Lesson video

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

I really hope you're willing to try your best and even if it gets challenging, I'll be there to support you.

So let's make a start.

So our lesson Outcome today is to be able to use our knowledge of trigonometric ratios on 3D problems. So Keywords that we'll be using in the lesson today, but are not new to you, you will have used them previously in your learning, are here on the screen.

Let's just go through them together now.

So the hypotenuse is the side of a right-angled triangle, which is opposite the right angle.

It's also the longest edge.

The trigonometric ratios are ratios between each pair of lengths in a right-angle triangle.

So we've got the tangent ratio, we've got the sine ratio and the cosine ratio.

And lastly, a cross-section is a 2D face made from cutting straight through any plane of a 3D object.

So in today's lesson on applying trigonometric ratios in 3D, we're gonna split the lesson into two learning cycles.

Our first learning cycle is gonna focus purely on cubes and cuboids.

And then when we get to the second learning cycle, we'll look at other 3D shapes as well.

Let's make a start on looking at trigonometric ratios in 3D problems, specifically cubes and cuboids.

So just like the Pythagorean Theorem, trigonometric ratios can be used on a flat surface or on a plane in three dimensional space.

So we've made use of the sine ratio, the cosine ratio, and the tangent ratio on right-angle triangles in two dimensions.

But today we're gonna look at how we can find two dimensional right-angle triangles in a 3D shape.

So here we've got a right-angle triangle where the adjacent is one centimetre, the angle is theta and the opposite edge is tan of theta.

And that's because the ratio of tangent is opposite and adjacent.

If we were using the opposite and the hypotenuse is that sine or cosine, just pause the video and think to yourself, if we know the opposite and we know the hypotenuse or if we want to work out one of those two, then which ratio would we be using, sine or cosine? Press play when you're ready to move on.

So if we've got the opposite and the hypotenuse, then that is the ratio of sine.

So each of them is a ratio between a pair of the three.

So you could have the opposite and the hypotenuse is sine, adjacent and hypotenuse is cosine, and opposite and adjacent is tangent.

So let's look at that cuboid.

That cuboid we know has got rectangular faces and rectangles have right angles.

So there will be right-angled triangles on each of the faces of a cuboid.

But the one here that's with the dotted lines is not on a face, but is instead on a cross-section of the cuboid.

And it's the cross-section would be rectangular.

If you imagine this as a block of cheese and you cut it across that diagonal, then the face that you would have created would be rectangular.

And therefore, there would be a right angle triangle.

So Jun says, "We might choose to use trigonometric ratios when angles are provided." So if there are right angle triangles, then it potentially it's Pythagoras's Theorem that you need.

But if you've also got an angle included, you need to work it out or you've been given it, then it's going to be the trigonometric ratios instead.

So on this cuboid now, we've got a length of the cuboid, 11 centimetres, and that's the length of that face as well.

We've also been told an angle between the length and its diagonal, which is 40 degrees.

So Jun says, "What approach could we use here to find the length of the dotted line?" Is it Pythagoras's Theorem, is it trigonometry, or can we not do anything? Pause the video and think about that for a moment.

If you wanted the length of that dashed line, the diagonal across that face, can you get it from the information given, and what mathematical skill will you use? The one you would use is trigonometry.

You'd use trigonometry, because you don't have two of the three edges.

We haven't been told that this is a square face, and therefore we do know two of the three edges, even if it's not both labelled.

So we haven't got enough information to use Pythagoras's Theorem, but with an angle and an edge we can use trigonometry.

In particular, which trigonometric ratio is it that you will use? Pause the video and think about that again.

So if you want to calculate that diagonal, let that diagonal be X.

Which of the three trigonometric ratios are you going to make use of? Press play when you're ready to check? So it will be cosine.

Cosine of 40 degrees is equal to 11 over X.

We can use the opposite edges on a rectangle are equal, so that 11 centimetres that has been marked there is equal to the edge opposite it.

And that then is the adjacent to the 40 degree angle.

Adjacent, we want the hypotenuse, and that's the ratio for cosine.

So cosine of 40 degrees is equal to the adjacent over the hypotenuse.

And for this particular triangle, it's 11 over X.

We then rearrange it to make X the subject and we do that by doing 11 divided by cosine 40 on the calculator would give us our value.

The hypotenuse is what we are calculating.

We know the hypotenuse is opposite the right angle and also the longest edge.

So being greater than 11 is what we would expect.

So Jun says, "How might we find the volume of this cuboid?" So just pause the video and remind yourself, think about how would you calculate the volume of a cuboid, any cuboid, not this particular one.

