Lesson video

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

So the lesson today is focusing on being able to choose the correct trigonometric ratio when you are trying to solve a problem that involves a right-angled triangle.

On the screen are the trigonometric ratios.

It might be that you wish to pause the video now to re-familiarize yourself or write them down before we get going in the lesson.

So we're gonna make a start of really being able to identify which of the trigonometric ratios we can use.

In a triangle, the proportional relationship between each pair of sides is set by the angles.

And each of the trigonometric functions, so that's sine, cosine, and tangent, find the multiplier between the two sides of a right-angled triangle given one of the other angles.

The first one, where you have the opposite and the hypotenuse, is the ratio or the function of sine.

Hypotenuse multiplied by sine of theta will give you the opposite.

The second one, adjacent and hypotenuse, is the trigonometric function of cosine.

Hypotenuse multiplied by cosine of angle theta will give you the adjacent.

And lastly is the tangent function, and that is the adjacent multiplied by the tangent of theta gives you the opposite.

So if we've got this right-angled triangle here, which of the trigonometric functions can we use to calculate the value of x? Jacob says he can see that sine would work.

The six centimetres is the opposite to the 58 degree angle and x is the hypotenuse.

And we just saw that we know that the sine is the multiplier between the opposite and the hypotenuse on a right-angled triangle.

Andeep says, "No, it's cosine.

The 6 centimetres is the adjacent to the 90 and the 32 degree angles and x is the hypotenuse." And we know that cosine is the multiplier between the adjacent and the hypotenuse.

So who do you agree with? Well, they're both correct.

They've just chosen to use a different focus angle.

So that 6-centimeter edge is the adjacent to the 32-degree angle, but it's the opposite to the 58 degrees.

So remember that when we label our edges opposite and adjacent, it does depend on that focus angle.

So here's Jacob's working out using the sine function, that sine 58 degrees is equal to 6 divided by x.

So rearranging that, that means that x is equal to 6 divided by sine of 58, which gives him 7.

1 centimetres to one decimal place.

And x is the hypotenuse, which is the longest edge, and 7.

1 centimetres is greater than 6 centimetres, so it feels like that that is an accurate answer.

Andeep's working is very similar except from using the cosine function with the angle of 32 degrees.

And unsurprisingly, they both get the same answer to one decimal place.

So here is a check for you.

If the 26-degree angle is the focus angle, which trigonometric function will you use to calculate the value of x? Pause the video whilst you'll make your decision.

Press play when you're ready to check.

Cosine.

If 26 degree is our focus angle, then the x, the edge x is the adjacent to the 26 degrees.

So then we are going to use the adjacent with the hypotenuse, which will be cosine.

So given the information that we have on that right-angled triangle there in that diagram, what methods could be used to calculate the value of x? Well, Alex says, "Well, I'm gonna use sine to calculate the length of the hypotenuse, which is h, and then you sine again to calculate x." So sine of 58.

6 is the opposite to the 58-degree angle, so sine of 58 equals 6 over h.

h is the hypotenuse.

Rearranging it and solving it, we get 7.

1 to one decimal place.

And Alex said he was then gonna use sine again, and that's because we've now got the hypotenuse as a length and x can be the opposite of the 32 degree angle.

So he's changed the focus angle that he's using.

So sine of 32 degrees is the opposite divided by the hypotenuse.

And in his previous working he worked out the hypotenuse.

Rearranging: hypotenuse multiplied by sine of 32 gives you the opposite, that is the multiplier in our sine function.

And so x is 3.

7 to one decimal place.

Aisha is going to do the same question but use a different method.

So she has said, "I'm gonna use cosine to calculate the length of the hypotenuse and then use sine to calculate x." And she is using the focus angle of 32 degrees, so that's why it's cosine, because 6 is the adjacent.

She gets the same value for the hypotenuse: 7.

1.

And then she's doing the same second part of the working as Alex with 32 staying as her focus angle, and therefore x is now the opposite, and she gets 3.

7 to one decimal place.

Jun says he could use cosine or sine to calculate the length of the hypotenuse, but then he's gonna use Pythagoras' theorem.

So cosine of 32 is 6 of h.

So this is the same initial working as Aisha.

And so h is 7.

1 to one decimal.

