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 use our knowledge of trigonometry to solve problems. On the slide, we can see many keywords, and all of these you've met before.

So let's read through them together before we start our lesson.

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 at point P on the triangle formed inside the unit circle.

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

And the point P can be seen on the diagram on the slide.

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

The tangent of an angle is the Y coordinate of point Q on the triangle which extends from the unit circle.

And once again, that point Q is on the diagram.

So I still would encourage you to pause the video so that you can refer to that diagram and check that alongside these definitions to make sure that you are confident with that before we start.

So this lesson is problem solving with non right-angle trigonometry, and we're gonna split the lesson into two learning cycles.

In our first learning cycle, we're gonna think about abstract problem solving, and then when we get to the second learning cycle, it's going to be more contextual.

So let's make a start looking at problems that are a little bit abstract and making use of our non right-angle trigonometry skills.

So Andeep says, "It's time to solve some more challenging questions using trigonometry." So we are here, we have a diagram.

Andeep says, "What is the area of this rectangle?" So just pause the video and think about this for a moment.

Think about what knowing it's a rectangle indicates, what implicit information that also gives us, alongside the fact that we know that that edge is nine centimetres and the angle between those two diagonals is 54 degrees.

Press play when you're ready to move on.

So I'm hoping you've thought about the fact that you know opposite edges in a rectangle are equal in length, that you know there are four right angles, that you know we have two sets of parallel sides, and these diagonals then allow us to think about alternate angles being equal, co-interior angles summing to 180, vertically-opposite angles are equal.

So there's quite a lot of information we can get.

There's other information that I've not stated.

So Andeep says, "Let's start by seeing what dimensions and measures we can find and then go back to our question." So here you can see that we've got the 54-degree angle that was given and stated, and then we've got opposite angles are equal, angles around a point, some to 360, angles on a straight line, some to 180.

So we can get all four angles around that middle point.

Then we need to think a little bit further about properties of a rectangle.

So we've got these diagonal lines, and these diagonals will meet and bisect each other.

So at the point of intersection, they bisect the diagonals, that's a property of the rectangle.

So if those two edges are equal, then these are all isosceles, and that's how we can then calculate the base angles of the four isosceles triangles.

You don't have to do all of them because we have got congruent isosceles triangles.

There's two pairs of equal triangles, two pairs of congruent triangles.

We can use the fact that it's a 90-degree angle, the interior angle of a rectangle.

So if we get the 63-degree angle, for example, then we know the other side of the diagonal would be 27 degrees.

So we can work out all of those angles within the diagram.

So there are two pairs of triangles making up the rectangle, as I said, but how many triangles do we need to find to the area for? Remember, we're trying to find the area of this rectangle.

So pause the video and think about that.

The diagram might support you with that thought as well.

Press play when you're ready to move on.

So if we think about splitting the diagram in half along that line of symmetry, then we can see that it's the whole area of the purple triangle, and then we've split the isosceles triangle in half, but we've got both halves, so it's one of each of those.

So it's the whole triangle that is purple plus one of the whole other triangles.

But if we split it vertically along the line of symmetry, so again in half, then we can think about it in a different way.

Andeep says, "Each one is a quarter of the area of the rectangle," so make sure you're confident with that, that the way that it's been divided up, we've got two pairs of congruent triangles and each one of those triangles is a quarter of the area.

So if we can work out the area of this triangle, for example, by multiplying it by four, we'll have found the area of the rectangle because that is equivalent to a quarter of the area.

So Andeep started to use the sine rule because we do have a pair of corresponding sides and the angle opposite, the nine centimetre and the 54 degrees, they were opposite each other and are corresponding.

So the sine rule can be used to work out the length of one of the other sides.

And so A is 9.

9121, there's still some more decimals after that.

So we're gonna try and use the most exact and we'll use our calculator for that.

And then we can make use of the sine formula for the area because we now have two adjacent sides and an angle between them.

So Andeep's remembered half AB sine C.

So if we use the nine, the 63 degrees, and the edge that we just calculated using the sine rule, then we can work out that the area of this triangle is 39.

7 approximately, and we know that this is one quarter of the area of the rectangle.

So if we multiply it by four, we get the full area, which is 158.

97 to two decimal places.

So here is a check for you.

Which equation is true for the length X for this rectangle? So pause the video, consider the three options, look at the diagram.

When you are ready to check, press play.

So it's equivalent to C.

We needed to think about what X was.

2X was the whole base of the isosceles triangle, so X would've been half of it, and therefore, X would be opposite 20 degrees because the altitude of that triangle would bisect the 40-degree angle that was given.

The six comes from the fact that that altitude is half of the length of the rectangle.

So if we know that the length of the rectangle is 12, then we know that that altitude or that perpendicular height is six, and the angle that that is opposite is 70.

