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 are looking forward to learning something new.

I'll be there to support you as we go through the lesson.

So our learning outcome today is to be able to apply the trigonometric ratios to practical situations, including angles of elevation and depression.

Here on the screen, there are some keywords that I'll be using during the lesson and you have met before, but let's go through them now together.

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

Again, refer to the diagram to support you with that definition.

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

And you can see the point Q on the tangent to the circle that is in the diagram.

You might still want to pause the video, read the definitions once again at your own speed and refer to the diagram to support you.

But when you're ready to move on, press play.

So our lesson today on apply trigonometric ratios in context is going to be split into two learning cycles.

In the first learning cycle, we're gonna identify the trigonometric ratio from context.

So we're gonna look at a few different contexts and which ratio is going to be useful for the context of that question.

In the second learning cycle, we're gonna look at angles of elevation and depression.

So let's make a start at looking at identifying the trigonometric ratio from the context that the question is in.

So the properties of right-angled triangles are applicable to real-life situations from many different vocations.

Before I reveal a few, can you think of any? Pause the video and think about when trigonometry might be used.

Press play when you're ready to move on.

So did you think of sailing or navigating? That doesn't have to be a yacht, it could be a cruise ship, it could be a speedboat, but navigating on open water.

Or did you think about film? Thinking about perspective.

What about satellites? These are only a few examples of where trigonometry is used and I'm hoping you came up with a few of your own.

So one area where trigonometry is used extensively is navigation.

So here we've got two boats.

We've got our north line to indicate the direction.

If we're given this further information, then what information can you deduce from the diagram? So pause the video and think about the information you can deduce now you have all of this information.

Press play when you're ready to go through it.

Okay, so we can say that the second boat is on a bearing of 70 degrees.

We can use the north line there and the angle that's within that right-angled triangle.

The distance between them is 400 metres.

That's the direct distance, which is a hypotenuse.

And we could use trigonometry to calculate the horizontal distance between the two boats as well as the vertical distance in this plan view.

If we've got the hypotenuse of that 400 metres, then we'd be using the sine ratio and the cosine ratio to get those two distances.

We might have also said that the angle missing in the triangle is 20 degrees using the interior angles of a triangle adding to 180.

You may have said that the bearing of the second boat to the first boat, the purple boat to the grey boat, you may have calculated what that was using all of the information that we have here and you should have got 250 degrees.

So there's quite a lot of information that we can deduce if we have that information.

So it's not always easy to identify how trigonometry fits into the problem, but drawing a diagram usually helps the situation.

And that diagram is probably then going to identify a right-angled triangle.

So Lucas says to Laura, "I'm trying to find you, where are you?" Laura gives him some information.

"I travelled 300 metres east of the lighthouse, then 150 metres north." So Laura has given some distance and also direction.

Lucas has then sketched a diagram of this.

So the lighthouse, she went 300 metres east and then 150 metres north.

Lucas says, "I'll try to work out what angle to turn to get to you quickly." So this is now our simplified diagram.

It's got the information that we need.

We've got the 300 metres east and the 150 metres north.

They are perpendicular directions, which is where the right-angled triangle appears from.

The angle that Lucas is trying to calculate is marked there as theta.

So can you see how Lucas is going to work out the angle? So here's the check, which calculation will find theta? Is it calculation A, arctan of 150 divided by 300? Is it B, tan of 150 divided by 300? Or is it C, arctan of 300 divided by 150? Pause the video and when you're ready to check, press play.

So the answer is A, and if you evaluate that using your calculator to three significant figures, it's 26.

6 degrees.

So the tangent ratio is the opposite divided by the adjacent.

The opposite of the angle theta is 150 and so the numerator needed to be 150.

So we're up to the first task of the lesson and here is question one.

Laura keeps getting lost after setting off from the lighthouse.

Help Lucas work out the angle to two decimal places to set off from to get to her quickly.

Lucas always starts facing the way Laura went when she left the lighthouse.

So there's four parts to this question.

Each one is a new question and a new diagram necessary.

Do draw yourself or sketch yourself that diagram to make sure you're working out the correct angle for the given distances.

So pause the video and then when you're ready to move to the next question, press play.

On question two, we've got a boat and the boat is trying to reach a swimmer in the sea.

Using the diagrams, how far away is the swimmer? So the direct distance from the boat.

So pause the video and then when you're ready to go through the answers to task A, press play.

So here's question one, part A: Laura went 200 metres east and then 75 metres north.

So the 200 metres is the adjacent and the 75 is the opposite.

So the arctan of 75 over 200 gives an angle of 20.

56 degrees to two decimal places.

On part B, Laura went 500 metres west, then 600 metres north.

So west is your adjacent and 600 is your opposite.

Arctan of 600 over 500 is 50.

19 degrees to two decimal places.

On part C, Laura went 780 metres west, then 150 metres south.

They are perpendicular directions.

