be able to calculate the area.

So in this particular way of sectioning up the regular octagon, we've created three different polygons and hopefully you can see them.

Firstly, we've got some right angled triangles and we've called these A.

Secondly, we've got some rectangles and we've called these B, and lastly there is one square in that centre part.

And so I've added all of the 10 centimetres that are in these polygons.

So the right angled triangles will have a hypotenuse of 10 centimetres.

They're the slanted edge in the way that the octagon is orientated on the screen.

The rectangles will have the length being 10 centimetres and you can see that on the octagon, and finally the square is 10 by 10.

We know that the right angled triangle will have a 45 degree angle, and we know this because the interior angle is 135 degrees in a regular octagon.

And polygon B is a rectangle, then we know that interior angle is 90 degrees, which leaves an angle of 45.

By being a 45 degree angle in a right angled triangle, then the other angle would also be 45 degrees.

So this is also an isosceles triangle.

And so we are going to give the width of the rectangle, which is also the base and the perpendicular height of the right angled triangle, this letter of X, because currently we don't have the length of it, something we need to calculate.

And so we can calculate using cosin, X is an adjacent to the 45 degree angle that is marked on that right angled triangle, and we know that 10 is the hypotenuse, and cosin is this ratio between the adjacent and hypotenuse, and so we can calculate X to be 7.

07 to two decimal places.

So that's rounded, we're gonna try and use the non-rounded value in our further calculations.

Was there any other way that we could have calculated that length of X? Well, I'm hoping you have thought about the fact that you could have used the sin function because the base of that triangle is X, which is the opposite of the labelled 45 degree, and we know the hypotenuse, so you could have used the sin function, but because it was an isosceles triangle then we could have actually used Pythagoras' theorem as well.

So now that I know what X is, I've calculated X, that was part of the dimensions of the triangle but also the width of the rectangle, so now we can calculate the area of each of these polygons.

So the right angled triangle area is 25 square centimetres, base times perpendicular height divided by two.

The rectangle would have an area of 70.

71.

that continues, and the square, hundred square centimetres.

So individual polygons, because they are our sort of polygons that we have got formula to work out the area, we can do that, and then we can total up our areas to get the area of this octagon.

So there are four of those right angled triangles, so four times 25, there are four of those rectangles, so four times the 70.

7 and one of the squares, and when you find that total, it's 482.

84 square centimetres to two decimal places.

So that is not the exact area, but it's a fairly accurate representation of the area.

The other way that they suggested was that we could surround it, and in this case we could surround it by a square.

So we're gonna use the same lettering, so use the X for that length of the right angled triangles, and we obviously know that the edges of the octagon were 10 centimetres.

So the length of the square that is surrounding the octagon would be 10 plus 2x.

So there are two of those X's along the length plus the 10, and because it's a square that's in the both directions.

Going back to the right angled triangle, and this time I'm showing that you could use it with sin function, it'd be 10 times sin 45, and that gives us that 7.

07 again.

So the length of the square substituting that into our expression is 24.

14 approximately.

And to find the area of a square, you can square it, because the length and the width are equal, the area would be 582.

84, and that's larger than what we calculated for the octagon.

And that's what we would expect because those right angled triangles have added area to the octagon.

So the area of the octagon is the area of the square minus the area of those four congruent triangles.

So if we do that, and remember to find the area of the triangles is half times based on perpendicular height, we get the exact same answer as previous.

That's what we would expect, we just did it in a different method.

So for your check, this is a regular decant, it's got 10 edges, and each edge is eight centimetres, and it's been surrounded by a rectangle.

I would like you to match the calculations, so there are three calculations there, to the correct length on the diagram.

There are three lengths marked with arrows.

So there's length A, length B, and length C.

So I would like you to decide which of the calculations would be calculating which of the lengths.

Pause the video whilst you match those up, and then when you want to check, press play.

So the first one, eight multiplied by sin of 18 degrees would be finding the length of A, and that's creating a small right angled triangle where 18 degrees is the focus angle, so A is the opposite, and we have the hypotenuse because it is the edge of the original decagon, which we know is eight.

The second calculation matches with C, and so this one is eight times cosin of 36 degrees.

Well, this is having to use the idea that there is reflective symmetry in this decagon and that 36 degree angle that is marked at the top left of the diagram would also be down in that area there where the C is, and C is an adjacent to a 36 degree angle, which is why we'd be using cosin.

And which means by process of elimination that the bottom one is for the length of B.

54 degrees is not marked onto the diagram.

However, it would be part of the right angled triangle with the 36 degrees and a 90 degrees, there would be 54 degrees would be the third angle.

So B matches there.

The alternative to get that length B would be eight sin 36 degrees.

You've got a task now about calculating areas.

So question one, calculate the area of each triangle and give your answer to one decimal place.

Make sure you're not rounding too early within your working out.