Pythagoras' is theorem: x squared, the shorter side, is equal to hypotenuse squared minus the other shorter side squared.

Substituting the unrounded value value of h, we get x squared as 14.

05662145.

Square rooting it, x is 3.

7 to one decimal place.

Sophia says, "I'm just gonna use tangent." So tan of 32 is x over 6.

x is the opposite edge to the 32 degree angle and 6 is the adjacent.

And the tangent function is the multiplier between the opposite and the adjacent.

And x is 3.

7.

So here we've had four different methods, four different solutions to the same problem.

Which one's more efficient, do you think? Well, the tangent was the most efficient way to calculate x on this right-angled triangle.

So are all of these methods equivalent? So still focusing on this right-angled triangle here, pause the video whilst you consider that.

And then when you press play we'll carry on.

Well, the middle one is not correct because if the focus angle is 32 degrees, the opposite is x and the adjacent is 6.

The last one is also correct, but it's using a different focus angle; that time the opposite is 6 because the focus angle is 58 degrees.

With the tangent function, it is really important to make sure the opposite for your chosen angle is divided by the adjacent.

So a check, complete the equation.

Pause the video.

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

So hopefully you went for 71 degrees because x is the opposite.

So x is the numerator and the adjacent is our denominator, the focus angle needs to be 71.

So for every right-angled triangle, given either two edge lengths or one edge and one of the other angles, all of the trigonometric functions can be used.

However, there is one that is normally more efficient than the other two, and we always want to be as efficient as possible.

So here on this right-angled triangle, what's the most efficient of the three trigonometric functions to use? Well, this one would be sine because you have the opposite to the angle theta and you also have the hypotenuse.

So we know that the sine function is a multiplier between those two edge lengths, so the sine function would be the most efficient.

On this one, which of the functions would be the most efficient? This one would be tan.

We've got the opposite edge and we want to calculate the adjacent.

And we know a focus angle of 29 degrees, so the opposite and the adjacent are in the tangent function.

And this one is going to be the cosine function.

So here we've got the adjacent of 9.

1 and we are trying to calculate the hypotenuse with a focus angle of 31 degrees.

So a check.

Which trigonometric function would be the best to use for this question? So pause the video, make your decision out of the three options, and then press play when you're ready to check.

This one, hopefully you went for tangent because you have the opposite and you would like to calculate the adjacent and you know one of the other angles that's not the right angle.

Which one is the best to use, the most efficient to use for this given triangle? Pause the video and then when you're ready to check, press play.

This one's cosine.

And what about this one? Pause the video and then when you're ready to check, press play.

This one is also cosine.

So we now onto the task.

So question one.

You need to sort the six triangles into the most efficient trigonometric function to use.

Pause the video whilst you're sorting those into the table.

And then when you're ready for question two, press play.

Question two, I want you to match the equation to the correct triangle.

Press pause whilst you do that.

When you press play, we're gonna go through our answers to question one and question two.

Here are the answers for question one.

You need to sort the six triangles into the most efficient trigonometric function to use.

So the sine function would've been triangle C and triangle E; cosine, B and D; and tangent, A and F.

Question two, you needed to match the correct equation to the correct triangle.

So one was for sine, one was for cosine, and one was for tangent.

So did you have the opposite, did you have the adjacent, did you have the hypotenuse? We are now at learning cycle two, which is now making use of the ratios.

So the trigonometric functions allow either an angle or an edge length to be calculated.

So this question: Calculate the length of t to one decimal place.

If we use the definition triangle, where we've got the unit circle, so we've got the hypotenuse of 1, we know that the adjacent for the angle of 34 degrees is cosine 34 degrees.

And so using a ratio table, we can see that on our triangle that we're trying to solve, the adjacent is t, our hypotenuse is 10.

So our hypotenuse from the definition triangle has been multiplied by 10, therefore to keep it a similar triangle, the adjacent needs to be multiplied by 10 as well.

Within the triangle, we can see that there is a multiplier of timesing by cosine 34.

The hypotenuse multiplied by cosine of 34 gives us the adjacent length, and we can use this to calculate t.

So t would be 10 times cosine 34, which is 8.

3 centimetres to one decimal place.

Another way of using these trigonometric functions is for using the relationships in a formula.

