Lesson video

Hello there, you made a great choice with today's lesson.

It's gonna be a goody.

My name is Dr.

Rolandson, and I'm gonna be supporting you through it.

Let's get started.

Welcome to today's lesson from the unit of trigonometry.

This lesson is called scaling the right-angled triangle from the unit circle, and by the end of today's lesson, we'll recognise the right-angled triangle within a unit circle, and use proportion to scale it to similar triangles.

This lesson will introduce some new keywords.

These words will be explained in more detail during the lesson, but if you like to pause the video and read them now before we start, feel free to do so.

Here are also some previous keywords that will be useful during today's lesson.

If you need to remind yourselves what these words mean, feel free to pause the video while you read them again.

This lesson contains three learning cycles, and we're going to start by learning to label the sides of a right-angle triangle.

The longest side of a right-angled triangle is called the hypotenuse, and we can see that labelled on the diagram here.

How could we refer to the other two sides in the triangle? Well, Yun has some ideas.

Yun says, "We could call one the 'vertical edge' and the other the 'horizontal edge,'" like we can see in the diagram here.

What problems could there be with Yun's idea? Pause the video while you think about this, and press play when you're ready to continue.

Laura says, "That won't work because triangles can be in different orientations.

None of the edges in this triangle are horizontal or vertical." So what other ideas does Yun have? Well, as the hypotenuse is the longest edge, Yun says, "Could we call one the 'middle length' and the other the 'shortest length' like we can see in the diagram here?" What problems could there be with Yun's idea this time? Pause the video while you think about it, and press play when you're ready to continue.

Laura says, "That won't work because the angle affects which one of the perpendicular lengths is the shortest." Currently, we can see that the vertical edge is the shortest one, but now it's not.

Now it's the horizontal edge, which is the shortest one.

"The two perpendicular edges can also be the same length when it's an isosceles right-angled triangle." So what else could we do? Well, Yun says, "The hypotenuse is opposite the right angle.

Could we use the other marked angle to refer to the other two edges?" Laura thinks about this and says, "Hmm, one of the edges is opposite the marked angle, and the other edge is adjacent to both the angles we can see the right angle and the 35 degrees." And this is pretty much how we go about labelling a right-angled triangle.

In this triangle here, we can see the right angle and one of the acute angles has been marked.

We can use these angles to label all three of the sides.

The hypotenuse is always the longest side, and the hypotenuse is always opposite the right angle.

We can refer to the other two sides of the triangle based on their position in relation to the marked angle, the acute one.

The side which is opposite the acute interior angle that is marked in the triangle is called the 'opposite,' and the side which is next to both the right angle and the other marked interior angle is called the 'adjacent.

' Now the hypotenuse is always the longest side, but the length of the other two sides do not affect which way round the opposite and adjacent are labelled.

These labels only depend on the position of the acute angle that has been marked.

So with the diagram on the left, here's our opposite and adjacent, and with the diagram on the right, here's our opposite and adjacent.

It doesn't matter that for one of these diagrams the opposite is short of the adjacent, and for the other diagram, the opposite is longer than the adjacent, length does not affect which way round they go.

Also, the orientation of the triangle does not affect which way round the opposite and adjacent are labelled.

These labels only depend on the position of the acute angle that has been marked.

So with the diagram on the left, the adjacent is horizontal and the opposite is vertical, it won't always be like that.

With the diagram on the right, they're all diagonal.

The important thing is the position of the acute marked angle 'cause the position of the marked angle does affect which way round the opposite and adjacent are labelled.

Switching the acute angle that is marked causes the labels for opposite and adjacent to switch around too.

So with the diagram on the left, the acute angle which is marked is in the bottom left-hand vertex.

But with the diagram on the right, it's a top-right vertex and let's see how that affects the opposite and adjacent, they're the other way around.

So let's take a look at how to do this together.

Label the sides of this right-angled triangle.

Alex is gotta help us with this.

Alex says, "The hypotenuse is the longest side, the side not touching the right-angle." So this is the hypotenuse.