Because if that's an isosceles triangle, then 180 minus 40 and dividing it by two works out that that is a 70-degree angle.

So for the first task on abstract problem solving, both questions are on the screen here.

So for question one, calculate the area of each of these rectangles, and give your answer to one decimal place.

So in each case, you're gonna need to consider the information you've been given, what further information you could calculate, whether you need the sine rule or the cosine rule or potentially both, and you're trying to work out the area of the rectangle ultimately.

On question two, you need to use the gridded background and the diagram to work out the size of the angle shown.

So pause the video to work through task A and when you finish those two questions, press play and we'll go through our answers.

Question one, all of the areas are on the screen.

So 53.

6 square centimetres, 75.

2 square centimetres and 2,181.

6 square centimetres, all to one decimal place.

Part A was very similar to the explanation that we went through, so in terms of what processes to do.

For part B, you definitely would've need to add things onto the diagram, so for example, the angle that is 18 degrees is part of an isosceles triangle, so the other angle would also be 18 degrees and that would mean that the non-equal angle would be 144 degrees.

An edge of the isosceles would be eight because the fact that the diagonals do bisect each other, and so your isosceles triangle would have two eight-centimeter edges, and then you can go on further working out the area of that triangle using half AB sine C, and then timesing it by four.

There are, however, other ways of you calculating that area.

So that's not the only method that you needed to use.

For part C, you were given the length of the rectangle, you had the 132-degree angle, which means you could work out the top angle of the other isosceles triangle, 'cause angles on a straight line sum to 180.

You could work out the base angles of that isosceles.

You would have the pair of corresponding edges and angle.

So you could use the sine rule.

Again, lots of different methods that you could go to use.

You could use right angle trigonometry because you could drop the perpendicular of that isosceles, and with the angle and the length, work out the perpendicular heights and then times it by four once you've got the area.

So as I say, question one has many methods to get the same answer.

So as long as you've added things to your diagram, thought about what you have, what you can do with the information you have, and have a bit of a process and a strategy to get the area.

On question two, the size of the angle is 135 degrees, but we're now gonna look at that in a little bit closer in detail in terms of how you could have got to that answer.

So here we have the diagram larger so we can discuss it, and we've got that angle that we're trying to work out.

We know it's 135 degrees, but how can we show that and work that out? Well, if we call one unit the distance of five of these little squares or between the bold lines, then this is a trapezium, and the top parallel edge is two units and the bottom edge is three units.

We can see that this trapezium is constructed by a rectangle and a right-angled isosceles triangle.

So by splitting this, thinking it as like a composite shape, we can think rectangle and triangle.

So this diagonal, which would be the hypotenuse because it is opposite of right angle, would be root two of units.

And we can use Pythagoras' theorem.

We know it's one unit by one unit, so one squared plus one squared and then square rooted.

There is a large right-angle triangle with short edges of one unit and three units.

So we can use Pythagoras' theorem again to calculate the hypotenuse as root ten, one squared plus three squared square rooted.

So now we've got three edges of the other side of that diagonal, which is not a right-angle triangle.

However, there is another right-angle triangle on that diagram, and that's with two short edges of one and two units.

They are perpendicular to each other.

So we can use Pythagoras' theorem to calculate that length of root five.

If we consider this triangle that's now highlighted in purple, we've got three edges, but because it's a right-angle triangle, we don't need to use the cosine rule, although we could.

Instead, we can just use the tangent ratio.

The opposite is three and the adjacent is one, so the angle that's marked as theta is 71.

6 degrees, roughly to one decimal place.

If we then go back to using this right-angle triangle, again, you could use the cosine rule because you've got all three edges of the triangle, but we don't need to, we can use the tangent ratio.

Two is the opposite and one is the adjacent.

Now, the angle marked delta is 63.

4 to one decimal place.

So if we know theta and we know delta, we've calculated both of them, then they will sum to 135 degrees, and that is the angle we're trying to calculate.

It may be that you remember that the exterior angle is the sum of the two in interior angles that's not at the same point.

That's true for all triangles.

Alternatively, we can say that we know the angle that's next to the 135 degrees will be equal to 180 minus the sum of delta and theta, and then the angle that we're trying to calculate will be 180 minus 180 minus delta plus theta, and so that leaves us delta plus theta, which is 135.

So we're now up to the second learning cycle where we are gonna continue working with non-right angle trigonometry, but this time, we're looking at problem solving in context, so contextual problem solving.

So Andeep says, "Let's now try solving problems within a story." So we've got some gridded paper here, and we're gonna say that one square equals one unit, and that's the more dark, bolded lines.

So Andeep says, "I want to navigate between these three islands." What are the distances and angles between them.

So we can join them up and that then makes a triangle.

Andeep says, "What steps might we take towards a solution for this problem?" So pause the video for a moment and think about this.

From the information we have, what can we do? Press play when you're ready to move on.