So we know that cosine of theta is equal to the adjacent divided by hypotenuse for any right-angled triangle.

So substituting our angle, which is 34 degrees, our hypotenuse of 10, and our adjacent, which is what we're trying to calculate as t, we can then rearrange it to make t the subject.

So t is equal to 10 times cosine 34 degrees, which is 8.

3 centimetres to one decimal place.

So on here we're going to do another explanation from me and then a check that you will be doing.

So find the value of x to one decimal place.

Well, x is the opposite to the 41-degree angle and 6 is the hypotenuse so it will be the sine function.

So we can go back to our definition triangle.

And we know that if the hypotenuse is 1, the opposite is sine of 41 degrees.

On this triangle, our opposite is x and our hypotenuse is 6.

So using the ratio table, we can see we need to multiply by 6; alternatively, we multiply by sine of 41.

So 6 times sine 41 will give us our value of x, which is 3.

9 centimetres to one decimal place.

So pause the videos, press play when you're ready to check your answer.

So you should have been using the tan function because it was the opposite you wanted and you knew the adjacent.

And 4 multiplied by tan 29 degrees would have given you the value of y, 2.

2 centimetres to one decimal place.

So here's another right-angled triangle.

And want to calculate x, which is the opposite, and I have the hypotenuse, so it's another sine function.

This time I'm going to use the formula.

So sine of 43 degrees is equal to x divided by 7.

I want to calculate x, so I need x to be the subject.

So I'm going to multiply both sides by seven.

And so seven multiplied by sine of 43 will give me the value of x, which is 4.

8 centimetres.

So this time I would like you to do this triangle by pausing the video and having a go using the formula.

So tan for you.

You wanted the opposite, you knew the adjacent.

Your angle was 30 degrees, substituting it in because there is this proportionality between the opposite and the adjacent, and multiplying by 5 To get y as the subject, your answer is 2.

9 centimetres to 1 decimal place.

Each of the trigonometric ratios includes two edges, and how we calculate them will depend on which edge it is.

So here, if I want to find the length of m to one decimal place, m is a hypotenuse, so it would be a sine function because the opposite is 6 centimetres.

So I'm using similar triangles, which is a 1-centimeter hypotenuse right-angled triangle with the same angle of 41 degrees, then the opposite is sine 41.

If I use my ratio table, I know that 1 multiplied by sine 41 gives me my opposite of sine 41 degrees.

If I wish to go from the opposite to the hypotenuse, then I would divide by sine 41, I'd use the inverse.

And so m is equal to 6 divided by sine 41, which I can calculate on the calculator and round to one decimal place: 9.

1 centimetres.

If I use the formula, then we substitute in what we have.

So we know the angle of the three is 41 degrees in my question, I know that my opposite is 6, and my hypotenuse is labelled as m, then I'd need to multiply by m on both sides.

And to make m the subject, I would then divide by sine 41.

We end up with that same fraction of 6 divided by sine 41, which we can calculate as 9.

1 centimetres.

So similar idea, I'm gonna go through one, you are gonna try one.

This one we're gonna use this definition triangle and the ratio tables.

So we're trying to calculate the adjacent when I know the opposite, so it's a tan function.

My opposite is 5 and my adjacent is x.

So to get from 1 to 10 of 29, we can multiply by tan of 29.

And so to go from the opposite back to the adjacent, I would use division.

And so x is equal to 5 divided by tan of 29.

So the side labelled x is 9.

0 centimetres long.

So pause the video whilst you calculate y using a ratio table.

And when you're ready to check, press play.

9.

6 should have been your answer.

We are gonna go through it using the formula instead.

So I've got an opposite of 6 and I want to work out an adjacent of x, so it is tan.

Tan of 31 is equal to 6 over x.

I'm gonna have to multiply by x, and that makes 6 the subject, and then divide by tan 31 in order to make x the subject.

I can calculate that, 10.

0 centimetres.

Your turn.

Use the formula, calculate the value of y.

So cosine.

Adjacent was 7, the hypotenuse is y.

Substitute it in and rearrange.

8.

5 centimetres.

The angle is another part of the trigonometric functions that we can calculate if the two edges are known.

So if we want to find theta to the nearest degree on this right-angled triangle, so I've got the opposite and I've got the adjacent, so this makes it tan.