And then press play when you're ready to consider this one in particular.

So you should have thought about the fact that the volume is length times width times height, or the area of the cross-section times by the length because it is a rectangular prism.

So what do we have here? Well, we could work out the area of that rectangular cross-section.

4 x 10 that would be 40, but we haven't got the length of the prism, so we need to do a little bit more work in order to be able to calculate the volume.

Jun says, "We need to find a way to calculate the length AB." So we've got two vertices labelled here, so we can discuss which length we are talking about.

And AB would be the length of this prism.

So can you see a way of calculating that length? Jun says, "It can be done but in two steps.

First of all, we're gonna have to use trigonometry to find this diagonal length." So did you consider trigonometry? We know that there is an angle and we can see a right-angle triangle, so I'm hoping you were thinking of trigonometry from the information you were given.

So we're gonna work out this diagonal length first.

So check for you which of the calculations on the screen is how we would work out the length of that line.

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

So we would be using the calculation A, because it is involving tangent.

Four centimetres is the opposite edge to the 18 degrees and the edge that we are trying to calculate is the adjacent to the 18 degrees.

So tangent is the correct ratio.

The next step, remember Jun said, we're gonna have to do this in two steps.

So our first step was using trigonometry to work out that diagonal across the rectangular face.

And now we're gonna use Pythagoras's Theorem, because the diagonal we just calculated is equal, is equivalent to this diagonal, because of the property of cuboids, the opposite faces are congruent.

So now that we have that length, remember that was 4 over tan(18), then we can use Pythagoras's Theorem, because that is a right-angle triangle.

We have the hypotenuse and a short side, so we can calculate the third side using Pythagoras's Theorem and then we'll have enough information, 'cause we will then have the length, the width, and the height of the cuboid to calculate the volume.

So here is the first task for you.

On this question one, there are actually three parts, there is only two on the slide here.

So pause the video and work through part A and part B.

And then when you press play, we'll move on to the last part of question one as well as question two.

So in both of these you are calculating the volume of the cuboid.

So press pause and then when you're ready for the next part, press play.

Here is question one, part C and also question two.

So question one, you're still working out the volume of that cuboid.

And on question two, you're finding the length of the dotted line AB.

Pause the video to work through those last couple of bits.

And when you press play, we'll go through the answers to Task A.

So question one, part A, find the volume of the cuboid.

Very similar to the one that Jun was speaking us through.

So once again, we need the depth or the length, depending on how you visualise this cuboid, in order to be able to calculate the volume.

So we're gonna use trigonometry first of all to get the length of the diagonal, that's one of the dashed lines across the face.

And you're going to use tangent once again, because three is the opposite and you are trying to calculate the adjacent.

That fraction is the exact value, and so we want to try and use that as much as possible.

So to then work out the depth, this was stage two, we were then using Pythagoras's Theorem, because the diagonal was the hypotenuse, and 12 is a short side of a right-angled triangle.

So square root of our hypotenuse, squared minus 12 squared gives us the value of the depth or the length.

This is the third dimension that we need, the third perpendicular length, in order to work out the volume.

So our volume is 12 x 3, so length times width, and then we need to times it by the depth.

You can see that our calculation includes the most exact value.

And that's what you should be trying to do on your calculator.

You can make use of the answer key, so that you don't have to retype everything, the most error-free answer in terms of rounding.

So the volume for one, part A, is 179.

48 cubic centimetres to two decimal places.

This is cuboid B.

You were given the length and the width of the cuboid and it was the height that we needed in order to work out the volume.

You were also given a right-angle triangle on one of the faces, and the angle.

So the height of that would be the opposite to the angle.

And we do have the adjacent, and that's because of the properties of rectangles and the fact that cuboids are made of rectangular faces.

Using the tangent ratio once again, because it's the opposite we would like, and we have the adjacent, then the height of our cuboid is 9tan(35).

And then to get the volume, which is what's on the screen, it will be 8 x 9 x 9 tan(35).

And that's 435.

73 cubic centimetres to two decimal places.

On part C of question one, it was a slightly more challenging question.

So what information were you given? Well, you were given that that angle comes from the centre point, and it's the centre of a square face.

We know it's a square face from the dimensions that we're given.

So the diagonals, those dotted lines, that are subtending that angle, create an Isosceles Triangle with a base of 10 centimetres.

So we can drop the perpendicular and know that that will split the Isosceles Triangle into two congruent right-angle triangles, and that therefore, will also split or bisect, because the angle that was given and it also bisects the base, that's where the 5 has come from.