Pause the video whilst you're working out those areas, then when you press play, you've got question two and three.

Question two is to calculate the area of this rhombus to one decimal place, and question three is to calculate the area of this parallelogram to one decimal place, given that it's perimeter is 54 centimetres.

So pause the video whilst you try questions two and three, and then when you're ready for question four, press play.

Question four, you've got a regular hexagon with edge length of five centimetres, and I'd like you to calculate the area of it to one decimal place.

So pause the video whilst you decide on which method you're gonna go for, so potentially section it up into polygons, or you could surround it.

Press pause whilst you do that, and when you press play, we're gonna go through the answers to these four questions.

Question one had four parts.

So this is part A.

So it was an isosceles triangle, you knew that because of the hash marks along the edges.

Firstly, you needed to calculate the perpendicular heights so that you could use the formula half times base times perpendicular height, and so your area is 64.

0 square centimetres to one decimal place.

Part B was another isosceles triangle.

You were given it with the 30 degree angle, which when you drop that perpendicular, that altitude, it is the line of symmetry so it would bisect the angle.

So when you are calculating the perpendicular height, you would be using a 15 degree angle.

It may be that you calculated the perpendicular height using a 75 degree angle and your calculations should have then been three multiplied by tan 75.

However, we would've both got the same answer of 11.

196 for the perpendicular height and then substitute that into the formula, and the area of the triangle is 33.

6 square centimetres.

Triangle C, so this was not an isosceles triangle, you needed to create a right angled triangle in order to get a perpendicular height.

So you can see where mine is on the diagram.

I've calculated that using the sin function and then I've substituted it into the formula with the 12 because it needs to be two perpendicular lengths.

17.

1 square centimetres is the area of that triangle.

Lastly, this is triangle D.

This one we needed the perpendicular height still, but the perpendicular height be the altitude that was the line of symmetry was not as useful because you would've then had to go on to do further calculations to get the base of that triangle.

However, because it is isosceles, the 12 centimetres that is marked is also 12 centimetres on the other side, so I'm finding the perpendicular height to that 12, and then I can calculate my area as 36 square centimetres.

Well done if you managed that one.

Question two was to calculate the area of this rhombus to one decimal place.

So being a rhombus that meant all the edges were seven centimetres.

You need to get the perpendicular height.

Remember that the perpendicular height is constant throughout, so placing it in an appropriate place to create a right angled triangle and then base times perpendicular height.

The area to one decimal place was 46.

9 square centimetres.

Question three, area of this parallelogram to one decimal place, given that it's perimeter is 54 centimetres.

So we need to recall what that means.

The perimeter is the sum of the edges and therefore we did need to calculate what that length of the parallelogram or the base of the parallelogram is and then we can make use of that for the area.

So the perimeter would be the base plus the 11 plus the base plus the 11.

If we walked, imagine yourself walking around the boundary of the parallelogram, we know it equals 54, which means that that base has to be 16.

Now that you have the base, you're gonna use that in the area calculation.

But what else do you need for the area? You need the perpendicular height.

So we're going to calculate the perpendicular height using a right angled triangle and sin of 66 and then do the product.

So 160.

8 square centimetres to one decimal place.

Finally, question four.

So this is a possible method.

This is not necessarily the method you went for.

So the final answer is 65.

0 square centimetres to one decimal place for the area of that regular hexagon.

You probably wish to pause the video and read through my method, if it isn't the method that you did or if you didn't get this quite correct.

So we're now up to the second learning cycle of the lesson and that is now calculating angles.

So still thinking about how trigonometry can help us with that to calculate angles.

So in this learning cycle, we're going to think about how to calculate angles.

So without the use of a protractor, we know a protractor is a very useful geometry tool which allows us to measure the size of a turn or an angle, how could we establish the size of this angle theta? Jen says, if we were to know two edges of a right angled triangle, then we can calculate the angle.

So can you see a right angled triangle that we'd be able to use? Perhaps you saw this one or maybe this one or this one or this one.

Maybe you saw a different one.

Which one of the ones that I have highlighted would be the most accurate to use? Well, it's gonna be the second from last one that I showed, and that's because the base edge is an integer and the vertical edge is an integer, the others, the base edges, I was sort of decreasing by one square each time, but the vertical height of that right angled triangle was not on a corner of a square and therefore I'd had to have estimated its height.

And then once you've estimated something, then you're gonna increase the chance of an error.

So this is the one where we've got a integer value for both the base of the right angled triangle and the height of the right angled triangle.

Jen says, if we use these two edges, the ones that are now marked with arrows, then this is part of the tangent function and that's because it's the opposite and the adjacent.

So the tan of angle theta is four over three.

So the opposite of theta is four units and the adjacent is three units.

So there is some value of theta which is in the ratio four to three, and hopefully you recall that to find an angle using the trigonometric ratio, we use the inverse of that trigonometric metric ratio.