"The opposite is a side opposite the acute angle that is marked." So this one's the opposite.

"And the adjacent is a side that is next to both the right angle and the marked or focus acute angle." So this is the adjacent, but what if we mark the other acute angle instead, like this? How would this affect how we label the sides? Well, Alex says, "The hypotenuse hasn't changed, but the other two may have changed.

The opposite is the side which is opposite the acute angle that is marked." So now this one's the opposite.

"And the adjacent is the side that connects the right-angle with the acute angle that is marked." So now this is the adjacent.

Alex says, "Switching the acute interior angle that is marked switches the opposite and the adjacent." So let's check what we've learned.

The opposite is always the shortest side of a right-angled triangle.

Start by pausing the video, choosing true or false, then press play to continue with this question.

The answer is false.

Let's now look at some possible justifications for that answer.

A, the side which is labelled opposite is always opposite the acute angle that is marked.

And B, the side which is labelled opposite is always opposite the right angle.

Which of those is the crept justification? Pause the video while you make a choice and press play when you're ready to continue.

The answer is A, the side which is labelled opposite is always opposite the acute angle that is marked.

Here we have a right-angled triangle with the edges labelled A, B, and C, which side is the opposite in this triangle? Pause the video while you choose and then press play when you're ready for an answer.

The answer is C.

Something about this triangle has change now.

So which side is the adjacent in this triangle? Pause the video while you choose and then press play when you're ready for an answer.

C.

Which side is labelled in the right-angle triangle below? Pause the video while you write the word down and press play when you're ready for an answer.

The answer is hypotenuse.

Okay, it's over to you Now for Task A, this task contains one question and here it is.

We have eight right-angled triangles, and you need to label the sides of each triangle with the words "opposite," "adjacent," "and hypotenuse".

In parts (g) and (h), you'll notice that one of those sides is labelled for you.

, so in those questions, you need to mark the appropriate angle with θ.

Pause the video while you do this, and press play when you're ready to go through some answers.

Let's see how we got on with this.

Each time I display the labels, check 'em against your own.

Part A, Part B, Part C, Part D, Part E, Part F, Part G, and Part H.

Great, now let's learn how to use sine and cosine for triangles within the unit circle to find the missing lengths on similar triangles.

The diagram shows a right-angled triangle inside a unit circle, and here we have Jacob who has drawn a similar triangle.

How does Jacob know it's similar? Pause the video while you think about this, and press play when you're ready to continue.

Jacob says, "They are similar because they have the same angles." We can see that both of these triangles have a right angle and an angle that is 37 degrees, The third remaining angle of each triangle would also be the same.

The two right-angled triangles are similar.

So how could Jacob use the unit circle to find the missing lengths on his triangle? Well, let's do this together.

Jacob says, "For the triangle in the unit circle, the hypotenuse has a length of 1 unit." We can see that here.

"The length of the opposite is 0.

6." We can see that here.

"The length of the adjacent is 0.

8," which we can see here.

Now as we have two similar triangles here, Jacob says, "I could find the missing lengths on my triangle by using a scale factor." We have the length of the hypotenuse in each of these triangles, so we could use that to get the scale factor of 5.

So let's apply that scale factor to the other two sides of the triangle.

0.

6 times 5 is 3, and 0.

8 times 5 is 4, and we're working with centimetres on Jacob's triangle.

So that's one way we can do it by using the scale factors between the triangles, or we could find the missing lengths by using their own proportions.

With the triangle on the left, to get from the hypotenuse to the opposite, we multiply by 0.

6.

So if we do the same with the triangle on the right, the length of the opposite would be 3, and we're working with centimetres.

With the triangle on the left, to get from the hypotenuse to the adjacent, we times by 0.

8.

So if we do that with the triangle on the right, we get 4 centimetres.

For any right-angled triangle, a similar triangle can be made inside the unit circle, like so, and we don't necessarily need the whole circle because the first quadrant is the only part of the unit circle where θ is equal to the interior angle of the triangle.

So we can just use this part.

