Lesson video

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

Today's lesson we're gonna have worked out how to use the cosine ratio to work out missing angles as well as missing side lengths in a right-angled triangle.

On the screen, there are some keywords that we'll be using throughout the lesson.

It might be that you wish to pause the video so you can refamiliarize yourself with those definitions before we start using them during today's lesson.

So the lesson is in three learning cycles.

The first learning cycle is to formalise the cosine formula.

The second learning cycle is to calculate angles using that formula.

And lastly, we're going to look at problems using the cosine formula.

So let's make a start at formalising the cosine formula.

So we know that these two triangles are similar.

The reason we know that is 'cause two of the angles in one of the triangles is equal to two angles in the other.

And we can find the length of the side labelled X by using a ratio table.

So these two hypotenuse, there's one centimetre hypotenuse on one of the triangles and then the similar triangle has a 15 centimetre hypotenuse.

And so, we can see that that has been multiplied by 15.

The scale factor of the enlargement of these similar triangles is 15.

The 0.

68 is the adjacent of the smaller of the two triangles.

And internally, we can see that the hypotenuse multiplied by 0.

68 gives you 0.

68.

So the proportion between those two edge lengths is 0.

68.

And that would be the same in our similar triangle.

To find the edge marked X, we can do 0.

68 multiplied by 15, and that is 10.

2.

So we now know that on a similar triangle to the original one, the adjacent would be 10.

2 centimetres.

And we can go further to generalise this for any right-angled triangle that has an angle of theta where the side adjacent to the theta and the right angle has a length of cos theta.

Back to our ratio table, we've got a hypotenuse of one and a hypotenuse that we've labelled as h because this is our generalised triangle.

So that h could be any number.

The scale factor of the enlargement is h.

One multiplied by h gives you the h.

And therefore, cosine theta multiplied by h would also give you this adjacent length on the larger of the two triangles.

The ratio in the triangle is that the one, the hypotenuse of one multiplied by cosine theta gives you cosine theta.

And that would be the same on the second triangle, that the hypotenuse h multiplied by cosine theta is the length of the adjacent, which we could as h times cosine theta, an expression for the edge length.

What does this tell us? Well, it tells us that the length of the hypotenuse, h, multiplied by the cosine of the angle theta equals the length of the side adjacent to that angle and the right angle.

And this is true for any right-angled triangle.

So using the ratio table, we can generate a formula that shows this relationship between the adjacent side and the hypotenuse.

And actually, we can represent that in three different ways.

So let's look at that now.

If we take a hypotenuse of any right-angled triangle and multiply it by cosine theta, then that would calculate or be equal to the adjacent edge.

So that's one form, that's one formula that shows the relationship between the adjacent side and the hypotenuse.

If we start with the adjacent, then working with an inverse, we can divide by cosine theta and that will be equal to our hypotenuse.

So this is a second form of the formula.

And lastly, if we start at the adjacent again, we can divide by the hypotenuse and this will be equal to cosine of the angle theta.

So here are three versions of the formula.

We can also derive the two other versions by doing a division.

So here is that first one that we formed.

The hypotenuse multiplied by cosine theta is equal to the adjacent.

If we divide both sides of the equation by the length hypotenuse, then on the left-hand side, we'll end up cancelling out the hypotenuse, that will be equal to one.

And so, all that will be left on that side is cosine theta.

On the right hand side, there is nothing to cancel down.

So we can say that cosine theta is equal to the adjacent divided by the length of the hypotenuse.

If we start with that same formula again, but this time divide by cosine theta on both sides, on the left-hand side, the cosine the divided by cosine theta will cancel down to one.

So it'll now have that the hypotenuse is equal to the adjacent divided by cosine theta.

And so, we've derived and come up with the same three versions of that formula.

When we're working with a non-generalized triangle, so this one here, we've got a hypotenuse of B, an adjacent of 21 and an angle theta of 72 degrees, we can substitute those into these three formulas.

B times cosine 72 degrees equals 21, or 21 divided by cosine 72 equals B, or 21 divided by B equals cosine 72.

We've just substituted the relevant part into our generalised formula.

So here is a check.

One correct equation for the right-angled triangle shown here is 20 multiplied by cosine 38 degrees is equal to the length of the adjacent.

So which of these are also correct to show the relationship between the hypotenuse and the adjacent side? So pause the video whilst you decide between A and B.

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

So dividing both sides by 20 will lead us to the formula or the equation B.

