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, the learning outcome today is to be able to use our knowledge of right-angled trigonometry to solve problems. On this screen, we've got keywords that will be used during the lesson.

They're not new keywords, you've met them before, but we'll go through them now so that we are familiar before we make a start on the lesson.

So, trigonometric functions are commonly defined as ratios of the two sides of a right-angle triangle for a given angle.

And those trigonometric functions that we're talking about here are sine, cosine and tangent.

The sine of an angle is the y-coordinate of point P on the triangle formed inside the unit circle.

You can see with the diagram on the slide the P that it is referenced into.

The cosine of an angle is the x-coordinate of point P on the triangle formed inside the unit circle.

Again, that point P is on that diagram.

A tangent to a circle is a line that intersects the circle exactly once.

And the tangent of an angle is the y-coordinate of point Q on the triangle, which extends from the unit circle.

And again, that is referring to the point Q on that diagram.

It may be that you do wish to pause the video, read that and refer to the diagram to make sure you're happy before we move on with the lesson.

So, this lesson, which is advanced problem solving with right-angled trigonometry, is split into two learning cycles.

The first learning cycle is using trigonometry to solve problems in 2D.

And then, we're gonna move into the second learning cycle where we're going to use trigonometry to solve problems in 3D.

So let's make a start at looking how trigonometry, specifically right-angled trigonometry, can help us with problems that are in the two dimensional space.

So, Andeep has said, "It's time to solve some more challenging questions using trigonometry.

I hope you're ready for this." So, here we have a tilted square, ABCD, and Andeep says, "Let's assume the grid is unit squares.

We can find the area of the square ABCD, but what makes this difficult?" So, pause the video and think about that.

What challenges might there be to find the area of this tilted square? Press play when you're ready to move on.

So, the challenges are that it's hard to measure the length of AB because it doesn't sit nicely on the grid lines.

And by measure, we mean to calculate rather than use a ruler.

Andeep has given some helpful advice, that when we're working with a geometry question, it's often helpful to draw extra lines and it will help you see more information and clues.

So these extra lines have been drawn onto the gridded background.

They have boxed in the tilted square, ABCD.

So, does this help when trying to find the area of the square? Pause the video and think how this additional information can help you find the area of ABCD.

Press play when you want to look at a solution.

So, one solution is to use this boxed square.

Because this square is orientated along the grid lines, it's very easy for us to count and work out the edge length when we've assumed that the grid is unit squares.

So, we have a square of six by six, which would have a total area of six squared, which is 36.

But what we are trying to actually calculate is the area of the tilted square which is within it.

So, we need to remove some additional area, and that additional area is created by four congruent right-angle triangles.

And we know how to calculate the area of a triangle, it's base times perpendicular height divided by two.

So, what's the area of one triangle? Well, four times two divided by two.

But we have four congruent right-angle triangles in this diagram, so that's why we'd multiply it by four.

And so, the area of the four congruent right-angle triangles is 16.

And so, therefore we can subtract that from the area of the boxed square and that gives us 20.

So, the area of ABCD is 20 square units.

So in this check, which two methods can help find the area of square ABCD? So, there is another method that we could have used instead of the one we just spoke through.

So, pause the video and when you're ready to check, press play.

So, A and C.

A is the method we just looked at where we boxed in ABCD, found the area of that larger square and then subtracted the area of the surrounding triangles.

But alternatively, because we've got perpendicular lengths and we've got these right-angled triangles, we could have calculated the hypotenuse of one of them, which is equal to the length of the square ABCD, and then squared it to find the area.

And that would've given us the same value of 20 square units.

Here's another example with a tilted square and that's square that's boxing it in.

But Alex says, "In this example, we've got different information.

Can we find the area of this square?" So, pause the video and think about what the different information that you have here compared to the previous example with Andeep.

Press play when you're ready to move on with working out the area.

So differences, it's not on a gridded background, but we do have an edge length of the box square, but we don't have the full edge, we've got part of it.

We also have an angle and that angle is from the horizontal edge to one of the tilted square's edges.

We can see that that is 56 degrees.

So, how can we make use of the information we have here to work out the area of that grey ABCD square? Well, Alex says, "We have the length of the side opposite the angle." So, we can see that that's 3.

316 centimetres.

"And depending on the approach, either EB or AB would be useful." So what Alex means in terms of the approach is the method to find the area.

So, if we were going to find the length of EB, how would we do it and what approach would you be needing that length for? So, pause the video and think about those two questions.

How would you calculate the length of EB? And then, which of the two approaches would it be necessary to know the length of EB? Press play when you're ready to listen to the answer.