Also, the first quadrant is the only part of the unit circle where both the x and y-coordinates are equal to the lengths of the triangle because they are positive here.

And let's now remind ourselves a few properties of this triangle inside the unit circle.

The length of the hypotenuse is always 1 unit.

The sine of the angle, sin(θ) is the Y-coordinate of the point where the radius has been rotated through that angle.

So for the triangle inside the first quadrant of the circle, the length of the opposite is equal to sin(θ).

The cosine of the angle, cos(θ), is the x-coordinate of the point where the radius that's been rotated through that angle, and for the triangle inside the unit circle, the length of the adjacent is equal to cos(θ) This is helpful because with these two triangles being similar, we can use a triangle on the left to help us find missing lengths with the triangle on the right, so long as the angle θ is known, and one of the lengths on the triangle on the right.

That way, the other lengths can be calculated by using the same proportions or ratios as a triangle from the unit circle.

To take a look at that, let's look at Jacob's example again, where we had the hypotenuse of 5 centimetres and an angle of 37 degrees.

When we look at the triangle in the unit circle sine of 37 degrees is 0.

6, and cosine of 37 degrees is 0.

8, so we can use that to help us find those missing lengths on the right-hand triangle, one of those missing lengths is the opposite, and the other one is the adjacent.

With both of these triangles, to get from the hypotenuse to the opposite, we can times by 0.

6, and with both of these triangles to get from the hypotenuse to the adjacent, we can times by 0.

8.

So let's check what we've learned.

We have two similar triangles.

What is the value of x? Pause the video while you work it out and press play when you're ready for an answer.

The answer is 8, we get it by doing 10 times 0.

8.

What is the value of y? Pause the video while you work that out and press play when you're ready for an answer.

The answer is 6, we get it by doing 10 times 0.

6.

And how about this one, what is the value of x? And you've got four options to choose from.

Pause the video while you make a choice and press play When you're ready for an answer.

The answer is C, 5.

We can get that by doing 0.

5 times 10.

And what is the value of y? Pause the video while make a choice and press play When you're ready for an answer.

The answer is D, 0.

87.

We can get that by doing 10 times 0.

87.

So let's now look at an example of how to use the unit circle to find missing lengths on right-angled triangles.

A diagram shows the sector of the unit circle that lies in the first quadrant.

The points around the circumference show the angle of rotation (θ) from the horizontal axis to the radius at each point.

And if you have access to this slide deck, there is a link on this slide to an interactive version of that quarter circle.

Let's use it now to find the value of x in this triangle.

First, we're going to use the angles to create a similar triangle inside the unit circle, like so.

it's got a right angle and 30 degrees.

Now let's compare these two triangles.

The 8-centimeter length on the triangle on the right is the hypotenuse, the hypotenuse for the triangle in the unit circle is 1 unit.

The side which is labelled x is the opposite, and the opposite in this triangle in the unit circle is 0.

5.

We can use the same proportions now to find the value of x, with the triangle inside the unit circle to get from the hypotenuse to the opposite, we times 0.

5, so let's do the same on our triangle on the right.

8 times 0.

5 is 4.

Let's look at a second example.

It's the same triangle as last time, however, it's the adjacent that is labelled x this time.

So, the length of the adjacent in the unit circle is equal to cos(θ), which is 0.

87.

We can then use the proportions to find the value of x.

To get from the hypotenuse to the adjacent in the triangle in the unit circle, we times by 0.

87, so we need to do the same thing again with our triangle to get the value of x.

8 times 0.

87 is 6.

96.

And how about this third example here? The same triangle again, apart from we now know the length of the adjacent, and we're trying to find the hypotenuse.

Well, we can use this triangle again, but use the inverse operation to find the value of x.

To get from the hypotenuse to the adjacent, we times by 0.

87.

So to get from the adjacent to the hypotenuse, we can divide by 0.

87.

8 divided by 0.

87 is 9.

2.

And here's a fourth example, which is a triangle in a different orientation.

