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.

Today's lesson outcome is to be able to consider which approach is useful by considering the properties or restrictions of a problem.

On this keyword slide, we've got the sine rule and the cosine rule.

I would suggest that you pause the video, read back through them, and potentially write them down on your workbook before we make a start.

Press play when you're ready to move on.

So this lesson is about considering the appropriate trigonometric rule, and we're gonna break the lesson into two learning cycles.

In the first learning cycle we're gonna focus on non right-angled triangles and problems involving non right-angled triangles.

When we get to the second learning cycle, it's gonna be mixed trigonometric problems, so it might include some right-angled triangles as well.

So let's make a start at looking at these problems involving non right-angled triangles and deciding which rule is appropriate.

So in this learning cycle where we're looking at problems involving non right-angled triangles, we're gonna consider strategies that we have available to us.

So there are a range of strategies available to finding missing measures in triangle geometry.

And the key is figuring out which tool fits best.

And this is true for most mathematical problems. That often you're gonna have a whole array of skills and you need to be able to decide what's the most efficient one to use for the problem that you're faced with.

And that comes with practise, and being exposed to many different style of questions, and recognising key parts that you understand.

So if we look at this triangle here, what do we know about it? Well, we know it's a right-angled triangle.

How do we know that? Well, because it has a right angle marker.

So then we need to think in our head, what else do we have available for right-angled triangles? So we know it's a right-angled triangle.

We also know it's a scalene triangle because two of the edges are different already, which means that the third edge will be different, as a right-angled triangle can be isosceles, but only if the perpendicular lengths were equal, in which this case they are not.

So for right-angled triangles we're likely to need the Pythagorean theorem, or the sine, cosine, and tangent ratios.

So when we are looking at triangle geometry, if we can see a right-angled triangle, if we can create a right-angled triangle, then our brain should start thinking about the use of Pythagoras' theorem.

If you know two of the three edges or sine, cosine, and tangent, which is when we have an angle involved as well.

So for this right-angled triangle, we've got sufficient information to use Pythagoras' theorem because we had the four and the three, so we could work out the length of the hypotenuse.

If we only had one of those edges, but we also then had the 37-degree angle, we'd be able to work out the other two edges by using the different ratios.

With the given information on this diagram, we can work out the third angle, the third interior angle because angles in triangles sum to 180 degrees.

If we were looking to work out the area of a triangle, because it's a right-angled triangle, those two edges that we have are perpendicular.

So we can work out the area of the triangle using half times base, times perpendicular height.

If we wanted to work out the perimeter, then we know we need the third edge to be able to sum the edges of the triangle.

And we have the skills in order to work that out.

But for non right-angled triangles, we are more likely to need to use the sine rule, the cosine rule, or the sine formula for finding the area of the triangle.

So in this triangle here we've got two edge lengths, and actually that's not sufficient information at this point to work anything else out.

If it was drawn accurately, if we knew it was drawn accurately, we could measure the angles to find an angle.

For any problem, but as we're working with these triangle geometry problems at the moment, it's useful to think about what information you're provided with as a starting point.

So Jacob says this problem involves adjacent sides and the angle between them.

So the nine centimetre and the four centimetre have a shared vertex.

And at that vertex we know the angle is 80 degrees.

So then your brain needs to start thinking, so what tools could I use with this information? So pause the video and just think for yourself for a moment.

We've already said that when it's non right-angled triangles, you're probably going to be looking at using sine rule, cosine rule, or the sine formula for the area.

So with this given information, what would we have available? Press play when you're ready to see what Jacob says next.

So here is the sine rule.

Is this useful here? Well, no.

For this to be useful, you need an angle and its opposite side.

So to use the sine rule, you will need a pair.

At this point in time with the information we have, we do not have the edge and its opposite angle.

So then let's think about the cosine rule.

Is this useful here? While we do have enough information to use the formula and that will work out the missing side length, so the side opposite the angle 80 degrees.

So with the side angle side with the adjacent sides and the angle between them, you can use the cosine rule.