Cosine of 38 degrees is equal to the adjacent divided by 20.

So another check for you.

Which of these correctly shows the relationship of the hypotenuse and adjacent for this triangle? So pause the video whilst you decide which of those five are showing correct relationships.

Press play when you're ready to check.

So if we start at C, C is correct.

This one shows us that the hypotenuse multiplied by cosine of the theta is equal to the adjacent.

B and E are also correct equations.

So which of these correctly shows the relationship between the hypotenuse and the side adjacent to the angle theta for this triangle? Again, pause the video and once you decide, press play so that you can check.

D is in that original form of hypotenuse multiplied by cosine theta equals adjacent and B and C are rearrangements of that.

So each version of the formula, why do we have three different versions? Well, that's because a different part of the triangle is the subject.

And so, it's helpful to use the version where the subject matches the part of the triangle that you are trying to find the value.

So if you look at this right-angled triangle here, we are trying to calculate what the length of P is.

Labelling the triangle, we can see that P is the adjacent.

So we would want to use the formula where P or the adjacent is the subject.

Which is the top one? So we are going to use this form because the adjacent is the subject.

We're gonna substitute in our values.

103 is our hypotenuse length.

The angle theta is 57 degrees and we can then calculate that on a calculator.

And P is 56.

098 centimetres to three decimal places.

So another check.

Which of these correctly shows a relationship between the hypotenuse and the side adjacent to the angle of 16 degrees for this triangle? Again, pause the video and when you're ready to check, press play.

So A, B and D are correct for this triangle.

Which of these is the correct equation to find the value of Q most efficiently? So they are the three that you just chose, A, B and D.

Which one is the most efficient to use if you are trying to calculate the value of Q? Pause the video whilst you're deciding and then press play to check.

Q is the subject on B.

So B is the one that you would use to be the most efficient.

So find the value of the side labelled Q to two decimal places.

Pause the video whilst you use your calculators to calculate that and then press play to check your answer.

So to two decimal places, it is 88.

43.

This seems like a very sensible answer since this is the hypotenuse and it should be greater in length than the 85.

So it's always worth to do a sense check just to check that the value has come out either greater or less than depending on what you expect.

So we're up to the first task of the lesson.

And on question one, you need to substitute the labelled sides and angles in each triangle into the cosine formula and find the length of its missing side.

Press pause whilst you're doing that and when you press play, we'll go onto question two.

Here's question two.

You need to firstly match the triangle to the relevant equation and then use that equation to get the answer to two decimal places.

So press pause whilst you are matching and calculating and when you press play, we'll move on to question three.

Question three, there are six right-angled triangles for you to use the cosine formula to find the lengths of the missing labelled sides.

Again, give your answer to two decimal places.

Press pause whilst you're calculating those missing edge lengths and then when you come back, there'll be question four.

Question four, you need to find the lengths of the distances marked A, B, C and D on this diagram rounding your answers to the nearest integer.

It might be that you want to draw these as separate triangles to help you and anything you do work out in a previous part move on to the next diagram.

Press pause whilst you're working through question four, be very careful and round at the very final stage of your calculation.

When you press play, we'll go through the answers to task A.

So question one, you needed to substitute from the right-angled triangle diagrams into the formula and then solve it to find the value of A and B.

So the value of A was 25.

19 and the value of B was 5.

56.

Question two, you needed to match and then calculate.

So the triangle A matched with the equation D, triangle B matched with the equation F, which left triangle C with E.

Then using a calculator, you can find the value of X, giving your answer to two decimal places.

Question three, there were six right-angled triangles and you needed to find the length of the missing side which was labelled with a letter for each one, giving your answer to two decimal places.

It may be that you wish to pause the video so that you can go through them one by one and check your answers.

Question four, A was 34 centimetres, B was 12 centimetres, C was 16 centimetres and D was five centimetres.

With this style of question, unfortunately, if one of your first answers had an error that may have moved down the rest of your working.

So it might be that you find that you've got A correct but you've got B incorrect and then that means that C is incorrect and D is incorrect.

So I would pause the video if you have made an error and go through your working out.

The second learning cycle is calculating angles using the cosine formula.

So we've been working with lengths and now we're moving on to angles.

Jacob and Izzy have both been saying that they can calculate the lengths of sides because they have an angle and they also have one of the edge lengths.

