Lesson video

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

So as the title of the lesson suggests, today we're going to be working with right-angle triangles and calculating missing side lengths, and making an informed decision on the most appropriate method on how to get to that missing side length.

On the screen of the trigonometric ratios.

It may be that you wish to pause the video so that you can refamiliarize yourself before we move on, 'cause we are definitely going to use them in today's lesson.

The lesson's broken into two learning cycles.

The first one is to decide whether actually we need to use trigonometry or not.

So think about other parts of maths that we could use on a right-angle triangle.

When we get to the second learning cycle, we're going to look at practical uses of trigonometry, and when we might need to find the missing side length of a right-angle triangle in our everyday life.

So we're gonna make a start with looking at finding the missing side lengths using an appropriate method, and deciding whether it is trigonometry or not trigonometry.

Alex and Aisha are discussing how they would find the perimeter of this triangle.

Alex has just said what the perimeter is, which the perimeter is the sum of the edges.

Aisha says, "Well that means we need to calculate the third edge first," because the information that's been given and labelled on that right-angle triangle is only two of the edges.

Alex has let the third edge be X, and that is opposite the given angle, or the focus angle of 53.

1 degrees.

And because he knows that the length of the hypotenuse is 15 centimetres, he's gonna use sine.

So X would be 15 multiplied by sine of 53.

1 degrees, which is 12.

0 centimetres, and therefore the perimeter would be 36 centimetres, if you find the sum of the three edges.

Now that we know what the third edge is, Aisha said she would've used Pythagoras's theorem to calculate the length of X.

So 15 squared is equal to nine squared plus X squared.

Rearranging that to get X squared as the subject, means that X squared is 144, once you evaluate those squares and do the subtraction.

Square routing gives her 12 centimetres, and again the perimeter would be 36 centimetres.

So they've both calculated the perimeter, both got the same perimeter, but they've done it by using a different method.

And this just highlights that sometimes you've got enough information to make a decision on what to do.

So this check, with the given information, what could you do? So A, B, or C.

Pause the video, and then when you're ready to check, press play.

You could do A or C.

So A is to use trigonometry to calculate the other two edge lengths.

So have 16 centimetres, which is the hypotenuse.

So you could use cosine and sine to work out the adjacent and the opposite.

Part C, you could work out the third angle of the triangle.

Well yes, you've got two of the three, and you know that they sum to 180 degrees, so you could work out the third angle.

Similar to the question that Alex and Aisha were answering, if we want the perimeter of this triangle, we need the third edge length.

We have two of the three, so we can't get the perimeter yet.

Pythagoras' theorem and trigonometric functions both require right-angle triangles.

So where can we create a right-angle triangle? So by extending the base edge that nine centimetres and dropping a perpendicular line, we have a right-angled triangle, and this shows you the right-angle triangle.

So then sine function can be used to calculate the opposite, which is the vertical height of this right-angle triangle, because the 16 centimetres is the hypotenuse, and we have that 35-degree angle.

So H would be 16 multiplied by sine of 35, which is 9.

177, and that continues.

So we're trying to use the most accurate value of H in our further calculations.

Pythagoras' theorem, the cosine function or the tangent function could then be used to work out X, which is the third edge of this right-angled triangle.

So here using cosine, 16 multiplied by cosine 35 gives you 13.

10.

And again that's not the most accurate value.

If we know that that is 13.

10, and we knew that part of it is nine centimetres, then that additional extension of the base edge, is 4.

1 roughly, centimetres.

Now hopefully you can see that slightly smaller right-angle triangle, of which we've just calculated its two edge lengths, which means that we can go on and calculate what Y, which is our third edge of our original triangle, is by using Pythagoras' theorem.

So Y squared, because it's the hypotenuse, is equal to the sum of the square of the two shorter sides.

So the 4.

1, but again we're gonna use the most accurate version by using our calculator, plus 4.

1 squared plus 9.

17 squared, is 101.

084 which we can then square root, and that's 10.

05.

And so our perimeter will be the sum of that 10.

05, the 16 that was originally given, and the 9, 35.

05 centimetres to two decimal places.

So in this check, which additional lines are most useful to calculate the length of the third edge of the triangle? So if you were to have a question that was asking for the perimeter, you would need all three edges.

So which of these additional lines is gonna be most useful to calculate the third edge, which if you needed to get the perimeter, you could then go on to do? Pause the video whilst you make a decision on that, and when you wanna check, press play.

It would be B.