Now this makes it a little bit trickier to directly compare to the triangle inside the unit circle, but we can still use that triangle inside the unit circle, so long as we label the sides of our right-angle triangle.

By first labelling the sides of the right-angle triangle, we can see which edges correspond between the two triangles.

We have the opposite, and we're trying to find the hypotenuse.

So we know the hypotenuse of the triangle inside the unit circle is 1 unit, and the opposite is equal to sine of 30 degrees, which is 0.

5.

We can then select an appropriate calculation to find the value of x.

To get from the hypotenuse to the opposite, we times by 0.

5.

So to go the opposite direction, we would divide by 0.

5.

6 divided by 0.

5 is 12.

So let's check what we've learned.

We have a right-angle triangle on the right here, where one of the lengths are known and is labelled x.

We've set up a similar triangle inside the unit circle, which we can see on the left-hand side of the screen, and we've read off some values.

Use what you can see here to find the value of x, pause the video while you do it, and press play when you're ready for an answer.

The answer is 4.

We take the 8 and we multiply it by 0.

5, which is the value of cos 60 degrees.

How about this time when x is the hypotenuse, what is the value of x here? pause a video while I work it out, and press play when you're ready for an answer.

The answer is 16.

We want to go from the length of the adjacent to the length of the hypotenuse, so we divide by 0.

5, which is the value of cos 60 degrees.

This time, we have a triangle in a different orientation.

Can you find the value of x? Pause the video while you work it out and press play when you're ready for an answer.

The answer is 6.

96.

We do 8 times 0.

87, which is the value of sine 60 degrees.

Okay, it's over to you now for Task B, this task contains two questions, and here is Question 1.

Use this sector of the quarter circle to find the missing lengths on each triangle.

You can either use the version of the quarter circle that you can see here or if you have access to this slide, you can click on the link to load up an interactive version of it.

Pause the video while you do it, and press play when you're ready for Question 2.

And here is Question 2.

Once again, you've got four right-angled triangles, but this time, you need to find some specific unknowns.

Pause the video while you do this and press play when you're ready to go through some answers.

Let's see how we got on with that.

With Part A, the opposite is 5 metres, and the adjacent is 8.

7 metres.

In Part B, the opposite is 8.

7 metres, and the adjacent is 5 metres.

It's the same triangles in Part A, it's just in a different orientation, and rather than displaying the 30-degree angle, the 60-degree angle is displayed instead.

With Part C, both the opposite and the adjacent are the same length, they are 5.

7 metres, and with Part D the opposite is 3.

1 metres, and the adjacent is 11.

6 metres.

And then with Question 2.

A is the hypotenuse, and that is 7.

8.

B is the hypotenuse, and that is 9.

3.

C is the adjacent, and that is 14.

1, and D is the angle, and that is 30.

Great work so far.

Now let's move on to using a tangent to a unit circle to find the missing lengths on a right-angle triangle.

Here we have Andeep and Izzy, and a right-angle triangle.

Andeep and Izzy are trying to find the value of x in this triangle.

They start by labelling the sides of the triangle, like so, and they then create a similar triangle inside the unit circle, like this.

But then Izzy says, "We're not using the hypotenuse this time." So this makes it a little bit different to some of the previous examples we've seen in today's lesson.

So how could they go about finding the value of x? Pause the video while you think about this and then press play when you're ready to continue.

So let's hear what they think.

Andeep says, "We could find the hypotenuse first and then use that to find the length of the opposite." So the value of cos of 40 from the unit circle, we can see is 0.

77.

We could divide five by 0.

77, and then the value assigned 40 from the unit circle is 0.

64.

So we can multiply the hypotenuse by 0.

64 to get the length of the opposite.

Izzy says, "We would have to be careful with rounding mid-solution here 'cause we could lose a bit of accuracy." Andeep says, Alternatively, we could find the multiplier from 0.

77 and 0.

64 and then apply the same multiplier to the 5 centimetres.

We could do that by dividing 0.

64 by 0.