So Jacob, "I can calculate the length of the side adjacent to the 40 degree angle because I also know the length of the hypotenuse." And Izzy said she can calculate the length of the hypotenuse because she knows the length of the side adjacent to the 59 degrees.

So here is Jacob's example.

If he wants the adjacent, he knows that six times cosine 40 will give him the adjacent, and that is 4.

60.

For Izzy, she's trying to calculate the hypotenuse.

She says she can do that with the length of the adjacent.

She would use this form of the formula because she's trying to calculate the hypotenuse.

So the hypotenuse is three divided by cosine 59 and that is equal to 5.

82.

Jacob is wondering how do we use this form of the cosine formula to find the angle if we know the length of the hypotenuse and the adjacent to that angle.

We can substitute in those two values.

We can substitute the adjacent and the hypotenuse, which is approximately 0.

565 when you evaluate that fraction to a decimal.

So how can you find the size of the angle theta? Well, on a unit circle, we can find the length of a side adjacent to the angle 30 degrees by applying the cosine function to that angle.

So our angle is 30 degrees and if we apply the cosine function to it, it gives us a value of 0.

866.

And that is the value along the x-axis.

So that's 0.

866 units and that's the length of the adjacent side on a unit circle.

The cosine function has an inverse function called arccosine or arccos for short.

And you may see it as cosine to the power of minus one on your calculator.

So how we apply this is if we do arccos to cosine of 30 degrees equals 0.

866, that's an equation.

So we're gonna do it to both sides of the equation.

Then the length of the adjacent side on the unit circle is 0.

866, and that gives us a 30-degree angle.

So finding the cosine of an angle outputs a ratio of the side adjacent to that angle and the hypotenuse.

And that's usually written as a decimal with respect to a hypotenuse of one, we can interpret the decimal as a set of equivalent fractions.

So cosine 30 degrees is 0.

866 over one, which is telling you that the adjacent is 0.

866 in length on a triangle with a hypotenuse of one.

But this is equivalent to 1.

732 divided by two, which is a triangle where the adjacent of 1.

732 has a hypotenuse of two.

Or even 86.

6/100.

This right-angled triangle would have an adjacent of 86.

6 and a hypotenuse of 100.

So the inverse function, the arccos function, takes a ratio of the side adjacent to the angle and the hypotenuse and gives you the size of the angle.

So arccos of 13/23.

Well, this is a triangle where the adjacent has a length of 13 and the hypotenuse has a length of 23, which we can evaluate as a decimal instead.

Our compute would give out an output of 55.

6 degrees.

So we would now know that a right-angled triangle with an adjacent and a hypotenuse of 13 and 23 respectively have an angle of 55.

6 degrees between them.

And we can use our calculator to find this angle.

So here, we've got another example of a right-angled triangle.

We've got an adjacent between theta and 90-degree of 38 centimetres and a hypotenuse of 89 centimetres.

So cosine theta is equal to 38/89.

And using arccos, we would work out the angle of theta.

So let's use our calculator to do this.

To write this onto the calculator, first of all, you need to press the Shift key, the arrow.

Press the cosine button and this will bring arccos onto your screen.

We're now up to the point of writing the ratio between the adjacent and the hypotenuse.

So press the fraction key, type 38, move down using the arrow key, type 89 and then press to the right so that you come out of the denominator.

Otherwise your cursor is still on the denominator.

And then, close the bracket.

Press execute.

This has now returned the angle of 64.

72 degrees or 64.

7 degrees to one decimal place.

So here is a check.

Which of the following equations are correct when finding the size of angle n in the equation? So cosine of n degrees is equal to 0.

86.

Which of the four are correct? Press pause whilst you work through that and when you're ready to check, press play.

B and C are correct.

So B is using the notation that you see on your calculators.

So cosine with like a power of minus one and part C is using arccos, which stands for the inverse cos.

Another check, which of these are equivalent to this equation? Once again, pause the video and then when you're ready to check your answers, press play.

A, C and D.

So A is equivalent because 8/18 is equal to 4/9.

4/9 is the simplest form of 8/18.

C is using arccos on 4/9 and D is also using arccos on 4/9, except 4/9 has been written as an equivalent fraction of 40/90.

Another check, which of the following equations are correct when finding the size of angle u in this triangle? A, B, C or D.

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

A and D.

So A is written as cos of u degrees is equal to the ratio of the adjacent and the hypotenuse, whereas D is solving to find u.

So we're using the arccos function.

B is not to using the arccos function, so that's not gonna calculate the angle.