A follows what we did in the example just before, but the angle of 21 degrees is in a different place to where that 35 degrees was in the original, in the example previously.

And because of that, that part A is not useful, and that's why it's part B, because by putting that altitude of the triangle on, the 21 degrees is then opposite the perpendicular height of the triangle.

So in continuing at looking at triangles, this learning cycle is all about whether it's trigonometry or not.

So this diagram contains three triangles.

Can you find them? So you may wanna pause the video, and try and work out where the three triangles are.

So here is one, here is the other one, and lastly it's the full diagram.

So if we focus on this larger one, the full diagram, we can calculate the length of the third edge by using Pythagoras' theorem, so I've labelled that as A.

And because we know the hypotenuse and the other short side, then knowing two edges on a right-angle triangle allows you to use Pythagoras' theorem 14 squared equals nine squared plus A squared, rearranging it to make A squared the subject, and evaluating and calculating, we know that A squared is 115, so that would be the area of the square along that edge.

So what length is it? It would be the square root, and that's 10.

72.

Now using the right-angle triangle at the base of this diagram, the tangent function can be used to calculate the length of the opposite, because the nine centimetres which was given on the diagram is the adjacent, it is between the 31-degree angle and the right angle, and so we can use the tangent function, because we know that that's a ratio between opposite and adjacent, to calculate what I have labelled as B.

So B is equal to nine times tan of 31 degrees, which is 5.

41 centimetres, to two decimal places.

But you can see on the diagram I've left that as 5.

40, dot dot dot, to be accurate.

And that then allows us to get X by using a subtraction.

The full height is 10.

72, that part of the height is 5.

4, and so therefore X is 5.

316 to three decimal places.

A check, I've labelled an edge there as A.

So which of these calculations are valid to find A? Pause the video, and then when you're ready to check press play.

Well actually, astonishingly all four of them calculate that length A.

It's a right-angle triangle and you know two of the edges, so therefore Pythagoras' theorem can be used.

Part B, A is the adjacent to the 65-degree, and we have the hypotenuse, so we could use cosine.

Part C, A is the opposite to the 25-degree angle.

The 25-degree angle is not labelled, but we could calculate it, because we have two of the three angles within the triangle.

And part D, we can use the tangent function.

So on to the task.

Remember this learning cycle is trigonometry or not, so you won't always need to be using trigonometry.

So do assess the situation that you have, and make a decision on everything you know up to this point, all the armoury you have for mathematics and what could be used.

So question one is calculate X in each triangle.

Pause whilst you do that, and when you press play you move on to question two.

So question two, there are two parts, A and B.

You need to calculate X in each triangle.

Give your answers to one decimal place.

So pause the video whilst you have a go at those two parts, and then when you press play you've got to question three.

Here's question three, again it's about calculating X and giving your answers to one decimal place.

So assess the situation each time, and make a decision on what's the most efficient way to get to X.

Press pause whilst you work through questions part A and B.

And then when you press play, you're onto the final question of this task.

So finally question four, work out the perimeter of this triangle, given your answer to one decimal place.

Press pause whilst you get to that perimeter, and when you press play we're gonna go through the answers to one, two, three, and four.

Here is question one.

So one A, you needed to use trigonometry.

You didn't have two of the edge lengths, so you couldn't use Pythagoras' theorem.

But you did know it was the right-angle triangle.

You did have a focus angle of 30 degrees.

It was the adjacent and you wanted the hypotenuse, so cosine is the trigonometric function you should have been using.

The answer was 46.

2, to one decimal place.

On B, you had two of the three edges, you didn't have any further angles except the right angle.

So this one was Pythagoras' theorem, X was the hypotenuse, so X squared stays as the subject, and X would be 50 centimetres.

Part C, you were told that the perimeter was 1.

2 metres.

Perimeter is the sum of the edges, so if you have the sum and you have two of the three, then the easiest way to do this was to use a linear equation to set up the perimeter.

You did need to be mindful that the perimeter was given in metres, and the edges were in centimetres.

So you did need to do a unit conversion at some point within your calculation, and X was 50.

This is part A of question two, you needed to work out X.

I've added an additional edge which is Y, 'cause that's part of my working out.

So I've used the tan function to calculate Y first of all, and then I've used pyrosis theorem on the largest right-angle triangle at the whole diagram, to get the full height of the triangle, and subtracted Y to leave me with the X.

I haven't used the eight degrees in this calculation, but you may have, because you may have used the big right-angle triangle and used trigonometry.