Using the definition triangle, we know that if the adjacent is 1 centimetre, then the opposite is tan of theta.

So 1 is our adjacent, and on our question the adjacent is 11.

Tan theta on the definition triangle is our opposite, and in our question it is 6.

So how do we get from 1 to 11? Well, we multiply by 11.

So if we've multiplied by 11, we need to do it on the opposites as well.

So we can say that 11 tan theta, or 11 multiplied by tan theta, is equal to 6, which we can then divide by 11 to get tan the equals 6 over 11, a fraction.

And then remembering we need to use the inverse tan function.

On your calculator, you're gonna have to probably press the Shift before you press the tan key to make that come up on your screen.

We are trying to go backwards and find out which angle gives us that ratio, and it's 29 degrees to the nearest degree.

If we use the formula where we know there is this relationship between opposite and adjacent, substituting what we have, which is the opposite is 6, the adjacent is 11, we would then use our arctan on that ratio to find the angle that gives the ratio of 6 over 11, which is 29 degrees.

So find the size of theta to the nearest degree.

First of all, which ratio or function am I using? Well, I've got the adjacent and I've got the hypotenuse, so it's cosine.

So on our triangle we know that our adjacent is 9 and our hypotenuse is 12.

So how do we get from a hypotenuse of 1 to a hypotenuse of 12? Well, we multiply it by 12.

So we need to do the same on the adjacent to keep it in ratio, to keep it proportionately the same.

12 cosine theta equals 9, therefore cosine of theta is 0.

75, is 3/4.

And then using our inverse cosine, or arccos, we would do arccos of north 0.

75 to find which angle has a ratio of 3/4, which is 41 degrees to the nearest degree.

So here's one for you to do.

Pause the video, use the ratio table for this.

And then when you're ready to check, press play.

Sine was the function you should have been using because the 3.

5 was the opposite and the hypotenuse was 7.

Sine theta would be 1/2, and therefore arcsine of 1/2 is 30 degrees.

So your answer was 30 degrees.

If we do this using the formula, substitute what I know.

So cos of some angle theta is equal to the adjacent divided by the hypotenuse, this ratio between the adjacent and the hypotenuse.

Arccos of 7 over 15 is 62 degrees to the nearest degree.

So one for you to do, so pause the video, using the formula.

And then when you're ready to check, press play.

So sine, again, for you, because it was the opposite and the hypotenuse were the two edges that you had, and then 46 degrees was the answer to the nearest degree.

So your task for this learning cycle.

First of all, you need to calculate the length of the edge marked x and give your answers to one decimal place.

Pause the video whilst you work through those seven questions.

And then when you press play, you've got question two.

Here's question two.

This time you are working out the size of angles marked theta.

Again, give your answers to one decimal place.

Pause the video whilst you work through those seven questions.

And then when you press play, you've got one more question in this task.

Question three, you need to calculate the size of the angle or the marked edge, and again, giving your answer to one decimal place.

Pause the video whilst you do those four questions.

And then when you press play, we're gonna go through our answers.

So here are the answers for question one, where you are calculating the length of the edge marked x.

It may be that you wish to pause the video and go through them one by one.

Again, here's question two.

This one was working out angles within the triangle, so you should have been using the arccosine or the arcsine or the arctangent and pressing the Shift key on your calculator.

Pause the video so that you can check through those answers.

Question three was a mixture between finding angles and finding edge lengths.

Again, you may wish to pause the video so that you can check through each of those questions.

So we're now up to the last learning cycle of the lesson, which is considering the necessary information when faced with the problem that trigonometry is going to help with.

So we have an isosceles triangle on the screen.

And is there a way of calculating the length of the equal edges? And those equal edges have been marked as a.

Sam says, "Well, Pythagoras' theorem and trigonometry rely on right-angled triangles, so we cannot use them." There is no right angle within this isosceles triangle.

Jacob said, "We can use trigonometry.

Look at this.

And so the altitude of a triangle is perpendicular to the edge, so now there is a right-angled triangle." So Jacob has drawn on an altitude to the isosceles triangle.

Sam has then said, "But we have no known edge lengths, so we cannot use trigonometry.

We might have a right-angled triangle, but we don't have any edge lengths." And Jacob said, "No, actually we do.

