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

They're nothing new, you've learned them before, but let's just take a look at them together before we start the lesson.

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

And those trigonometric functions are normally sine, cosine and tangent.

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

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

So you can see the point that it's referencing on the diagram on the slide.

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

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

Again, point Q is on that diagram.

So I would suggest you to pause the video and reference that diagram to the information that I've just read out before we move on with the lesson.

So this lesson on problem solving with right-angled trigonometry is going to be split into two learning cycles.

The first learning cycle is looking at using Pythagoras' theorem to solve problems, and then we'll move on in the second learning cycle to using cosine, sine, or tangent to solve problems, so the trigonometric ratios.

So let's make a start at looking at problems where Pythagoras' theorem is going to be really helpful.

So, Andeep says, "It's time to solve some more challenging questions using trigonometry!" So here we've got a shape on a grid background.

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 just pause the video and look at that diagram.

How would you calculate the area of ABCD, knowing that it's a square, and that the grid background is unit squares? Press play when you're ready to continue with Andeep.

So, Andeep says, "It's hard to measure the length of AB, because it doesn't sit nicely on the grid lines." It's not horizontal, nor is it vertical.

Remember, because it's a square, AB is equal to BC, and CD, and AD.

So, although Andeep is discussing the line segment AB, he could be discussing any of the other three as well.

So, "In geometry," Andeep says, "It's often useful to draw in extra lines to help if you can see more information and clues." So he has done that just here.

Does this help to work out the area of ABCD? Pause the video and have a look.

Do you have a method or a strategy now to work out the area of that square? So did you think about doing this? Andeep has calculated the length of the line segment AB, because that is the edge of the square.

And he said, "Keeping my answer as a surd makes finding the area of the square easier, but also more accurate," because we know that a surd is irrational, so the only way we can write that as a decimal is if we round it, and we want to try and be as accurate as possible.

So he's used Pythagoras' theorem, because what he added, the extra lines that he added, were on the grid lines, and those grid lines are perpendicular.

So he is constructed right-angled triangles, and made use of Pythagoras' theorem to get the length of AB.

So what's the area of the square? The area of the square, ABCD, is therefore the square of the length, AB.

Two root five all squared is 20.

So if you had given a rounded value for AB, then your area would be different, even though the area is actually exactly 20 square units.

So that's another reason to really ensure that we try not to round too early in our solutions.

Andeep says, "Often there are multiple approaches to a problem.

Can we solve this without finding AB?" So without finding the length of AB, the length of the square, can you still find the area of the square ABCD? So pause the video and really ponder that.

Maybe that's the method you were thinking of anyway.

How can we get the area of the square ABCD without calculating the edge length? Press play when you're ready to have a look.

With the given information, the additional lines that Andeep drew, we can also say that this larger square that sort of surrounds ABCD, has an edge length of six, and we're just using our gridded background to get that value.

Andeep goes on to say that "The four triangles are equal," they are congruent right-angled triangles, "And we have their measurements for their base and the perpendicular height." So by having the base and the perpendicular heights of the triangles, we can work out the area of each of those four right-angled triangles, base times perpendicular height divided by two.

But because there are four congruent, four identical, then we'll multiply it by four to get the total area of the four triangles.

However, those four triangles are not what we're trying to calculate.

We're trying to calculate the area of ABCD.

So what's this calculation 6 squared minus 16? The value is 20, the answer is 20, and we know that that is the area of ABCD, but why has Andeep subtracted 16 from 6 squared? Pause the video, and can you see what that calculation is doing? Press play when you're ready to carry on.

So 6 squared is the area of the larger square that boxes in or surrounds ABCD.

He has then subtracted the area of the four triangles and that will leave, or the remaining area will have to be square ABCD.

So here's a check: Which two methods can help find the area of square ABCD? Pause the video, read through your options, and make a decision.

When you're ready to check, press play.

So A, where we box in ABCD and subtract the surrounding triangles, or C, box in ABCD, and use Pythagoras' theorem to find a side length for the square, and then you'll square it to get the area.

So we're onto the first task of this lesson, which is using Pythagoras to solve problems. So there's one question, but two parts.

So on Part A, you need to find the area of the square, ABCD, and you can see it on a grid background.