7.

1 was the answer that you should have hopefully finished on.

Part B only needed one calculation.

It may have been that you took more calculations to get to 16.

0, and that's absolutely fine.

However, we do always want to try and evaluate the situation to get the most efficient, and the way to be the most efficient in this diagram was to recognise that the smaller right-angle triangle, the sort of bottom of the diagram, you had two of the three angles, so you could calculate the third angle, and then combining that angle with the 19-degree angle, you had the full angle for the right-angle triangle, which was 37 degrees, and you could then use cosine to calculate the hypotenuse of X.

Question three, again you may have used trigonometry.

You had enough information that you could have used trigonometry.

However, because it was isosceles triangle, it may have been easier to use Pythagoras' theorem.

You knew it was isosceles, because two of the angles were given, the 45-degree and the 90, which allowed you to calculate that the third angle would also be 45 degrees, therefore you've got two angles that are equal.

It's an isosceles triangle.

So on both of mine I have used Pythagoras' theorem.

Part A, X is 9.

9 and part B, X is 19.

8.

Lastly, you needed to work out the perimeter of the triangle giving your answer to one decimal place.

A, the perpendicular height had been marked onto this diagram.

So we're starting with that right-angle triangle where I've labelled it as 15 centimetres, A centimetres and B centimetres.

I've used cosine to get the length of A, and I've used sine to get the length of B.

It was necessary to work out A, because we cannot assume that this is an isosceles triangle.

So although we know that length is 24 centimetres, we need to know the distance between the vertex and where the perpendicular height meets, the altitude meets that base.

And we can do that by subtracting A from 24.

When we have that, we can then use Pythagoras' theorem on the second right-angle triangle that the triangle has been split into to find X, and X comes out as 11.

3499, and we can then use that to find the perimeter.

24 centimetres plus the 15 centimetres, plus that value of X we've just calculated, means the perimeter is 50.

3 centimetres to one decimal place.

We're up to the second learning cycle, where we're gonna take a look at practical uses of trigonometry.

So this time we are not gonna worry about whether we could use an alternative method, like Pythagoras' theorem.

We're going to use trigonometry in more contextual-based problems. So during this learning cycle, we're gonna be looking at practical uses of trigonometry, and trigonometry as a topic of mathematics is used in a whole host of different careers and contexts.

So this is just one of them.

A ramp needs to be built outside a shop for wheelchair access.

The incline of the ramp cannot exceed 4.

8 degrees.

The bottom of the doorway is 34 centimetres high.

How far away from the shop will the start of the ramp need to be? And so this is part of the building regulations for wheelchair access in a public place.

There are other restrictions as well, but one of them is to do with the incline.

So you can see here on the diagram that 34 centimetres is marked.

We've got the 4.

8 degrees, which is our maximum angle of incline.

And what we're trying to work out, is that horizontal distance from the wall of the shop to the start of the ramp.

We are making an assumption, and often with practical uses of trigonometry there is gonna be an element of assumption, or you're going to have to simplify the problem, the model, in order to be able to use the mathematics.

So assuming the wall of the shop is perpendicular to the ground, the tangent function can be used to determine the adjacent length.

So if you look, that D centimetres would be the adjacent, and we would use tangent with the 4.

8 degrees, and the 34 centimetres is the opposite.

And so 404.

9 would be D to one decimal place.

So the start of the ramp is approximately 4.

05 metres away from the shop.

A check, same concept, but this time the bottom of the doorway is 30 centimetres high, and the question is asking you to calculate the length of the ramp.

So look to see where D is now labelled on the diagram.

Pause the video whilst you calculate that length, and when you're ready to check, press play.

So you wanted to work out the hypotenuse of this right-angle triangle, again assuming that the ground that this ramp is being built on is perpendicular to the wall of the shop, and 358.

51 would be in centimetres the length of the ramp, which means it's about 3.

59 metres long.

Another practical use of trigonometry is being able to estimate the height of unknown objects, tall objects, buildings, in this case a traffic light.

So how could you do this using trigonometry? So Sam says if we know Alex's height and measure the angle to the top of the traffic lights from his eye line, we can use trigonometry.

And a tool that you could use is a clinometer, and that can measure the angle.

So we know Alex's height is 156 centimetres, the angle of elevation, so the angle from the horizontal of his eye line up to the very top of the traffic light is 52 degrees.

We are looking to calculate H.

So the three metres that Alex is stood horizontally from the traffic light is the adjacent of this right-angle triangle.