77 to get that long number there, and then if that's the multiplier to get from the adjacent to the opposite with the triangle on the unit circle, we use the same multiplier with the triangle on the left, and you get the value of x that way.

Izzy says, "The benefit of the unit circle, though, is that one of the lengths is 1 unit.

So this doesn't seem as helpful." so Andeep has another idea.

He says, "We could use the unit circle to create a triangle where the adjacent has length 1 unit, like this." So now we have a length in this triangle that is 1 unit, and Izzy says, "The opposite in this new triangle is now a tangent to the unit circle." "We could use this to find the length of the opposite for the triangle in the unit circle, which is 0.

84." "And then we could use these proportions to find the value of x." 1 times 0.

84 is 0.

84.

So 5 times 0.

84 is the value of x.

The tangent of an angle between 0 and 90 degrees is the length of the opposite side, which lies on the tangent to the unit circle at 1, 0.

So with this diagram here, the length which is the opposite on this triangle is part of the tangent, and that length is tan(θ).

The tangent can be used to find either the opposite or adjacent in a right-angle triangle, when the other length and an angle is known.

So let's take a look at some examples.

Here we have a right-angle triangle, and let's find the value of x.

We're gonna start by labelling the sides.

X is the opposite, and the 6 centimetres is the adjacent.

We're using the opposite and the adjacent, so with our unit circle, let's draw a tangent at x =1, like this.

We can then use a tangent to create a similar triangle, and we need that triangle to have an angle of 40 degrees, so it would look like this.

We can check which edges correspond between the triangles.

So we've labelled the opposite and adjacent on our triangle, let's do it with the triangle in the unit circle, it's these ones.

The adjacent in the unit circle has a length of 1 unit.

The length of the opposite in this triangle is equal to tan or (θ), which in this case, tan 40 degrees is 0.

84.

We can then use the same proportions to calculate the value of x.

So 1 times 0.

84 gives 0.

84.

So let's do 6 times 0.

84 to get the value of x, which is 5.

04.

And here's the second example.

This time, we're trying to find a length for the adjacent.

Well, we have the same angle so we could do all the same things we did last time, apart from, we could use the inverse operation to find the value of x instead.

From the triangle unit circle, we can see that to get from the adjacent to the opposite in both of these triangles, we times by 0.

84, so to get from the opposite to the adjacent, we could divide by 0.

84, and that would give 5.

04.

So let's check what we've learned.

Here we have a right-angle triangle with one of the lengths labelled X, and we've created a similar triangle with a unit circle and a tangent.

What is the value of x? Pause the video while you work it out and press play when you're ready for an answer.

The answer is 5.

7, which we get from doing 10 times 0.

57.

What about this time, what is the value of x here? Pause the video while you work it out and press play when you're ready for an answer.

The answer is 17.

5.

We get that by doing 10 divided by 0.

57.

Okay, it's over to you again for Task C, this task contains one question, and here it is.

We have four right-angled triangles, and we need to use this sector of the unit circle to find the value of each unknown.

Now, you can either use the version you can see on the screen here, or if you have access to this slide, you can click on the link to load up an interactive version of it.

Either way, pause the video while you work it out and press play when you're ready for answers.

Let's see how we got on with that.

We can start this question by drawing a tangent to the unit circle, and then using that, we can find the values of the unknowns.

A is 11.

9.

B is 8.

4, C is 3.

3, and D is 6.

9.

Fantastic work today.

Now let's summarise what we've learned in this lesson.

There are two key right-angled triangles that we've used throughout this lesson.

One of them is within the unit circle, and the other is with the tangent to the unit circle.

There are specific ratios within the two key right-angled triangles defined by the unit circle.

One of the key right-angled triangles has a hypotenuse of length 1 unit.

The opposite and adjacent sides of this triangle have lengths sin(θ) and cos(θ) respectively.

The other key right-angled triangle has an adjacent side of length 1 unit, and the opposite side of this triangle has length tan(θ), and you can use similarity between any right-angled triangle, and either the key right-angled triangles in the unit circle to find the missing lengths.

Great job today, thank you very much.