So inverse tan or arc tan of four over three, would use our calculator, you need to press the shift key on most calculators to find the inverse tan is 53.

1 degrees to one decimal place.

And in fact we've been ended up being more accurate than if we'd used a protractor where you'd only been measuring to the nearest degree.

So we check which right angled triangle would be the most accurate to use to calculate the angle theta marked on these diagrams. Pause the video, and when you're ready to check, press play.

C would be the most accurate, and that's because the vertical height is on the corner of one of our squares on the the grid paper.

Whereas on A and B, the first one you would say it's like 3.

4 and B, you might say, oh, it's 2.

7, but you were guessing how tall that vertical edge is.

So thinking about this further, can we use a similar method for the obtuse angle theta, and Jim says, he thinks you can still create a right angled triangle.

So do you agree with Jim? Can you see where a right angled triangle could be usefully drawn in order to calculate angle theta? This is one option and there are more than one place than just this one.

But with this right angled triangle, we wouldn't be calculating theta.

If we label that alpha, we'd have to calculate alpha and then subtract it from 180 to get to theta because we know that angles on a straight line sum to 180 degrees.

So using that, can you calculate the angle theta? Pause the video, and then when you're ready to check, press play.

So you are gonna use tan and more importantly arc tan to calculate the angle of alpha, which is 45 degrees, and then subtracting that from 180 degrees will give us theta, 135 degrees.

Well done if you did that one.

Alternatively, you may have thought about drawing a right angled triangle here, and this right angled triangle is creating another angle alpha that is part of theta.

So in order to calculate what theta is, we'd get alpha plus 90 degrees.

So can you do it using that method? Press pause whilst you're working out theta, and then when you're ready, press play.

So here the triangle has an opposite of five and an adjacent of five, so we still end up with five over five, which is one, and therefore arc tan of one is 45 degrees, 45, add 90 is 135 degrees.

It is the same diagram, we've just calculated it in a different way.

It's not always going to be the case, however, that the triangle below the line and the triangle above the line give you the same angle.

So don't just assume that.

So let's think about angles for more context.

A yacht leaves a port and travels 60 kilometres due east and then 40 kilometres due south.

So that's 60 kilometres due east and then 40 kilometres due south.

If another yacht was to leave the same port starting by facing east and travels directly towards the first yacht, what angle should it turn through? So what should angle theta be in order for the second yacht to sail directly to where yacht one ended up? Well, we can see a right angled triangle, and the reason we know it is right angled because east and south are compass directions that have a 90 degree turn between them.

So our right angled triangle would have an opposite of 40 and an adjacent of 60.

So tan of this angle theta is in the ratio 40 over 60.

Using arc tan or inverse tan, we end up with 33.

7 degrees to one decimal place.

A check, press pause, read through it, and then when you're ready to check your answer, press play.

So you should have started with tan of theta is 50 over 30, so the opposite is 50 and the adjacent was 30 and you should have ended up with 59.

0 degrees to one decimal place.

So the final task of the lesson, calculate the marked angles on each diagram.

Pause the video, whilst you're doing that, when you press play, you've got question two and three.

So question two, a yacht is given the instruction to sail 20 kilometres due east and then 60 kilometres due north.

If the yacht is already facing east, what angle should it turn to be facing directly towards its final position? Question three, a helicopter travels 30 kilometres due north and then rotates theta degrees before travelling 50 kilometres in that direction.

The helicopter finishes directly due west of the starting point, what angle of turn did they make? I would suggest for both of these questions that you draw yourself a diagram, it doesn't have to be accurate.

So sketch yourself a diagram that represents what you've just been told in both question two and three.

Pause the video whilst you're doing that, and then when you come back, we're gonna go through the answers to questions one, two, and three.

Question one, I have not put any working out onto the screen, but the answers are there, all given to one decimal place.

So if you rounded it to more accurate, then hopefully if you rounded it to one decimal place, you'd agree with me, or if you've rounded it to a less accurate answer, so potentially the nearest degree, then once again, if you round to mine to the nearest degree that you would, we'd be in agreement.

Question two, I've drawn the diagram that is most useful for this scenario.

So it went east and then it went north.

You were asked for the angle to turn, so it was facing directly towards the final position and so that would be 71.

6 degrees to one decimal place.

Question three was asking for an angle that was not an interior angle of the triangle, so you needed to sort of use what you had done on question one for an obtuse angle, and that's to find alpha, in this case that I've labelled, and then subtract it from 180 degrees.

So the answer for this one is 126.

9 degrees.

To summarise today's lesson, which was about problem solving with trigonometry.

Trigonometry can be applied in many different problems in mathematics.

When faced with a problem, you should always begin by just considering the information you've been given and what can be found from it first, and not to worry too much about what you need to try and work out for the question.

I hope you've enjoyed today and I look forward to working with you again in the future.