We are gonna calculate X, so that is the distance from the eye line, the horizontal of his eye line, up to the top.

And because we have his height, we can add that on to get the full height of the traffic light, and that comes out as 5.

4, 5.

4 metres.

Again, there is an assumption that Alex is stood on a ground that is perpendicular to the traffic lights, that the traffic lights are vertical, that there is no slant to them.

Here's a check, a different set of traffic lights, and he is two metres from them this time.

So fill the blanks to estimate the height of this traffic light.

Press pause whilst you're doing that, and when you press play, we'll go through the answers.

First of all, it would've been a two.

So it was three metres that Alex was stood on the last traffic lights, but he's now two metres horizontally.

Tan of 52, and that calculates to be 2.

55, and that's why the height of the traffic lights was 2.

55 plus 1.

56, 1.

56 is his height.

The 2.

55 is the distance between the eye line and the top of the traffic light, which means an estimation of the height is 4.

12 metres.

So continuing to look at other practical uses of trigonometry, Sofia is standing on top of a cliff which is 60 metres above sea level.

She can see two boys straight in front of her.

The angles of depression, so this time this is the angle from the horizontal of her eye line down to the line of sight towards the boys, are 47 degrees and 32 degrees respectively.

Sofia wants to estimate how far apart the two boys are from each other in the sea.

And so we've got a diagram here that shows that information.

What does the 1.

35 metres represent here? Pause the video whilst you think about that.

That's the height of Sofia to her eye line.

So 60 metres was told to us about the top of the cliff to the sea level, but we've also going up to her eye line, so we do need to increase that distance by that height.

We were told that the angles of depression were 47 and 32 degrees respectively.

So we can easily see the 32 degrees, and that's the angle of depression to that second, the boy that's further away from the cliff side.

Why is it 15 degrees rather than 47? Well, 15 plus 32 gives you 47 degrees.

So the angle between the two lines is 15 degrees.

And the simplified version of this diagram will be useful here.

So with a mathematical model you do not need to have a realistic image with the sea and the cliff, et cetera.

We just need to get the parts to be able to do the calculation.

So something like this is much more useful, and we've now got right-angle triangles appearing.

But what has changed is we've suddenly now got a 43-degree angle.

We had a 15-degree angle previously, but we've now got a 43-degree angle, and that's because from the horizontal to the cliff would be 90 degrees, assuming we're using a model where the cliff is vertical.

And so that part there from the cliff edge up to the line to the first boy is 44 degrees.

I've also labelled distances along that sea line.

So the full distance from the cliff to the second boy I've labelled as X.

The distance from the cliff to the first boy is Y, but the question wants the distance between the two boys, which is our D.

So if we focus focus on X, which is the distance from, the horizontal distance from the cliff to the second boy, we can use the height as our adjacent, and we would like the opposite to the 58-degree angle.

Well where did that 58 degree angle just come from? Well, that's the 43-degree plus the 15.

And so we're going to use tan, opposite and adjacent, and X is 98.

18 metres to two decimal places.

So the second boy, or the boy further away from the cliff edge, is that far away horizontally.

Now we're going to use this right-angle triangle to calculate the distance to the first boy, or the boy closest to the cliff.

And again we're gonna use tan, because it is the opposite edge that we are trying to calculate knowing the adjacent, and that's 57.

21 metres to two decimal places.

Going back to our full diagram, then the distance between the two boys can be calculated by finding the difference in those two distances.

40.

97 metres to two decimal places.

So the boys are 40.

97 metres apart.

On another day, Sam is stood on top of a cliff 60 metres above the sea level.

They can see two boys straight in front of them.

The angles of depression of these boys are 46 degrees and 32 degrees respectively.

So what is the size of X? Pause the video whilst you decide on that.

It might be that you want to go back and go through the example with Sofia again, and then come back to this check.

But when you're ready to check the answer, press play.

So 14 degrees, because the angle of depression to that very first boy is 46 degrees.

So if you've already passed through 32 degrees to the line of the second boy, then you need to turn a further 14 degrees to get to 46.

So Sam wants to calculate the distance D between the two boys.

Which pair of triangles will they use in their calculations? So pause the video again and when you're happy with your decision, press play to check.

So they would be using the pair of triangles in C.

46 degrees gets us to the line of sight to the first boy, which means that there would be 44 degrees left to go all the way to the edge of the cliff.

And 44 plus 14 gives us the 58 degrees of the larger right-angle triangle, where we'd be able to calculate the horizontal distance to the second boy.

