Lesson video

Hello, I'm Mrs. Lashley, and I'm gonna be talking you through the lesson today.

I really hope you're willing to try your best, and even if it gets challenging, remember, I'm here to support you.

So our learning outcome today is to be able to appreciate the range of values of the trigonometric functions.

Some keywords I'll be using during the lesson are on the screen here.

Let's spend some time going through them together so that we're happy before we start the lesson.

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

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

And you can see that on the diagram.

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

And once again, that's referring to the diagram.

A tangent to a circle is a line that intersects a 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.

So you still may wish to pause the video so that you can look at that diagram carefully and reference it to those definitions, and then when you're ready to make a start, press play.

So our lesson on checking and securing understanding of trigonometric ratios is split into two learning cycles, and our very first learning cycle is looking at finding side length by using the trigonometric ratios.

Later on, we'll look at finding the angle.

So let's make a start of finding side lengths by using the trigonometric ratios.

So we're first gonna look at the sine ratio, and the sine ratio is the ratio between the opposite and hypotenuse of a right angle triangle.

And so we can substitute the values for any right-angle triangle into the sine formula to represent that relationship between the opposite of the angle and the hypotenuse.

And on the left hand side, you can see the three rearrangements of the sine formula, so the top one says that the hypotenuse multiplied by sine of theta is equal to the opposite length.

So if you were trying to calculate the opposite and you know the hypotenuse as well as angle theta, then that would be the rearrangement of the sine formula that you would use because the opposite is the subject.

And you can see for this particular triangle, the substituted values, so B, which is the hypotenuse, multiplied by sine of 70 is equal to 20.

As this triangle is B, the hypotenuse we're trying to calculate, this is not the preferred form that I would use.

On the second row, we can see a rearrangement of that first one, which is opposite divided by sine of theta is equal to the hypotenuse.

So the opposite edge length divided by sine of the angle theta will give you the length of the hypotenuse.

And so for this question, the one we can see on the screen, this is the form that we'd want to use because the hypotenuse is the subject.

So 20 divided by 70 will give us the length of the hypotenuse B.

And lastly, the bottom row is our rearrangement, where we have sine of theta as the subject, so the opposite length divided by the hypotenuse, and this shows us the ratio for sine.

Again, if you substitute the values for this particular triangle into it, we have B on the denominator of that fraction and we'd like to calculate B, so we'd need to rearrange it and that rearrangement would go back to the middle row.

If we look at this example here, we have a right-angle triangle, so we know that we can use the three trigonometric ratios of sine, cosine, and tangent.

We also have the other two angles in the triangle.

E is the hypotenuse, and we know that's the hypotenuse because it is opposite the right angle.

And then we have the nine centimetre.

So the nine centimetre is opposite the angle of 38 degrees and adjacent to the angle of 52 degrees.

So if we're trying to use the sine ratio, then we're going to be using the 38-degree angle.

So we now have nine as the opposite to the 38-degree angle and E is the hypotenuse.

So using the ratio, we can say that sine of 38 degrees is equal to the opposite over hypotenuse, which is nine over E.

What we are trying to calculate is E, the hypotenuse, so this form needs to be rearranged to make E the subject.

So we can multiply both sides by E and then divide both sides by sine of 38 degrees to make either subject, and then we can calculate that, making sure that our calculator is in degree mode.

And here I've rounded the answer to two decimal places.

So the hypotenuse of this right angle triangle is 14.

62 to two decimal places.

So here's a check for you.

Which of these correctly shows the relationship between the hypotenuse and the side opposite the 80 degree angle for this triangle, rounded to one decimal place? So pause the video and look at the five equations and which equations are showing the correct relationship.

Press play when you're ready to check.

So C is a correct relationship.

The hypotenuse multiplied by sine of the angle is equal to the opposite, but A and D are also correct, they're rearrangements of this, so we've got opposite over hypotenuse equals sine of 80 degrees, and then we have opposite divided by sine of 80 degrees is equal to hypotenuse.

So A, C, and D are all correct relationships between hypotenuse, opposite, and the angle.

So if we now go onto the cosine ratio, the cosine ratio is the ratio between the adjacent and the hypotenuse.

So the hypotenuse multiplied by cosine of theta is equal to the adjacent.

And once again, we can substitute the values for the given triangle into that form, and we can see the equation.

So C times cosine of 82 is equal to four.

The second row shows a rearrangement of this, which is the adjacent divided by cosine of theta is equal to the hypotheses.

And then with the correct values, we can see that is the most efficient rearrangement to use for this particular question because C is the subject, so we'd be able to calculate the left hand side and that would be equal to our hypotenuse.

And then the bottom row, similar to what we saw for the sine adjacent divided by hypotenuse is the ratio of cosine.

So adjacent divided by hypotenuse is equal to cosine of theta, and for this particular triangle, that means that four divided by C equals cosine of 82.

So when given the angle and the adjacent side or hypotenuse, then it's possible to determine the remaining side.

So here, we've got a right angle triangle angle of 71 degrees, the adjacent edge is unknown, but labelled A, the hypotenuse is known to be 120.