And that's our right angle triangle.

The 780 is your adjacent and the 150 is your opposite.

So arctan of 150 over 780 is 10.

89 degrees to two decimal places.

On part D, there I've put the diagram here.

There were three stages to her directions.

So Laura went 300 metres east, then 150 metres north and then 50 metres east.

So in total she was 350 metres east of where she started and 150 metres north of where she started.

So the arctan of 150 over 350, her total horizontal distance, and that came out as 23.

20 degrees to two decimal places.

Really well done, especially on that part D if you manage that.

On question two we were looking at how far away is the swimmer from the boat.

So on part A, we were given the adjacent as 30 metres and the angle of 35 degrees.

We were looking to find the hypotenuse with a direct distance from the boat to the swimmer.

And so we needed to do 30 divided by cosine of 35 and that equals 36.

62 metres to two decimal places.

The degree of accuracy wasn't stated in the question.

So if yours rounds to 36.

62 to two decimal places, then you are correct, check the calculation as well.

On part B, again, you're looking for the direct distance from the boat to the swimmer.

This time we've got the opposite is 35 metres and the angle is 30 degrees.

So we need to do 35 divided by sine of 30 and that came out as 70 metres and that's an exact value.

On part C, we've got the adjacent is 35 metres and the angle is 35 degrees.

So it's a very similar question to part A, but the values are different so we need to work out the hypotenuse.

So 35 divided by cosine 35 gives you 42.

73 metres to two decimal places.

So we're now up to the second learning cycle where we're gonna look at the trigonometric ratios in context and in particular, angles of elevation and depression.

So trigonometry is also useful to identify angles of elevation, which is things above you and depression, which is things below you.

So if we've got our camera here and a person in front, then the angle from the horizontal up to the view to the top of the person is called an angle of elevation.

That the angle we are going up from the horizontal.

Whereas here, if we've got a helicopter maybe doing a search and rescue mission and a swimmer in the sea, then the angle of depression is the angle below the horizontal because the swimmer is below the helicopter.

Using parallel lines, the horizontal here, then we can also think about alternate angles in a parallel lines being equal and making use of the angle of depression in a right-angled triangle to calculate either the height of the helicopter or the horizontal distance of the swimmer from the helicopter depending on what information we've been given.

So what angle does the radar on the ground need to use to point directly at the satellite? So Lucas says, "What information would make this calculation possible?" So have a think about that, pause the video.

How are we going to work out the angle that the radar needs to use to point at the satellite in the sky? Is it an angle of elevation or is it an angle of depression? Press play when you're ready to move on.

So if we look at a right-angled triangle with our horizontal distance along the floor and our vertical distance up to the satellite.

So if we had two sides of this triangle, let's say we had the hypotenuse and the adjacent or we had the opposite and the hypotenuse, or even if we had the adjacent and the opposite, we could work out the angle using the one of the trigonometric ratios or if we had at least two angles in the triangle, then we could definitely work out the other angle because angles in a triangle, some to 180 degrees, that would be the easiest.

So if you were given the measurements for these two sides, the ones with the arrows, which trigonometric function would you use to find theta? So think of this as our satellite and our radar.

If we knew the horizontal distance along the ground, the level ground, but we also knew the distance directly from the radar to the satellite and we wanted to work out that angle of elevation that's marked as theta, which of the trigonometric ratios would we use? Pause the video and when you're ready to check it, press play.

So we've got adjacent and hypotenuse, so we'd use cosine.

How about if we knew the horizontal distance along the level ground and the vertical distance of the satellite from the level ground? Which of the trigonometric ratios would we use this time? Once again, pause the video and then when you're ready to check, press play.

So this time we've got the opposite and the adjacent.

So it would be the tangent ratio that we would make use of to get the angle of elevation.

How about now? This time we're not looking at the angle of depression exactly because this is not the angle from the horizontal, but it's from the satellite to the radar if we are still in that context, but which trigonometric ratio would we use with these two side lengths and this be in the angle we're trying to calculate? Pause the video and then when you're ready to check, press play.

So you've got the opposite and the adjacent, so we'd use tangent again.

So the angle of depression can be found by working out the angle in the right-angled triangle and subtracting it from 90 degrees.

So the angle is the angle from the horizontal.

And so if we calculate the angle theta inside of this right-angled triangle, then we can calculate the angle of depression by doing 90 minus theta, because it would be a right-angle, the horizontal and that vertical would be perpendicular.

However, you could work out the other angle within the right-angled triangle, and as I said earlier, using equal alternate angles that would be equal to the angle of depression.

So let's have a look at two examples.

One with the angle of depression and one with an angle of elevation.

So what is the angle of depression from the lighthouse to the boat? So we've got an angle there marked as theta.

Is that the angle of depression? Well, no, that one is not the angle from the horizontal.

So once we've calculated the value of theta, we're gonna subtract it from 90 in order to work out the angle of depression.