If we then found that third edge, then we would have a pair, we would have the edge and its opposite angle.

So that would then allow us to use the sine rule.

For example, if we wanted to find out another angle within the triangle.

What about the sine formula for finding the area? Is this useful here? Well, yes, we can use this if we wanted to find the area of this triangle.

A and B would be our nine and our four, and this angle C would be our angle of 80 degrees, two sides that are adjacent and the angle between them.

So I'm gonna challenge you and Jacob's challenging you to find the perimeter, the area, and all the internal angles of this triangle.

So pause the video whilst you work through that.

You might wanna rewatch some of the bit we just spoke through about which formula is gonna be most useful when, which one can you start with with this information.

And then when you're ready to check the answer to finding the perimeter, the area, and the internal angles of the triangle, press play.

So the perimeter means that we need to work out the third edge.

And with the given information, the cosine rule is the only one we can use from sine rule and cosine rule.

So let's use the cosine rule to work out that third edge.

So we start with the generalised form and now we're gonna substitute our information into it.

So a squared is what we're labelling to be the unknown edge.

Then we've got 4 squared and 9 squared minus 2 times 4 times 9, which is why that's 72 times by cosine of 80.

Remember that when we evaluate those squares, we get 16 and 81.

When we sum them together and that gives us 97, we do not then subtract 72.

The 72 is part of a product.

So we need to use our calculator to evaluate what 72 cosine 80 is 'cause that is a product of multiplication.

And then we can do 4 squared plus 9 squared minus that value.

The calculator can do it all in one line if you so wish.

And so we now know that a squared is 84.

497.

I can estimate that our value for a is greater than nine because nine squared is 81.

So we're gonna square root it and it's 9.

192.

There's more values, that's not exact.

So now we've got an edge length for the third edge.

The perimeter is a case of just summing them.

So the perimeter to two decimal places is 22.

19 centimetres.

Did you manage to do that by yourself? I hope you did and really well done if you did.

But if not, have a look at this screen, make sure you're happy with what we did.

We used the cosine rule to find the third edge, added the three edges together.

Then to find the area, we'll go back to what the original information was.

We can use the sine formula for area.

So half ab sine C because we have two adjacent edges and the angle between.

So substituting that in and evaluating it, the area would be 17.

73 square centimetres to two decimal places.

Once again, well done if you managed to do that independently.

Then the angles, we've got a few options at this point.

So you could use the cosine rule because you've now got all three edges.

So we can use the cosine rule to work out one of the other angles and then 180 minus the sum of the other two.

Alternatively, we can use the sine rule because we now have a pair of corresponding angle and edge.

So if we use the sine rule, we can substitute in.

I've chosen to use the length of four, but I could use the length of nine.

If I use the length of nine then I'd be working out the other angle, I'd be working out the angle opposite to the edge that is nine.

So I've put in our length of a.

I'm gonna use the exact value from my calculator.

I'm gonna use the answer key, so I'm not including a rounding error.

That divided by sine of 80, so that's the angle opposite, it is equal to the four divided by the sine of the angle opposite that, which we're gonna call B.

So I need to rearrange this to get sine B as the subject and then use arcsine to find the angle.

And so the angle is 25 degrees to the nearest integer, the whole degree.

And we can just use angles in a triangle summed to 180 degrees to work out the other angle.

And that can be rounded to 75 degrees and those three angles add up to 180 as we would expect.

So once again you might wanna pause the video, especially the rearrangement from the second line of working to the third line of working.

So how did we make sine of B the subject, firstly, and then we used arcsine or inverse sine to work out the angle of B.

Press play when you're ready to move on.

So here is a check for you, What is the perimeter of this triangle? Giving your answer to two decimal places.

So pause the video, work that out, and when you're ready to check that you've got that correct, press play.

So to two decimal places it is part C, 34.

72.

So you'd used cosine rule to work out the third edge and then added that to 12 and 10.

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

It's got two questions of which both of them are on the screen here.