Because it's an isosceles, this particular altitude bisects the base edge." So that base edge, instead of being 8 centimetres, can be split in half to two 4-centimeter edges.

Sam says, "Well, that's great.

Now we have a right-angled triangle, an angle, and one edge length." And as you know, that means you can calculate the second edge length in the ratio.

So check.

Here is a nice isosceles triangle, what is the length of x? Pause the video, and then when you're ready to check, press play.

Again, isosceles, that altitude will bisect the edge because it's the line of symmetry.

And so 22 divided by 2 is 11.

So once the right-angled triangle has been established, then we can make use of our trigonometry.

So here is the right-angled triangle that we are going to use.

We have a hypotenuse of a, an adjacent of 4, that's why it's cosine.

So ratio table.

A hypotenuse of 1 would give you an adjacent of cosine of 80.

Our adjacent is 4, our hypotenuse is a.

So we have multiply it by cosine 80 and we need to do that to both parts.

And so a multiplied by cosine 80 equals 4.

And we can divide by cosine 80 to leave a as the subject because that's what we're trying to calculate, and that's 23.

0 to one decimal place.

The equal edges of that isosceles is 23 centimetres.

So here we are faced with another triangle, not isosceles or not known to be isosceles, but also not right angled.

One way we can always make a right-angled triangle is by using the altitude of a triangle because it will always be perpendicular to the edge that it meets.

So here is the altitude drawn onto this triangle.

And now you can see there are two right-angled triangles.

What can we calculate now that we have that additional line onto our diagram? Well, using the left right-angled triangle, the one with the 80-degree angle within it, we can calculate the length of the altitude, which we've labelled as x, using the sine function.

It's opposite that 80-degree angle and the 9-centimeter edge is the hypotenuse, so the x would be 8.

86, roughly, centimetres.

So now we have the perpendicular height, which is opposite the angle of theta, and 12 centimetres, which would be the hypotenuse of this right-angled triangle means we can use sine again.

Sine of theta is 8.

86, roughly, over 12.

We're gonna use the more exact value in our calculation.

And then we can use arcsine to calculate the angle of theta, which is 47.

6 degrees to one decimal place.

So check for you.

The altitude of the triangle will always bisect the edge it meets.

True or false? Pause the video whilst you make a decision.

And then when you press play, we're gonna justify that answer.

This is false.

So justification.

A, it only does if the altitude is also a line of symmetry of the triangle; or B, if it bisects the angle, then it also bisects the edge.

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

It's A.

The altitude will only bisect the edge if it is a line of symmetry of the triangle, so on an equilateral triangle or on an isosceles triangle.

Here we've got a stack of right-angled triangles.

So we don't need to find the right-angled triangles, they are all right angled triangles.

Some of the edges are shared and some of them are part of another edge.

And we want to calculate theta to the nearest degree.

So there are four right-angled triangles that I've labelled A, B, C, D in order to be able to communicate more effectively.

So theta is part of triangle D.

Another side length is needed before we can calculate it.

So we have got the adjacent on that triangle D, but that's the only edge length.

And we know to use the trigonometric functions to find the angle, we need to know the two edge lengths.

Triangle A has enough information to calculate the length of an edge.

And there is also a shared edge with triangle C.

So by working out the shared edge, we're also working out an edge on triangle C.

So focusing only onto that right-angled triangle, we know the hypotenuse is 4, we have an angle of 44 degrees, and the x, which is the shared edge, is the opposite to that 44 degrees, so we're gonna use the sine function.

That would be four times sine of 44, which is 2.

77.

We're gonna keep our most accurate answers in our calculations.

So now we have an angle and a side in triangle C.

So calculating the length of the hypotenuse seems a sensible next step.

We could calculate the length of the opposite in triangle C, but that is not shared with any of the other triangles, so it is less useful.

So here is triangle C.

We know that that adjacent is 2.

77, we calculated that on triangle A, and we are going to work out the hypotenuse, which I've labelled as y.

So this time we're you gonna use cosine.

y times cosine of 71 is equal to the adjacent, that is the relationship between adjacent and hypotenuse.

So if we divide by cosine 71, we get our value of y, which is 8.

53, approximately.

The hypotenuse of that triangle C is the sum of the adjacent on triangle B and the opposite on triangle D.