So then we can use trigonometry to work out the length of the hypotenuse, which is one of those dashed edges.

Both of the black dash lines have the length of 5 over sine of 42.

875.

So now we still are looking to get the height of the cuboid.

We cannot get the volume until we know the height of the cuboid, but we now have another right-angle triangle that we can make use of.

The hypotenuse is one of those dashed black lines.

The height is what we're trying to calculate.

And 5 root 2 is the distance or the length from the vertex of the square base to the centre, and you can work that out using Pythagoras's Theorem with the thinking about a square face.

So if you look at the diagonal of a square face, you would do 10 squared plus 10 squared, square root it and then half it.

And that's where the 5 root 2 comes from.

So now we can use Pythagoras's Theorem to calculate the H and then times it by 10 x 10, and that's 200.

1 cubic centimetres to one decimal place.

Really important that we're trying to use the most accurate answer.

So using our calculator to its full capabilities.

On question two, your task was to work out the length of AB.

It was told to you that three, three centimetre cubes were placed together and Pythagoras's Theorem or trigonometric ratios could be used here.

But we're gonna have a look at using both together.

So if we look at this triangle, this is thinking about the rectangular face that is making the base of this shape.

We know it's six centimetres long, because it was made of three, three centimetre cubes, and then it had a height or a length or width of three.

So the purple dash line you can see on the solid, but also on this right-angle triangle, we can calculate using Pythagoras's Theorem, and that then can be an edge length of the right angle triangle that has a height of six.

Think about it stood up, because once again you've got a stack of two, three centimetre cubes, the angle that was given, which is 48.

19 degrees, and we've just calculated that that opposite edge to that angle would be 3 root 5.

AB is what we're trying to work out, and therefore, we can use Pythagoras's Theorem on those two edges to get the hypotenuse, and that equals nine centimetres.

You could however have used trigonometry with the angle and the six to get AB straight away.

So we're up to the second learning cycle, where we're gonna still continue looking at trigonometric ratios in 3D, but this time other 3D shape.

So we can apply the trigonometric ratios to other 3D shapes such as cylinders and pyramids too.

Jun's there to remind us that, "Where there are right-angles, there are right-angled triangles." We need right-angled triangles in order to use sine, cosine or tangent.

So we're likely to need the trigonometric ratios where right-angle triangles and one further internal angle is provided and that's instead of Pythagoras's Theorem.

So if you see on here we've got a triangular prism, and we have the angle 42 degrees, which is an interior angle of that right-angle triangle.

And the opposite edge to that 42 degree angle is four centimetres.

We can't work out the volume for this shape without a measure of its width.

So this is a prism, it's a right-angle triangular prism.

So we do the area of the cross-section, which is this right-angle triangle, and then we'd multiply it by its length, or we might call it the width, in terms of the way we are looking at it.

So if we did know this width, if we were told that it was 12 centimetres, then we've now got sufficient information to calculate the volume of this right-angled triangular prism.

And what we need is to work out this distance.

We need three perpendicular dimensions when we're working with volume and we need this distance so we can work out the area of that triangle, which is the cross-section, base times perpendicular height divided by two.

And this is where we're gonna make use of trigonometry.

So what is the correct calculation to find the missing length of this triangular prism? Pause the video and look at the information you have, look at the options for the question, and when you're ready to check your answer, press play.

So we're looking to work out the adjacent on this triangle.

We have the opposite to the 42 degrees and it's the adjacent to the 42 degrees.

So that's why all of these options involve the tangent ratio.

But then it's which one's in the correct form? Well we know that tan(42) is equal to 4 over question mark, in this case or the unknown.

And so we'd need to rearrange it and that would be 4 over tan(42).

We're up to the last task of the lesson and we are going to on question one, find the volume of these 3D shapes to two decimal places.

So on part A, you've got a triangular prism and on part B, you've got a right pyramid.

So pause the video and work out the volume of those 3D shapes.

And then when you're ready for the next question of the task, press play.

So on question two, you need to find the surface area of these 3D shapes to one decimal place this time.

So remember the surface area, the total surface area is the sum of all faces added together.

So pause the video whilst you're working through question two and when you press play, we'll move through to the answers.

So question one, part A, was a right-angled triangular prism.

So we needed to work out the base of that right-angle triangle, in order to be able to get the area, which is the cross-section and times by the length.

So it was the adjacent that you were trying to calculate, knowing the opposite and an angle.

So that would be 5 over tan(36) gives us the length that is marked with a question mark, which is the adjacent.