And C, the fraction is the wrong way up, is inverted.

So we need to get adjacent divided by hypotenuse.

If you try to do C on your calculator, it will come up with a math error and that is because the value of cosine will never be greater than one.

69/16 is clearly a top-heavy fraction, an improper fraction, and therefore greater than one.

So we're onto the task of finding angles using the cosine formula.

So question one, you need to match the triangle to the equation and then to the size.

So there's three parts that you need to match up.

Press pause whilst you're doing that and then when you're ready for question two, press play.

Question two, there are six angles that you need to calculate using the cosine formula.

Part A and part B has got a little bit more structure for you, so you need to fill in the missing gaps and then you can continue through C, D, E and F.

Press pause whilst you're working through those six questions.

And then when you press play, we'll go through the answers to task B.

So task B, question one.

You needed to match the triangle, the equation and the value.

So A, the adjacent was three centimetres, the hypotenuse was four centimetres, cosine theta was equal to 3/4.

When you do arccos of 3/4, you get 41.

41 degrees to two decimal places.

B, adjacent was three, hypotenuse was five, and theta was the angle you were trying to evaluate and that came out as 53.

13 degrees using equation F.

So arccos of three divided by five.

And lastly C, the adjacent was four, the hypotenuse was five.

So cosine of theta is equal to 4/5.

And when you use arccos of 4/5, you get 36.

87 degrees.

Question two, you had those six questions where you needed to find the angle in each of those triangles.

A and B had a little bit of structure for you.

So part A, 32.

0 degrees was the answer to the angle A.

Part B, 28/71 would've been the fraction.

So adjacent over hypotenuse and then solving that on the calculator, 66.

8 degrees.

C, 29.

5 degrees, D, 64.

2 degrees E, 14.

8 degrees, and F, 29.

5 degrees.

So we're now at the last part of the lesson and the final learning cycle, which is exploring problems using the cosine formula.

So on the screen, we can see a right-angled triangle and cosine the equals 5/7.

Jacob says, "I wonder what the triangle this equation describes looks like.

Is there a way to find out?" So the cosine formula describes the ratio between the length of the hypotenuse and the side adjacent to the angle theta.

The five is the length of the side adjacent to theta degrees and seven is the length of the hypotenuse, which we know is the longest edge of a right-angled triangle, always opposite to the right angle.

So this example here, this right-angled triangle is what that equation is describing.

However, we can write equivalent fractions to 5/7, and the value of cosine theta will not change and therefore the size of the angle theta, and that's because the ratio of five to seven is in its simplest form.

So cosine theta equals 15/21, another triangle, but the angle of theta would be the same because the ratio or the proportion of the adjacent and the hypotenuse are the same.

2.

5/3.

5, also in the same ratio.

35/49, also in the same ratio.

Those four triangles are similar to each other.

They're enlargements of each other.

The ratio of their adjacent to hypotenuse is the same in all four.

And so, if we were to use the arccos, the inverse cosine function on 5/7, 15/21, 2.

5/3.

5 or 35/49, they will all output 44.

4 degrees to one decimal place.

So in order to draw a unique triangle whose equation of cosine theta is 5/7, one more piece of information is required.

So if we are told that the cosine equation for a triangle with a hypotenuse of 77 centimetres is cosine theta equals 5/7, what is the length of the side adjacent to theta? So we can answer this by sketching two similar triangles.

The first one having an adjacent side and a hypotenuse equal to the numerator and the denominator of the fraction that's given.

So this one here, the adjacent is five and the hypotenuse is seven.

Our question told us that we had a triangle with a hypotenuse of 77 centimetres.

So in our next drawing, our similar triangle will mark our hypotenuse as 77 centimetres.

So what would the adjacent be? Well, the hypotenuse is 11 times the size.

The scale factor of the enlargement between these similar triangles is 11.

And therefore, our adjacent is 11 times longer than the adjacent of the smaller one.

That was one way to do it, which is to think about how they were sort of an enlargement of each other between the two edges, but we could also do this using a ratio table.

So 5/7 means that our adjacent is five and our hypotenuse is seven.

In the question, we needed our hypotenuse to be 77.

So we'll place the 77 in the hypotenuse column underneath.

From here, we can see that, to get from the top row to the bottom row, we've multiplied by 11.

Alternatively, to get from the hypotenuse to the adjacent, we would times by 5/7.

And if we do that, you get 55.