So we can now calculate the length of the adjacent on triangle B with the given information.

We have the hypotenuse, which is 7.

2, and we have an angle, a focus angle of 32 degrees.

So if I call it z, the adjacent, I can calculate that as 6.

10.

So now I have part of the sum, which would mean that the opposite on triangle D is the hypotenuse of C subtract the adjacent on B, it leaves us that remaining length of 2.

42.

Now we have two edge lengths on the triangle, so we can calculate theta.

So tan would be 2.

42 divided by 4.

8, and therefore using arctan, we'd get 27 degrees, approximately, to the nearest degree.

So here's a check.

From the information given on this diagram of sort of stacked right-angled triangles, which hypotenuses could be calculated in one step? Pause the video whilst you make your decisions and then press play to check.

So you could, with the given information, calculate the hypotenuse of triangle A and the hypotenuse of triangle B.

Same diagram, but from the given information, which opposites could be calculated in one step? Again, pause the video and then when you're ready to check, press play.

So you could calculate the opposite of triangle A.

Same diagram, but which of the adjacents could be calculated in one step? Pause the video whilst you make your decisions.

And when you're ready to check, press play.

So you could calculate the adjacent of triangle B, you would use tan for that.

And you can also calculate the adjacent of triangle C, and that's because you can calculate the opposite on triangle A, which is the adjacent of triangle C.

So that one's a bit sneaky, it's a shared edge.

So actually it's not by using triangle C to get that edge, you would get it by using triangle A.

So we're onto to the last task of the lesson.

So question one: Calculate the missing angle or length in each of the isosceles triangles.

Pause the video whilst you have a go at that; and then when you press play, we'll move on.

Question two, I would like you to calculate the missing angle or length in each of the diagrams. Pause the video whilst you work on them.

And question three, I would like you to calculate the missing angle to the nearest degree and the missing edge to one decimal place in this triangle stack.

So press pause whilst you work through that question.

And then when you're ready to move onto the answers, press play.

Here are the answers for question one.

And that was all isosceles triangles.

So when you dropped the perpendicular or drew on the altitude to find right-angled triangles, you were by bisecting the base edge or the edge that it met.

Pause the video so you can check your answers.

Question two, this is part A only, and the next slide will have part B.

You needed to calculate the angle theta, and that came from doing a couple of steps on the way to it.

So again, you might wanna pause to go through and check against your solution.

44 degrees to the nearest degree.

This is triangle B.

You needed to draw on the perpendicular height or the altitude to calculate y.

And then you could go from that because it's a shared edge between the both of those right-angled triangles.

The side labelled x would be 8.

6 centimetres long, accurate to one decimal place.

Okay, so question three, there's lots of stages of working.

You needed to get both x and theta.

So using triangle D, because you had enough information on triangle D, you could calculate the adjacent, and that was 2.

7, roughly.

Triangle C, you were given the opposite, you had an angle, so you could work out both the adjacent and the hypotenuse.

So the adjacent, I've labelled as b, and that's 2.

63.

And the hypotenuse, which I've labelled as c, is 8.

51.

Both of those are really useful because the edge I've labelled b is a shared side on triangle A.

And the hypotenuse, which I've labelled c, is the sum of two sides on triangle B and D.

We're gonna continue on to the next slide.

So triangle B, we can calculate the adjacent is the full hypotenuse of C minus that adjacent on triangle D, so that equals 5.

82.

And from that, we can then use arccosine, or inverse cos, to calculate theta, which is 14 degrees to the nearest degree.

And then going back to triangle A, we have a shared side with triangle C, so that's that 2.

63, you have an angle of 43 degrees, and you want to find the hypotenuse, so it will be sine, and that is 3.

9 centimetres to one decimal place.

Really well done if you worked through that independently and got to those correct answers.

So to summarise today's lesson, which is about choosing the right trigonometric ratio, they are a proportional relationship between two lengths in a right-angled triangle based on the size of the focus angle.

Determining which ratio to use is the first step in answering any question in trigonometry.

Sometimes what you need to calculate is not achieved in just one step or one calculation.

So considering what you can calculate with the given information is always a correct first step.

Do consider your accuracy and do not round your answer until the very end of your calculation.

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