So, EB could be calculated with the given information by using the tangent ratio because EB is the adjacent edge to the 56 degree angle and 3.

316 is the opposite edge.

So we could use the tangent ratio to calculate EB.

If we calculate EB, then we're probably going to end up using the method of finding the area of the larger square and subtracting the area of the four congruent right-angle triangles to leave us the area of ABCD.

So now, let's think about AB.

The other approach is working out the edge length of the square and squaring it, and AB is the edge of the square.

So which trigonometric ratio will you use to calculate AB? Pause the video whilst you think about that, and then when you're ready to move on, press play.

AB is the hypotenuse of the triangle AEB, and 3.

316 is that opposite edge to the 56 degree angle.

So, opposite and hypotenuse are in the sine ratio.

So, we can use sine of 56 degrees to calculate the length of AB, the hypotenuse.

So AB equals 4.

0 to one decimal place.

And you can see that we've rearranged the formula for sine.

So, sine of 56 is equal to the opposite over the hypotenuse.

We have the theta, we have 56 as our theta, our angle, and we have our opposite length.

So, we rearrange that to make AB the subject because that's what we wish to calculate and it comes out as 4.

0 to one decimal place.

Alex says, "If AB, which is the edge of the square, is four centimetres, then the square must have an area of 16 square centimetres." Four squared is 16.

So on this check, where is a useful right-angle triangle to use trigonometry in this picture? Pause the video and think about that before you press play and I'll show you where I think a useful right-angled triangle would be.

So here, and often we call this dropping the perpendicular.

So from that vertex, we are dropping a perpendicular line to the base.

That happens to be the perpendicular height, and in this case it's an altitude.

How is this helpful for trigonometry? Well, we do need to be mindful here.

We haven't got enough information to be able to do much more with this diagram.

If we had the angle between the five centimetre and the eight centimetre edge, then we would be able to use trigonometry using one of those right-angled triangles to calculate the perpendicular height of the triangle and to also work out the adjacent edge, which is partitioning the eight centimetre.

What we shouldn't do is assume that this is an isosceles just from the way it is drawn.

We haven't got enough information to know that for certain.

However, if we did know it was an isosceles, then we would know that that perpendicular height, when we have dropped that, that altitude would be the line of symmetry and it would split the base edge into a four centimetre and a four centimetre.

And then, we would be able to calculate the perpendicular height.

So until we've got some further information, we cannot do much more.

But here is a way of finding some right-angled triangles which will be necessary for any trigonometry that we wish to do.

So up to the first task of the lesson, and in this task, we've got three questions.

So, this first question is on the screen, you've got two parts to it, part A and part B.

In both cases, you are calculating the area of the square.

So, pause the video whilst you work through question one, and then when you press play, we'll move on to question two.

Here's question two.

Find the area of this circle with diameter AB in terms of pi.

So, pause the video and work out the area, make sure you're leaving it in terms of pi.

And when you're ready for the final question of this task, press play.

So here, we have question three, the last question of this task.

So, find the combined area of these two shapes.

So, you can see this as a composite shape that's created from an equilateral triangle and a square.

So, pause the video and whilst you're working that out and then when you press play, we'll go through all of the answers to task A.

Question one was where we were dealing with the tilted squares.

Obviously there was a couple of methods that you could use and we discussed those in the learning cycle.

But for 1A, the area of ABCD, that square, is 17 square units.

The method that I'm making use of here is looking at a right-angled triangle, where the hypotenuse is the length AB.

I can look at how the vertical change between A and B and the horizontal change.

So, the square root of four squared plus one squared is equal to the square root of 17.

So that's used in Pythagoras's theorem.

Then we can square that to get our area.

On question part B, EFGH, E, F, G and H were given just with their coordinates.

So, we didn't have a diagram.

It may have been that you decided that you wanted to sketch that out, but that wasn't actually necessary.

So, you needed again to work out the length of a line segment between adjacent vertices.

GH is the one that Andeep suggests is the easiest one to calculate.

I think that's because both the x and y-coordinates are positive.

So, how did he get the one and how did he get the four? Well, he is looking at the difference between the x-coordinates, so that's one, and the difference between the y-coordinates, which is minus four.

But because we are looking at this as a length of a right-angled triangle, then we're gonna take the modulus of that.

And so, we'll take the positive four.

It's the change the length between them.

So, the square root of one squared plus four squared, we've seen before, is the square root of 17, and then we can square it because the line segment GH is one of the edges of the square.

It's important that you didn't find the length for line segment of non-adjacent vertices.

The square was labelled EFGH, so E would be linked to F, then F would be linked to G, then G would be linked to H and then H would be linked back to E.