We times that by 5, times it by 11 and times it by half.

You may not have done the product in that order, but remember that multiplication is commutative, so it doesn't matter in which order you did that product.

The answer is 189.

25 cubic centimetres to two decimal places.

On part B, you needed to work out the volume of this pyramid.

So that volume of any pyramid is 1/3 times the base area times the perpendicular height.

So we can work out the base area from the information that's given.

It's a rectangular base and so we can do 8 x 6.

We then need to get the perpendicular height, and that's where we're gonna use trigonometry.

So the perpendicular height is the opposite in this right-angle triangle to the 62 degree angle.

And we do have the adjacent length.

The adjacent length will be three centimetres, because this right pyramid means that the apex is directly above the centre and therefore, we can see that it would be along the midline of the six centimetre, which is why it's three.

So 3 tan(62) degrees gives us the perpendicular height of our pyramid and then we can apply the formula.

Once again, if you have written it as 1/3 x 8 x 6 x 3 tan(62), that is equivalent to what's written on the screen here.

The answer is 90.

27 cubic centimetres to two decimal places.

On question two, we were trying to calculate the surface area of the two, 3D shapes.

So part A, you needed to calculate a few different things using trigonometry and Pythagoras's Theorem in order to work out that surface area.

Because if we think about this triangular prism, it's got two congruent triangular faces, and we don't currently have enough information to work out those areas.

It then has three different rectangular faces.

Each one of those rectangular faces has a length of 8, which is the length of the prism, but the other dimension is the edges of the triangle.

So we've got 2 and we will be working out what we've labelled as A for the area of the triangle, but we also need the hypotonus, in order to get the front face.

That's sort of the slanted rectangle.

So A is calculated using 2 over tan(28) degrees, because it's the adjacent.

And B can then be calculated using Pythagoras's Theorem.

But you may have also used sine and have written it as 2 over sine of 28.

So that's an equivalent form for B.

Once you've done that, then you need to work out the surface area, which is the total of all of the faces areas.

So 2 x 2a over 2, that part of the calculation is the area of both of the right-angle triangles.

So base times perpendicular height divided by 2, gives you the area of one of those triangles and that's why we're doubling it, a pair of congruent faces.

The 8a is the base face, the one that the prism is sort of sat on.

8 x 2 is the perpendicular rectangle face, and 8 x b is that slanted, that largest rectangle.

Substituting the values of a and b into that calculation gets you 87.

7 to one decimal place.

There's a possibility that you've got a very close answer to 87.

7, but not exactly, and that's probably occurred by you rounding too much within the working out.

So really try to make use of the calculations, because they are the most exact value of a and b.

And then when we get to question two, part B, it was the surface area of this right pyramid that has a rectangular base.

So the perpendicular height is 4tan(52) degrees in the same way that we did for the volume, because it's the opposite that we're trying to calculate.

And using the 4, which is half of the 8 centimetre rectangle.

The surface area is then adding together all of the faces.

Well this is a rectangular-based pyramid, which means there are two pairs of congruent triangles.

The opposite triangular faces are equal, because they're coming from the same edge on the rectangle, and a rectangular base.

So the rectangular base is the 12 x 8, and then we have 12 times the square root of 4 squared plus H squared.

And this one is the face that we can sort of see along the 12 centimetre edge.

And then we have, plus 8 times the square root of 6 squared plus H squared.

If we look at the 12 times the square root of 4 squared plus H squared, that is using the triangle that we can see, the right-angle triangle, that the 4 is from the 8 centimetres and the H is our perpendicular height.

We are trying to work out the length of the altitude of that Isosceles Triangle, the hypotenuse for that right angle triangle.

And so we are using Pythagoras's Theorem.

We know that there are two congruent faces and therefore, we would do, 12 times the square root of 4 square plus H squared divided by 2 gives us the area of one of them, but if we're gonna double it, then the timesing by 2, and the dividing by 2 is equal to 1.

So that is why there is no halfing or doubling involved in this line of calculation.

The surface area for this pyramid is 237.

1 square centimetres to one decimal place.

So to summarise today's lesson on applying trigonometric geometric ratios in 3D, if angles are involved then you may wish to apply trigonometric ratios to the 3D problem.

This allows you to calculate the angle between the line and the plane.

And it also can be used to find missing lengths if the angle is known.

So we're looking for right-angle triangles.

And really think of Jun, "If you can find a right-angle, then you can find a right-angle triangle," and if there is some other angle involved, then trigonometric ratios are useful.

Really, well done today, and I look forward to working with you again in the future.