So five times 11 is 55, but equally 77 times 5/7 or 5/7 of 77 is 55.

Here's a check for you.

The cosine equation for a triangle with a hypotenuse of 70 centimetres is cosine theta equals 11 over 20.

Find the values of A, B, C and D in these similar triangles that satisfy this equation.

So pause the video.

It might be that you need to go back and re-watch a little bit before you try this one, but press play when you're ready to check.

So A was 11 centimetres, B was 70 centimetres, C was 3.

5, you may have written that as 7/2 as a fraction, and D was 38.

5 centimetres.

Another check.

This time we're using a ratio table.

So find the values of A, B, C and D in this ratio table.

Pause the video when you're working that out and then press play to check your answers.

So check against your answers for this.

What did you get for A, B, C and D? Continuing with this, the cosine equation with a hypotenuse of 90 centimetres is cosine theta equals 0.

45.

What is the length of the adjacent? So if the value has been written as a decimal, we can convert a decimal to an equivalent fraction in one of two ways.

So it's 9/20.

We could write it as 0.

45/1.

9/20 means that the adjacent nine, the hypotenuse is 20 and we want the hypotenuse to be 90.

So putting that into the ratio table, we can see that 20 multiplied by 4.

5 is 90, and therefore our adjacent would be 40.

5 centimetres.

If we'd use the other equivalent fraction, the adjacent of 0.

45 and our hypotenuse of one, then putting our hypotenuse of 90, that's an easier, more obvious multiplier of 90 and we'll still get to the same answer.

A check.

The cosine equation for a triangle with a hypotenuse of 72 centimetres is cosine theta equals 0.

79.

So find the value of A in the equivalent fraction.

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

That would be one.

If our numerator equals our decimal, then we must be dividing by one.

So carry on and calculate the side adjacent to theta in this triangle.

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

56.

88.

The multiplier is 72 and therefore 0.

79 times 72 will give you that adjacent edge on the similar triangle.

So onto the last task.

Match the pairs of similar triangles and find the size of all mixing angles labelled with a letter, giving your answer accurate to one decimal place.

So pause the video whilst you work through the six angles.

Then when you press play, we'll move on to question two and three.

So questions two is on the left-hand side, question three is on the right-hand side.

So the cosine equation for a triangle with a hypotenuse of 175 is given.

By using similar triangles, what is the length of the side adjacent to theta? On question three, again, you've got a equation for cosine and you need to have an adjacent of 81 centimetres.

Use the ratio table to the hypotenuse of that triangle.

Press pause whilst you're working through questions two and three.

And then when you press play, you've got question four and five.

Question four and five are on the screen.

Question four, you need to find the length of the adjacent in a triangle which has a hypotenuse of 63 centimetres and a ratio between the adjacent and hypotenuse of 0.

178.

Question five, you have the cosine equation for the triangle and you need to find the length of the hypotenuse and the size of every angle in the triangle.

Press pause whilst you do question four and five, and when you're finished and press play, we're gonna go through the answers to task C.

So question one, the answers are on the screen.

A and E actually have the same value.

B and C have the same value and D and F have the same value.

And it should be that you can see why.

And the reason why is because the adjacent divided by their hypotenuse is equivalent in those two triangles.

Question two and three are on the screen now.

So on question two, you needed to work out the length of the adjacent, which would be 80 centimetres.

And on question three, you needed to get the length of the hypotenuse, which would be 360 centimetres.

Question four and five, the answer of the adjacent for question four, 11.

214.

So question five, you needed to find the length of the hypotenuse and the size of every angle.

So the hypotenuse was 250 centimetres.

We can calculate the angle of theta as 71.

6 degrees using arccos, and then using angles in a triangle, you can calculate the third angle as 18.

4 degrees.

So to summarise today's lesson, the relationship between the hypotenuse of a right-angled triangle and the side adjacent to an angle theta can be written as a formula in three different ways and they're there on the screen.

This cosine formula can be used to find the length of the hypotenuse, which is the middle version, the length of the side adjacent to the angle theta, which is the right-hand formula, or the size of angle theta itself, which is using the left-hand formula.

In order to find that angle, the function inverse to the cosine function must be used and this is the arccos function.

On a calculator, we're gonna see arccos or arccosine usually written as cos to the power of minus one.

A triangle can be constructed from its cosine formula and either the length of its hypotenuse or the side adjacent to one of its angles in order for it to be a unique triangle.

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