That would be the edges of the square.

On question two, you need to find the area of the circle and you were told that the diameter was AB.

You were leaving it in terms of pi.

There were two edges that were four centimetres and they were marked with hash lines, so we knew they were equal to each other.

So the four centimetre is the adjacent edge to the 65 degree angle.

AB was the hypotenuse of a right-angled triangle and therefore we could make use of cosine, the ratio of adjacent and hypotenuse.

Once we calculate AB, that is the length of the diameter and you can see that on the screen, we would then need to half it to have our radius.

The radius is what we needed to find the area of a circle.

And then, pi times radius squared was 22.

4 pi square centimetres.

The rounding has happened at the final part in terms of the multiplier of pi.

On question three, you needed to find the combined area of these two shapes.

We knew that it was an equilateral because of the hash lines that indicate that all three edges of the triangle were the same length and it was also a square rather than a thrombus.

The edges were all equal, but we also had our right angle markers.

The only distance we were given was the diagonal distance of the square, and that was given as root two.

We need to consider a right-angle triangle.

So, the right-angle triangle that we are gonna consider is the diagonal of the square and the two edges.

So, it's actually a right-angled isosceles because of it being a square.

Then we can use Pythagoras's theorem to work out the edge length, and that was one because root two squared would be equal to x squared plus x squared, where x is the unknown edge length.

That would be the same as saying that two is equal to two x squared, therefore x squared is equal to one and x would be equal to one.

So, we now know that the edge length and every length on that diagram that's marked with a hash line is one, which means the area is one of the square, and then we can go to our equilateral triangle.

There's a couple of ways that you could have done this part.

So, our equilateral triangle has got all edges of one unit or one centimetre.

If we drop the perpendicular, if we put an altitude, because it's an equilateral triangle, that will bisect the edge that it meets and also the angle.

So, we can use Pythagoras's theorem to calculate the perpendicular height of the equilateral triangle, which is root three over two.

Can you remember where you've seen that value before? I'm hoping you remember that that is one of the exact trig values.

And the reason that we could have used the exact trig values here is because of the equilateral triangle, the interior angles are 60 degrees.

Then if we look at that right-angled triangle with the hypotenuse of one, there would be an angle of 60 degrees.

And so, the perpendicular height, which is the opposite to the 60 degree angle, means that sine of 60 degrees is equal to the opposite.

And sine of 60 degrees is equal to root three over two.

It's one of our exact trig values.

So we could use Pythagoras's theorem like you can see on the screen, but we could have just remembered and recalled or worked out and derived that the sine of 60 degrees is equal to root three over two.

So, that's our perpendicular height.

We need to get the area of the triangle.

So, our triangle has a base of one, a perpendicular height of root three over two, and now area of a triangle is half times base times perpendicular height.

So, this is root three over four.

Remember you were trying to find the combined area.

So, add them together, one add root three over four is 1.

433 if you took it out of exact form, but you could have also equated that to four plus root three all over four.

So, up to the second learning cycle where we're going to look at using trigonometry to solve problems in 3D now.

So, Andeep says, "Now we're working in three dimensions.

Can I find the volume of this cylinder?" So, a cylinder we can see on the screen and we've got a diagonal distance from the centre of the circular face to the circumference of the opposite circular face.

So firstly, how do you work out the volume of a cylinder? Pause the video and try and remember how we work out the volume of any cylinder regardless of this information.

Press play when you're ready to move on.

So, to work out the volume of any cylinder, you do pi r-squared, which gives you the area of the circular cross-section and multiply it by the perpendicular height.

So here, we've got a distance from the centre of the circle to the base of the cylinder diagonally.

So, do we have enough to work the volume out? Do we know the radius? No.

Do we know the perpendicular height? No.

So currently, we cannot work out the volume of the cylinder.

Andeep has given us some further information.

So, the radius has been drawn between the centre and a point on the circumference that's directly above the point on the circumference on the base and has given us the angle between that and the diagonal, 70 degrees.

So, can we now find the volume? Can we work out the radius from that information? Can we work out the perpendicular height from that information? Pause the video and think about that.

Maybe have a go.

Press pray when you're ready to move on with that.

So, if we label this ABC as a right-angle triangle, then angle BAC is 70 degrees.

The line segment AB is the adjacent to the 70 degrees.

AC is the hypotenuse.

BC is the opposite to the 70 degrees.

So, AB can be calculated.

Because it's the adjacent to the 70 degree angle and we have the hypotenuse, then we're gonna be working with the cosine ratio.

Nine times cosine of 70 is equal to 3.