So on question one you are calculating the perimeter of these triangles to two decimal places.

And on question two you're calculating the missing angles for the triangles.

Given the edge lengths, you need to work out angle A, angle B, and angle C, remembering that they are the angles opposite the edge A, opposite the edge B, opposite it's corresponding edge.

So pause the video, work through both of those questions.

When you're ready to go through your answers, press play.

So all the answers are on the screen.

It may be that you wish to pause the video and check how you've got on.

On question one, they will all have taken a very similar strategy to work out the perimeter, but how you did that will be different.

So on the first triangle, you'd use the cosine rule to work out the third edge and then add it onto the other two.

On the second it was an isosceles triangle, which means that we can calculate that the other two angles in the triangle were 25 degrees each.

By knowing that we then had a pair, well, we were given a pair anyway, 130 degrees on its opposite edge of 14.

But we would also know the angle, so we could use the sine rule to work out the edge lengths that are missing, and then once again add them up.

On the third triangle, you've got side angle side, you've got a pair of adjacent sides and the angle between, so we'd use the cosine rule.

And for the last one we've got a pair, 72 degrees and the opposite edge is 14.

So we can make use of sine rule with the 42-degree angle to get the edge opposite that.

And you can use angles in a triangle, sum to 180 to work out the missing angle, interior angle.

And then you had a choice, you could use sine rule once again or you could use the cosine rule 'cause you'd have then had two edges and the angle between them.

Once you've worked out all three, 'cause there's a lot of stages of working there, you'd need to add them together.

On question two you were working out the missing angles.

So you would be making use of cosine rule to begin with because you have all three edge lengths, and then you'd use arccosine to calculate the angle.

Once you've got that, you would have a pair, so you could then swap to sine rule if you so wished.

Or you could just use a different rearrangement of the cosine rule.

So the angles are on the screen to two decimal places for both triangle A and triangle B.

Now, we're up to our second learning cycle where we're going to look at mixed trigonometric problems. So on this learning cycle, we're now gonna look at mixed trigonometric problems. So it's important to be able to select the best tools to solve triangle problems, but often there's more than one way to solve them.

So this was one of the questions that you had in your task.

It was an isosceles triangle.

The base edge or the edge that we can see there is labelled as 14 centimetres and the angle opposite it is 130 degrees.

So we can find the perimeter of this shape using the sine rule to find each side.

And first you'd have to deduce the size of the missing angles, which would be 25 degrees, because we can do 180 minus 130 and then divide it by two, 'cause we know those two angles are equal.

And because you've got the pair, you've started with a pair, the corresponding pair, we can set up the sine rule where A is one of these missing edges.

We know that both edges are equal so they would both be A.

But you could also use our trigonometric ratios and our right-angled trigonometry using this right-angled triangle.

Once again because it's an isosceles triangle, we can drop that perpendicular, which is the altitude, which does bisect the base edge and that top angle.

It creates two congruent right-angled triangles.

So we would have then an adjacent with the 25 degrees or it's the opposite to the 65 degrees.

So you could use sine.

And we'd want to work out the hypotenuse because we're trying to get the edges.

And so cosine of 25 is equal to seven over the hypotenuse and we can rearrange that to calculate it.

So which do you think is the most efficient strategy and why? So pause the video and think about that.

If you're with somebody, discuss it.

You may have differing opinions and that's absolutely fine.

It's about articulating and explaining why you believe one method is more efficient than the other.

So pause the video to have that discussion, or to think about it if you're not with a partner.

And then when you press play, I'll give you my opinion.

So I personally don't feel that one is more efficient than the other.

There is still only one calculation that you need to do before you can add up the sides.

The first example with using the trigonometric ratios does need you to remember that the altitude will bisect the base edge and use the line of symmetry of an isosceles triangle.

Whereas the second one, all it is, is angles inside of a triangle, and we can do that fairly easy.

Both of the calculations are fairly basic.

Once you've used them quite a lot, you feel more familiar and your algebraic skills of rearranging are strong, but you just might prefer one over the other.