So if we're trying to calculate the adjacent, then this is the form of the formula that we would use because the adjacent is the subject.

So if we substitute our values in, so we know that our hypothesis is 120, our angle of theta is 71, and then we can calculate that to one decimal place is 39.

1.

So which of these calculations is correct? Pause the video, and when you're ready to check whether it's the first one, the second one, or the third one, then press play.

So the first one is correct.

That is the adjacent edge to the 75 degree angle given the hypotenuse, so we would multiply the hypotenuse and cosine of 75 to have the length of the adjacent.

On the second triangle, it's incorrect because the edge that is marked is the opposite to the 75 degree, and opposite and hypotenuse are the sine ratio, so you wouldn't use cosine necessarily to work that edge out.

And on the third triangle, it is the adjacent, that there's calculation of four, but it's not the correct calculation.

So 250 divided by cosine of 75 does not equal the length of the adjacent.

So then we've got the third trigonometric ratio, which is tangent, and tangent is the relationship between the opposite and the adjacent of a right-angle triangle.

So the adjacent multiplied by tan of theta will give you the opposite.

For this particular triangle, where you can see on the screen an adjacent length of B, an opposite length of 15, and an angle of 70 degrees, then when we substitute those into the relevant places, we would end up with the equation B times tan of 70 is equal to 15.

If we're trying to calculate B, then we want B to be the subject, so we'd divide both sides by tan of 70 and that leads us to the second row, 15 divided by 10 of 70 is equal to B, and that's because the opposite divided by tan of theta is always equal to the adjacent.

And then lastly, we can rearrange this to get tan of theta as the subject, and that would give us opposite over adjacent, and that's that ratio between opposite and adjacent that we've discussed.

So if we look at this example, we've got find the value of K.

Well first of all, what is K? Well, K is the opposite edge to the 17-degree angle.

And what is 105? 105 is the adjacent to the 17-degree angle.

So we are looking at opposite and adjacent, which tells us this is the tangent ratio that we're gonna make use of.

Tan of 17 degrees would be equal to the opposite, which is K, divided by the adjacent, which is 105.

We would like to calculate K, we'd like to find the value of K, so we want to make that the subject.

So we're going to rearrange this to make K the subject by multiplying both sides by 105.

So K is equal to 105 times tan of 17, which is 32.

10 to two decimal places.

So we then know that the opposite edge of this right-angle triangle is 32.

10 centimetres.

So here's a check for you.

Which of these equations helps find the value of Q most efficiently? So look at the diagram and look at the three options.

Which one is gonna get you the value of Q in the most efficient way? Press pause whilst you think about that, and when you're ready to check, press play.

The most efficient way is when Q is already the subject, so that would be option B.

So now use your calculator and calculate the length of the side labelled Q to two decimal places.

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

64.

01, and the units here are metres, to two decimal places.

So we're into the first task of the lesson, where we're looking at using the trigonometric ratios to find side lengths of right-angle triangles.

So calculate the length of the unknown side labelled with a letter for each of these triangles.

Pause video whilst you do that, and when you press play, we'll go through our answers.

All the answers are on the screen here.

So I'm just gonna tell you which ratio you should have been using to calculate these, because if you've chosen the wrong ratio, then that will affect your answer.

So on A, you had the opposite and the adjacent, so you should have used tangent.

On B, you had the opposite and the hypotenuse, so you should have used sine.

On C, you had the opposite and the adjacent, so you should have used tangent.

On D, you had the adjacent and the hypotenuse, so you should have used cosine.

On E, you could have used cosine or sine because you were given both of the angles, 22 is either the adjacent to the 43-degree angle or the opposite to the 47 degree, so you could have used sine or cosine depending on which angle of 47 or 43 that you decided to use.

And finally for F, it was the opposite and the hypothesis, so you should have used sine.

With any of these questions, you can always calculate the third angle in the triangle, and then you would be able to use a different ratio because you could use the different angle and then it becomes the opposite or it becomes the adjacent.

But in math, we always try to do things as efficiently as we can, so adding an extra step of a calculation to work out the third angle doesn't feel necessary.

We're now up to the second learning cycle where we're gonna find angles using the trigonometric ratios.

So now we're looking at finding angles using our trigonometric ratios.

So to find the angle we're gonna use arc sine, arc cosine, and arc tangent, which are the inverse trigonometric ratios to find the missing angle, provided that we have two sides in the right angle triangle.

And those two sides will determine which of the ratios you are using.

So if we look at this right-angle triangle, it's a generalised right-angle triangles, so we've got angle theta in that position, an edge length marked as B, and an edge length marked as H.

Well, we know that the edge length marked as H is the hypotenuse because it is opposite the right angle, and so what is the edge length marker as B? Well, it is opposite the angle theta, so we have the opposite and the hypotenuse, and so that tells us we are using sine.

So sine of theta degrees is the ratio of opposite over hypotenuse, so B over H in terms of the labels we've got on the edges here.

So then we would use arc sine, which has this notation of inverse sine notation, of B over H to give us the angle theta.