07 approximately.

What does AB represent on the cylinder? Well, AB is the radius and we know we need the radius to work out the volume.

BC equals nine times sine of 70 and that's 8.

45 approximately.

And that's because BC is the opposite to the 70 degree angle.

What's an alternative way that we could have calculated BC? Pause the video and think about that.

Press play when you're ready to check.

So, the alternative way having worked out AB was we could have used Pythagoras's theorem.

We would then have had two lengths of a right-angle triangle, and so we could work out the third.

If you were to do that though, it'd be really important to make sure you are using the exact value of AB, nine cosine 70, and not a rounded value.

So, we can now work out the volume of our cylinder because we've got the radius and we have the height, the perpendicular height.

So pi r-squared times by that perpendicular height, and you can see in the working that we are using the exact value.

The dot-dot-dot indicates we are not just using a rounded value to two decimal places or three significant figures, we're going to use the exact value.

We'll be able to do that by using the calculator and that is equal to 251.

7 cubic centimetres to one decimal place.

So, try and leave the rounding to the very final answer.

So here, we've got a check, we've got a cuboid.

What combinations of letters form right-angle triangles? So, look at this and think about which letters would create a right-angle triangle.

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

So, there are some examples on the screen.

There are quite a few right-angle triangles that could be formed.

So ADC is a right-angle triangle that is on the face, the top face of the cuboid.

ADF is a right-angle triangle on that left-hand face of the cuboid.

AFG is a right-angle triangle on the back face and it's making use of the fact that these are rectangular faces.

AFH is also a right-angle triangle going across the diagonal of the cuboid and that's making use of the rectangular cross-section AFHC.

But as I say, there were many more that you could have found.

But now up to the final task of the lesson, task B, and here is question one.

So, find the volume of the cuboid if AB equals 25 centimetres, AH equals 32 centimetres and angle FAH is equal to 60 degrees.

It may be that you wish to add those dimensions and the angle onto the diagram.

Pause the video and have a go at that question.

When you press play, we'll be going through its answer.

So, we couldn't calculate the volume from the given information immediately because to work out the volume of the cuboid, you need the length, the width and the height.

You need three perpendicular measures.

AB and AH are not perpendicular to each other.

So, this triangle AFH is the triangle that we're gonna need to start with in order to move forward in this question.

AH was given to us and it would be the hypotenuse.

Angle FAH was given as 60 degrees and FH is the diagonal across the base face.

So, what can we work out from this triangle? Well, you can work out AF and you can work out FH.

So AF would be 32 times cosine 60, which is equal to 60.

And that's because cosine 60 is equal to a half.

It's one of our exact trigonometric values.

And FH, we could then calculate using Pythagoras's theorem, but you could have also used the sine ratio to work that out.

And FH is 16 root three.

So now, we have AF and the diagonal across the base of the cuboid.

AF is one of the lengths that we do need to find the volume.

So now, we've got the height of the cuboid, we've also got the width of the cuboid AB.

So, what we are needing is one of the lengths that is the depth.

So GH or BC or AD or FE, they're all equal to each other, but we need to know that measurement in order to work out the volume.

So if we focus on getting BC, well, 16 root three is the length of FH.

There is another line segment that also has that length and that is AC.

Because of the congruent faces of a cuboid, then that FH is equal to AC and we can then make use of the right-angle triangle on that top face.

So, a 16 root three is the hypotenuse, 25 is one of the short sides and we are looking to work out the third edge.

So, we'd make use of Pythagoras's theorem.

So, 16 root three all squared.

And if you're using the calculator, it's really important that you are putting brackets around that.

<v ->25 squared</v> and then square-rooting in it gives us the length of BC.

We then have three perpendicular dimensions to calculate the volume of this cuboid.

And that would be AB times BC times AF.

AB was given, AF we calculated and BC we calculated.

To one decimal place, it's 4,783.

3 cubic centimetres.

So, to summarise today's lesson on advanced problem solving with right-angled trigonometry, there is times where the answer may be best left in an exact form and we especially should be trying to leave our answers in exact form as we work through a solution and try to only round at the very final stage.

When dealing with right-angled trigonometry, it's important to look at what information you have and to think about what you can deduce.

Because often, especially in a problem-solving style question, you're not going to do one step to the answer.

It might be that you need to work out another piece of information and then go forward from that piece of information closer to what you're trying to work out.

Consider whether Pythagoras's theorem or trigonometric ratios are more efficient to use.

So, because they are both applicable to right-angle triangles, sometimes you could use either of them and it might be more efficient to use one than the other.

Well done today and I look forward to working with you again in the future.