So Andeep said, "Does the grid help us?" Did you consider the grid? Because at the moment, we have no numerical values except from a scale to do with the grid, that we know that one square is one unit.

So if we look between these two islands, we can draw ourselves a right-angle triangle by considering the horizontal distance between the two islands and the vertical distance.

We can do the same between these two islands and the last pair of islands.

So what will this allow us to work out? Pause the video and think about what you can work out with these right angle triangles.

Press play when you're ready to move on with Andeep.

So from these, we can use Pythagoras' theorem.

Horizontal and vertical are perpendicular directions and so we can use our scale.

If we look at the top right-angle triangle, it's got a horizontal distance of five units and a vertical distance of one.

So we can use Pythagoras' theorem to work out the distance between the two islands.

Five squared plus one squared is 26, so that is root 26 in length.

And we can do that for all pairs of islands.

So we have root 26 between those two, root 10 between those and root 40 between these two.

Root 40 does simplify, so if you've got that written as two root 10, that's also correct.

So what can we now use to find an angle," says Andeep.

So the angle between the islands, what can we now use? Once again, pause the video and think about that.

From this information that we've just calculated, how do we get the angles? Press play when you're ready to see what Andeep thinks.

So Andeep thinks we should be using the cosine rule.

So the reason it's not the sine rule is because we do not have a pair of corresponding edge and its angle, so we can't use the sine rule, whereas the cosine rule, if we have the three edges of a triangle, we can use the cosine rule to calculate an angle.

So what is the value of angle A to one decimal place? So angle A is the angle between the root 10 edge and the root 26 edge, and is opposite the root 40 or the two root 10.

So pause the video and use the cosine rule to calculate the angle A Press play when you're ready to check.

So the angle is 97.

1 degrees, so you should have been substituting root 40, root 26, and root 10 into the cosine rule and rearranging it to make cosine A the subject and then using arc cosine or the inverse of cosine function to calculate that angle.

Okay, what about B? So pause the video, have a go at calculating angle B, and then when you're ready to check, press play.

Angle B is part A, 53.

1 degrees.

You did then have a different option.

So you could use the cosine rule again, but this time, you would be doing root 26 squared equals root 10 squared plus root 40, all squared minus two times root 10 times root 40 times cosine of B, rearranging it to make cosine of B the subject and then using arc cosine to calculate B.

However, because we just calculated the angle at A, we then did have a pair of corresponding side and angle, the angle A and the root 40, so you then could have used the sine rule.

The issue of using the sine rule with the answer that you just calculated is that you may use a rounded value and that would then increase the rounding error.

And if you had got the angle A incorrect, then you would get the angle B incorrect as well.

At times, you might want to be mindful about using the information you've calculated or the information you've been given.

Lastly, what is the value of angle C to one decimal place? And again, think about how you could get this angle having worked out angle A and angle B previously.

Press play when you're ready to check.

So this is 29.

7 degrees to one decimal place, and I'm hoping that you thought about, actually, I don't need to use the cosine rule nor the sine rule.

I know the angles in a triangle sum to 180 degrees and I know the other two, so by adding angle A to angle B, you can find the difference from 180 degrees, and that will be angle C.

Alternatively, you could have used cosine rule or sine rule.

On this Task B, you've got two questions, they're both on the screen.

So find the length of the root between each point on the maps.

And on question two, what are the angles between each point to one decimal place? So that is referring to the maps in question one.

So work out the lengths between each point and then work out the angles between each points.

So pause the video, have a go at that, and then when you're ready to go through your answers to task B, press play.

So question one, all of the lengths are on the screen.

So root 26, root 50, and six.

Root 26 doesn't simplify, but root 50 does.

So if you have written five root two, that's equivalent, well done.

Then we got root 17, root 41, and four, neither of those simplify, they're in their simplest form, and we kind of know that because of the prime numbers.

And then on the third diagram, it's eight centimetres, 6.

39, and five centimetres.

The 6.

39 was the one that you needed to calculate, and you could do that using cosine rule because you had two adjacent sides and the angle between them.

For question two, you then needed to work out the angles between the points on the maps, and so that was 56.

3, 78.

7, 45.

0 to one decimal place.

On the second one, it was 37.

3, 104.

0, and 38.

7 degrees to one decimal place.

And on the third one, you already had one of the angles, that was 53 degrees, so you needed to calculate that the other two were 38.

7 degrees and 88.

3 degrees.

So to summarise today's lesson on problem solving with non right-angle trigonometry.

Non right-angle trigonometry can be applied in various contexts, and we only looked at a couple during the lesson.

It has the potential to be applied whenever a triangle can be drawn.

And it's really important that when you're meeting a problem, that if you can find a right-angle triangle, then that gives you the options of Pythagoras or the trigonometric ratios, but if you can find any triangle, then you've got the sine rule and the cosine rule available to you as well.

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