So if we're trying to find the angle, then we're gonna use the arc sine when you have the opposite and the hypotenuse of a triangle.

If we look at this triangle, another generalised triangle, the A is the adjacent, it is adjacent to the angle theta, and the H is the hypotenuse.

So when we have adjacent and hypotenuse, then we are working with cosine.

Cosine of theta is adjacent over hypotenuse, so we'd use arc cosine, which is the inverse cosine function, to calculate that angle of theta.

And then the last pair that you could end up having is opposite and adjacent.

So the opposite is B, the adjacent is A on this triangle relative to that angle of theta, so tan of theta is equal to the opposite, B, over the adjacent, A, also which use the arc tangent function or the inverse tangent function.

On our calculator, in order to get that arc cosine, arc sine, or arc tangent function, you often have to use the shift key, or it might be called the second function key, and then, normally, it's the same buttons as sine, cosine, and tangent.

Look on your calculator to find that and make sure you are in the degree mode.

I'm gonna go for an example, and then there'll be one for you to try.

So use the arc sine function to find theta.

Give your answer to two decimal places.

So we're telling you that we're gonna use the arc sine, but why are we using arc sine? Well, we're using arc sine because 40 is the opposite and 70 is the hypotenuse, so opposite and hypotenuse are the ratio of sine.

So we can first of all state that ratio, that sine of theta is equal to 40 over 70.

And then we're gonna use the arc sine function to make theta our subject and to calculate theta.

So arc sine of 40 over 70 is 34.

85 degrees to two decimal places.

So can you calculate this one? Pause the video whilst you do that, and then when you're ready to check your answer, press play.

So arc sine again, 50 is the opposite, 80 is the hypotenuse, so your formula is sine of the is equal to 50 over 80, and therefore, arc sine of 50 over 80 gives us 38.

68 degrees to two decimal places.

Similar idea.

Let's look at this one.

So use the arc cosine function to find theta, giving your answer to two decimal places.

So it's arc cosine because 20 is the adjacent and 90 is the hypotenuse.

So cosine of theta is 20 over 90, therefore theta is the inverse of cos, or arc cosine, of 20 over 90, which is 77.

16 degrees to two decimal places.

So if you have a go at this, pause the video.

When you're ready to check your answer, press play.

So cosine of theta would be 30 over 100, and therefore, arc cosine of 30 over 100, or 0.

3, is 72.

54 degrees to two decimal places.

So now let's look at arc tangent.

So we're gonna use arc tangent to find the angle of theta for this right-angle triangle.

And again, we're giving our answer to two decimal places.

So the opposite is 16 and the adjacent is 12, so which means tan theta is 16 over 12, and therefore, we're gonna use arc tan or arc tangent on that ratio, that gives us 53.

13 degrees to two decimal places.

So here's your one to do, so pause the video, and when you're ready to check, press play.

So you should have returned that tan of theta is equal to 20 over 16, and therefore, arc tan of 20 over 16 is 51.

34 degrees.

Something that you need to be mindful of is that the tangent ratio can go above one, so not getting into the habit of always thinking that the fraction has to be proper.

It does have to be a proper fraction for sine or for cosine.

Otherwise, you'll get a mathematical error appear on your calculator, which will indicate to you that maybe you've got your fraction the wrong way up, but for tangent, that's not the case.

So it is just remembering that it's the opposite divided by the adjacent regardless of what values they are.

So here's another check for you.

Which of the following equations are correct when finding the size of angle U in this triangle? So pause the video, and then when you're ready to check, press play.

So equation A is correct, that is our sort of definition of cosine.

The cosine of the angle U is equal to the adjacent over the hypotenuse, and therefore, when we try to solve that to find the value of U, then we're gonna use our arc cosine function.

So D.

Throughout the last task of the lesson, you're gonna use your calculator to find the size of each angle marked with a letter, and give your answer to one decimal place.

So pause the video, and then when you're finished with that question, we'll go through the answers.

So the answers are on the screen for each of the angles.

Once again, I'm gonna go through which ratios you should have been using on each triangle.

So on triangle where the angle A, nine was the opposite and 30 was the hypotheses.

So you should have been using the sine ratio and using arc sine to find the angle.

On the second triangle, with the angle of B, 59 centimetres is the opposite to the angle B and 70 is the hypotenuse, so once again, you should have been using arc sine to find the angle.

Moving to the triangle with angle C, nine is the adjacent to the angle C and the hypotenuse is 13, so therefore, we should have been using the cosine function and arc cosine to get the angle.

Onto the one with D, 22 is the opposite to the angle D and 20 is the adjacent, so you should have been using tan and arc tangent to find the angle of D.

On E, the nine is the opposite, the eight is the adjacent, so again, that would be tangent or arc tangent for the angle.

And lastly for F, 0.

5 is the adjacent to the angle F and 0.

7 is the hypotenuse so we're going to be using cosine.

To summarise today's lesson on check an understanding of trigonometric ratios, you can read through the points on the screen, and that is the basis of what we've been using today.

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