So the cliff is 60 metres high, and it is 1.

63 metres to Sam's eye line.

The distance of the first boy is X.

Complete the calculation to find the horizontal distance to the nearest boy.

Press pause, and then when you know what should be in that blank, press play to check.

So we would be using the trigonometric function of tangent, because we would like to calculate the opposite and we will know the adjacent.

We know the height of the cliff plus the distance to Sam's eye line.

So the distance of the first boy is 59.

515, and the distance of the second boy is 98.

628.

What is the distance between the two boys to the nearest metre? Pause the video, once you've calculated that press play to check your answer.

39 metres to the nearest metre.

So that was the subtraction, because we know the horizontal distance to the the boy that is further out to sea, and we know the horizontal distance to the closest one.

So the distance between them is a subtraction.

So we're up to the task B in the lesson.

Question one and two are on the screen right now.

I would heavily suggest that the first thing you do is read through and then sketch yourself a diagram, a simplified diagram.

Remember it doesn't need to be an artistic interpretation of the context.

You need to find yourself a right-angle triangle.

So sketch yourself a right-angle triangle, and apply the given information onto it.

Press pause whilst you're doing that.

When you press play, we'll go onto question three.

Here's question three.

A seven-meter pole is secured by support wires.

So there's a diagram there to show that.

The wires are pegged in a straight line.

The outer wires make an angle of 31 degrees with the floor, and the inner wires make an angle twice the size.

How far apart are the wires? So you are trying to calculate that distance that is being labelled as D on that diagram.

Pause the video whilst you work through that question.

When you press play, we're gonna go through the answers to questions one, two, and three.

Here is question one.

So I really am hoping that you did sketch yourself a right-angled triangle.

So a ramp is being built outside a house.

The bottom of the door is 42 centimetres off the floor.

The angle between the ramp and the ground should not exceed 4.

8 degrees.

What is the minimum slope length of the ramp? So you were trying to calculate the hypotenuse.

You had the opposite of 42 centimetres, and the angle was 4.

8 degrees.

So the slope is a minimum of 501.

9 centimetres, or 5.

02 metres long.

Question two, a man standing on a riverbank notes a particular tree on the opposite bank of the river.

He then walks 60 metres along the riverbank in a straight line.

On looking back, he can see that the angle between the line of his path and a line to the tree is 32 degrees.

How wide is the river? So this triangle represents what's just happened.

Where the right angle is is where the man started, when he noted the tree on the other side.

He then walked 60 metres, which is what that edge length is.

And then on looking back, he can see that the angle between the line of his path and the tree is 32 degrees.

How wide is the river? So you were trying to calculate the opposite and you know the adjacent, so you would be using tan.

The river is 37.

5 metres wide.

This is clearly an estimate of the width of the river.

It's not exact, but it shows you that if you can create that right-angle triangle, then trigonometry is very useful.

Question three was actually very similar to the boy's question.

The only difference was where the angles were marked.

So using the sort of largest right-angle triangle with the 31 degrees, and the seven being the opposite, you needed to calculate the distance to where it was pegged in the ground, and that would be the adjacent.

So we can use tan to calculate that distance, and that was 11.

65 to two decimal places.

Then we're gonna use the smaller right-angle triangle with the inner wire, and you were told that it makes an angle twice the size of the 31.

So that means it's 62 degrees.

You're repeating using tan to find the adjacent, but this time you are using the 62-degree angle.

Please note, and hopefully you're aware of this, that just because the angle is doubled, doesn't mean that the distance is halved.

It's not inversely proportional in a linear fashion.

And so that distance would be 3.

72.

So the distance between the pegs in that straight line would be 7.

93 metres.

So to summarise today's lesson, which was about choosing an appropriate method for finding lengths of a triangle, and that really leads into this idea that trigonometry can be used in many practical situations.

Often it is used alongside Pythagoras' theorem, and sometimes Pythagoras' theorem will just be more efficient than trigonometry.

So do always assess the situation, make sure you're trying to find the most efficient way to get to the answer.

Trigonometry is often used to calculate or estimate the height of large objects, when measuring them is impractical.

So we saw that with a traffic light, the width of a river, the boys in the sea where it would struggle, unless you're in the boat on the sea to do it.

But larger objects, so buildings or even distances between places, trigonometry can be really useful to calculate that.

Really well done today.

There was some challenging elements to that lesson.

I hope to work with you again in the future.