For Part B, we've got square EFGH where you've been given the coordinates of each of those vertices.

So pause the video and work out the area for each of those squares.

When you press play, we'll go through our answers.

So the answer to A, is 17 square units, and the answer to B, is also 17 square units.

Let's take a look at Part A to begin with.

We can use one of the methods that Andeep spoke through.

We can box in this tilted square and use Pythagoras' theorem to work out the length of an edge so using the square background we can say four units by one unit is our right-angled triangle, so 4 squared plus 1 squared and then square rooted gives us the hypotenuse, and that is AB.

That's the edge of a square, so we can square that value to give us the area.

So the area of ABCD is 17 square units.

On Part B, you needed to once again find the length of the line segment between two of those vertices, because that would be the edge of the square.

It wouldn't matter which pair you chose as long as they were in the order of the vertices.

So EFGH means that it goes from point E to point F, then from point F to point G, then point G, then to point H, and then from point H back to E.

If you'd changed the order, then you would've got a wrong answer I'm afraid.

So, GH is the one that I've used, and why did I use that? Because it's the easiest length to calculate, according to Andeep.

So if you look at the difference on G to H, the x-difference is one, and the y-coordinate, the difference, the change there is four.

So, 1 squared plus 4 squared, and then we get square root 17 once again.

So if we square it, we get our area.

So we're now at the second learning cycle where we're still gonna work with problems, but this time we're gonna make use of our trigonometric ratios, sine, cosine and tangent.

So Alex says, "In this example we have different information." So we've got the same concept of a tilted square, and we've boxed it in with a sort of a square that's sat on a horizontal.

"Can we find the area of this square, the grey square?" Well, what information do we have? We have the angle EBA, which is part of a right-angled triangle, and we also have the length of the line segment EA, 3.

316, which is the opposite to that angle, 56 degrees.

So we have the length of the side opposite the angle, and depending on our approach, either EB or AB would be useful.

If we calculate EB, that's when we're going to look at working out the area of the surrounding square and subtract those four congruent right-angled triangles.

Whereas if we work out AB, which is the hypotenuse of the right-angled triangle, it's also the edge length of the square that we're trying to calculate the area for, so we can square it.

So depending on which method you use, depends on which edge you would like to calculate.

So if we look at finding AB, the edge length, then we can see that EA, the given length, is the opposite, and AB is the hypotenuse.

So we'd be using the trigonometric ratio of sin.

Sin of 56 degrees is equal to the opposite, which in this case is 3.

316, over the hypotenuse, which is the unknown, the thing we wish to calculate.

If we rearrange that, then AB, the hypotenuse, is equal to the opposite 3.

316 divided by sin of 56 degrees, and that comes out as 4.

0 to one decimal place.

So now we have the edge of the square.

As Alex says, "If AB is 4 centimetres, then the square ABCD must have an area of 16 square centimetres," because 4 squared is 16.

So in this check, where is a useful right-angled triangle to use trigonometry in this picture? Pause the video, and when you're ready to check, press play.

If we drop the perpendicular or the altitude, so a line that passes from the vertex to the opposite edge and it's perpendicular to it, then we have created right-angled triangles.

How do we go further with this? Well, we'd need some further information.

If we knew it was an isosceles triangle, then we could use Pythagoras' theorem to work out the length of that altitude, which would also be the perpendicular height.

If we were given an angle, an interior angle, the one that's between the five centimetre edge and the eight centimetre edge, then we could use trigonometry to work out the height, the altitude length, and we'd be able to work out where it partitions the base.

This diagram looks to be isosceles, it looks to be symmetrical, but we can't assume that, we don't have the information that tells us that.

So when we have found ourselves with a triangle, yes we can drop the perpendicular or draw on an altitude, but be careful that you do not assume that that is going to bisect the edge.

It will only bisect the edge if you know it is an equilateral triangle or if you is an isosceles triangle.

But, regardless, if you find yourself with the right-angled triangle, then you can think about using Pythagoras' theorem or one of the trigonometric ratios.

We're up to the last task of the lesson.

And so, Question One, find the area of this circle with diameter AB in terms of pi.

So look at the diagram.