So now let's look at the two lengths that we have.

It's one kilometre from the lighthouse to the sailing boat directly and it's 560 metres horizontally from the cliff or the lighthouse to the to the sailing boat.

So if we've got the hypotenuse and we have the opposite, then we're going to use the sine ratio.

What else is the issue here? Can you see another issue? Well, hopefully you spotted that we've got mixed units.

So we do need to make sure we're either using all of the lengths in kilometres or all of the lengths in metres.

So we're going to do 90 minus arcsine or inverse sine of 560 over a thousand.

So arcsine of 560 over a thousand is the angle theta, but because we're trying to calculate the angle of depression, then we're subtracting it from 90.

And so that will be 55.

94 degrees to two decimal places.

So here's a check for you.

What is the angle of elevation from the camera to the actor? So pause the video and then when you're ready to check, press play.

So you should have used cosine this time because 0.

8 is the adjacent to theta and 1.

2 is the hypotenuse and cosine is the ratio of adjacent and hypotenuse.

Angle of elevation is the angle that's marked as theta.

It's the angle above the horizontal.

So 0.

8 divided by 1.

2 and then you need to do the arccosine or the inverse of cosine and that gives you 48.

19 degrees to two decimal places.

So we're up to the final task of the lesson for you.

Question one is here on the screen.

A ramp is set at a 15 degree incline to get up a large 70 centimetre tall step.

How long does the ramp need to be? So that's part A, how far from the step does it need to be placed, is part B.

And part C is the council decide it is too steep and it needs to be at an eight degree incline instead.

What all the new dimensions have to be? So definitely sketch yourself a diagram here to make sure you can see where your right-angled triangle is to consider which trigonometric ratio you'll need to be using to calculate which lengths.

So pause the video and then when you're ready for the next question of task B, press play.

Question two, find the following.

So in part A, the height of the tree and hopefully you can see the tree there on the diagram.

Part B, the height of the building.

And part C, the distance between the tree and the building.

So the horizontal distance between them.

Pause the video and when you're finished with question two and press play, we'll go for our answers to task B.

So question one, the ramp question.

So part A was how long does the ramp need to be? So that is the distance from the floor to the top of the step.

So that's the hypotenuse.

So 70 divided by sine of 15 degrees is 270.

5 centimetres to one decimal place.

So 2.

7 metres approximately.

On part B it was how far from the step does it need to be placed? And so that's asking how far away is the start of the ramp horizontally.

So 70 divided by tan of 15 equals 261.

2 centimetres to one decimal place.

So in part A, you were calculating the hypotenuse of a right-angled triangle with a opposite of 70 centimetres.

And on part B you are working out the adjacent of a right-angled triangle with an opposite of 70 centimetres.

On Part C, the council changes the angle, it says it's too steep.

So this time you need to recalculate, but instead of having sine of 15 degrees and tan of 15 degrees, you should have been using sine of eight degrees and tan of eight degrees.

And you can see the measurements there on the screen.

Question two, you needed to work out three things.

So part A, the height of the tree.

The triangle is isosceles, so the height of the tree is nine metres.

You may not have done it by just looking at that and recognising that this would be an isosceles triangle.

We're simplifying and modelling the tree to be completely vertical and therefore perpendicular to the level ground.

So if you've got a right-angle and a 45 degree angle, then the third angle is also 45 degrees and hence we know it's an isosceles.

You could have done this using the tan ratio as well.

On part B, you needed to work out the height of the building.

So the larger triangle is an enlargement scale factor nine.

So the hypotenuse of each triangle is nine route two and 81 route two respectively.

So the height of the building is 81 metres.

So here we're looking at two similar triangles that are overlaid.

They have a shared angle of 45 degrees.

We can use the little triangle with the tree, we can use Pythagoras' theorem to get the hypotenuse.

Or you could use a trigonometric ratio to work out the length of the hypotenuse.

Which is from the point marked on the ground to the top of the tree.

And that is nine route two.

Then by adding it to the 72 route two, that is the full hypotenuse of the largest right-angled triangle.

And then we can see that there is this scale factor of nine, nine route two times by nine is 81 route two.

And hence you can then use that the height of the tree will be a ninth of the height of the building.

So nine times nine is 81.

The distance between the tree and the building.

Well if we know these are both isosceles triangles, then the height of the building is equal to the distance from the point to the bottom of the building.

And we know that the distance from the point to the bottom of the tree is nine.

So 81 minus nine is 72.

So to summarise today's lesson about applying trigonometric ratios in context, trigonometric ratios are used in many scenarios and we've only discussed a few during this lesson.

Trigonometric ratios can be used to calculate the height of a structure and they can also be used to calculate the angle of elevation, which remember, is the angle when you look up to something, the angle above the horizontal as well as the angle of depression.

So that's where we are looking at something below us and it's the angle below the horizontal.

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