You've got AB, knowing that AB is the diameter, how is that gonna be helpful to work out the area of the circle? So pause the video, and then when you're ready for the next part of Task B, press play.

So Question Two, the final question of Task B and this lesson, find the combined area of these two shapes.

So we can think of this as a composite shape that is created from an equilateral triangle and a square.

How do I know it's an equilateral triangle? Well, it's got the hatch marks that indicate that all three edges are equal in length.

And then we have our square, we've got all four edges are equal, and the right-angled markers to indicate it's a square rather than a rhombus.

We've also been given a distance across the square, so the diagonal, and we know that that is root 2.

So pause the video, and work out the area of this composite shape by combining the area of the equilateral triangle and the square.

Press play when you're ready to go through the answers of Task B.

So, in Question One, you needed to work out the area of the circle.

So first of all, how do you work out an area of a circle? Well, it's pi times radius squared.

We're gonna leave our answer in terms of pi, because that's what the question stated, and that's the most exact form.

So what I need is the radius.

Well, we can see from this diagram that AB is the diameter, and we have a relationship that the diameter is two times the length of the radius for that circle.

So we need to work out that length AB, and then we'll have the radius.

AB is the hypotenuse of the right-angled triangle with an adjacent edge of four centimetres to the 65-degree angle.

So we're gonna use cosine as our trigonometric ratio to work out the hypotenuse.

And so, AB, the diameter, is 4 divided by cosine of 65, that's 9.

46 roughly, but we're gonna try and use the most exact answer by using our calculator.

But the diameter is not what we actually need, we need the radius, so we're gonna divide that by two.

The radius is therefore 4.

73 approximately.

So the area of the circle is pi times the radius squared, which is 22.

4 pi rounded to three significant figures.

But in terms of pi.

On Question Two, we were looking to find the combined area of the two shapes.

So we start by looking at the square.

The square has the diagonal length of root 2, and we can make use of the fact that it is a square, and therefore this right-angled triangle is an isosceles right-angled triangle to calculate the edge length of the square.

So, we would know, let's call it A, A squared plus A squared equals root 2 squared.

So 2A squared equals 2, therefore A squared equals 1, therefore A equals 1.

So now we have the length of the edge of the square, but that is not only the edge of the square, it's also the edge of the equilateral triangle.

So the area of the square would be 1 times 1, which is 1 square centimetre.

So now we need to focus on the equilateral triangle.

While the equilateral triangle will also have edges of one, as I said, because they are joined along an edge, an equilateral triangle also has a 60-degree angle.

As we saw earlier, that with an equilateral triangle, when you drop the perpendicular, because it's along a line of symmetry, it will bisect the edge as well as the angle.

So we can make a right-angled triangle from that equilateral triangle, and get the perpendicular height.

Why is the perpendicular height important? Because we need to get the area, base times perpendicular height divided by two.

So this is one way of getting that perpendicular height, which is root 3 over 2.

I'm hoping that you actually notice that this is an exact trig value that we've met before, and that's because we've got that 60-degree angle within that right-angled triangle.

So we could have said that the hypotenuse is one, the angle is 60 degrees, and we are trying to calculate the opposite.

So sine of 60 degrees is equal to root 3 over 2.

It's one of our exact trig values.

So that's another way that you could have got the perpendicular height of the equilateral triangle.

So what's the area? Well, it's gonna be 1/2 times the base times the perpendicular height.

The base of our equilateral triangle is one, the perpendicular height is root 3 over 2, we've just calculated that.

So 1/2 times root 3 over 2 times 1 is root 3 over 4.

So now we have the area of the square, the area of the triangle, we can add the two together.

So the total area is 1 plus root 3 over 4, but that's equivalent to 1.

433 approximately.

So to summarise today's lesson on problem solving with right-angled trigonometry, sometimes an answer may be best left in an exact form, and often that's the case in the middle of your solution.

You really do not want to put rounding error in the middle of your calculation, so try and wait until the very end before you add any degree of accuracy.

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

And you can also consider whether Pythagoras' theorem or trigonometric ratios are more efficient to use.

So as we just saw in that last one, we could have used Pythagoras' theorem, but also an exact trig value was available too.

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