So here is an example which strategies would work to find a missing side length using the information given here.

So pause the video, look at the information that you do have, look at the options for the answer, and make a decision on which strategies would work.

Press play when you're ready to check.

So the sine rule, so 72 and 42, if you sum that together, it's 114.

And, therefore, the other angle is not a right angle.

So we can't use Pythagoras' theorem.

Pythagoras' theorem only applies on a right-angled triangle and this is not a right-angled triangle.

Cosine rule needs you to either have all three edges and you're trying to calculate an angle, or two adjacent sides in the angle between them.

We only have the length of one edge here.

So the sine rule by process of elimination is the one that we are going to use.

But the way to spot that sine rule is available is if you do have that pair of angle and corresponding side.

And we have that here, the 72-degree angle is opposite the 14 centimetre.

So on the last task you're gonna have to really go through those processes of thinking about which part of trigonometry that you know is gonna be the most efficient or available to use.

So you've got two questions once again on this task.

Both of them are here on the screen.

On question one, you're calculating the area of each of these triangles to two decimal places.

And on question two, you're then calculating the perimeter of each of the triangles to two decimal places.

So pause the video whilst you're working on those questions.

And when you press play, we'll go through the answers and the strategies that you may have used.

So question one on the first triangle.

It's our isosceles triangle that we've seen previously.

So I could try to use half times base, times perpendicular height to work out the area, by splitting the triangle into two right-angled triangles, using my trigonometric ratios to get that perpendicular height.

And then timesing it by 15 and timesing it by a half.

Alternatively, because I have a pair of corresponding angle and side, I can use the sine rule, and work out an edge, and then I can use the sine formula for the area, because I would then have two adjacent side lengths and the angle between them.

Either way the area is 39.

39 square centimetres.

On the second triangle, it's not an isosceles.

If I do 80 plus 40 and then subtract from 180, the other angle is 60 degrees.

So it's a scalene triangle.

So dropping the perpendicular and try and split this into two right-angled triangles is not a good method because I wouldn't know where it's going to partition that 12 metre edge.

What I do have though is a pair of angle and it's opposite edge.

So I will use sine rule.

And I'm gonna use the sine rule to find a length so that I can then use the sine formula for the area.

The area is 40.

70 square metres to two decimal places.

On the third triangle you may have overthought it, but this is a very basic triangle, find the area, because I do have the base and the perpendicular height.

So I can just do half times base, times perpendicular height to get the area.

So 45 square metres.

So the last triangle has got an area of 49.

18 square metres.

And one method to do this is to recognise that, that right-angled triangle with the 45-degree angle would therefore be isosceles.

And so if we use trigonometric ratios, the sine ratio to work out the length of the dash line, because that is the opposite to the 30-degree angle with a hypotenuse of 12, then we would know that, that dash line has a length of six.

If that has a length of six and it's an isosceles triangle, then I can use Pythagoras' theorem to work out the edge length, and that would be six root two.

Then I can use the sine formula for the area because I would have 12 metres that was given.

I've just calculated the other edge of the triangle is six root two in an exact form and the angle between them I could calculate as 105 degrees.

So I can use a half times 12, times six root two, times sine of 105 degrees to get my answer.

Like everything though, that's not the only method.

It may not have been the method you chose to use.

So check the answer first.

On question two, you needed to calculate the perimeter of each of the triangles.

So the first one we needed to work out our edge length.

It's the same triangle as question one.

So if you had worked out those edges, which you probably did, hopefully, you didn't repeat your working and you just made use of them.

So 33.

31 centimetres is the first perimeter.

On the second triangle, again, you will have needed to have worked out at least one of the edges.

So you need to work out the third edge as well to get the perimeter, so 30.

39 metres to two decimal places.

And on that last triangle, the way I did the area previously is I've worked out already that one edge was six root two, the other, the longest of the edges, the one opposite the 105-degree angle is made up of a six.

And then I can use my trigonometric ratio to work out that adjacent edge knowing that cosine 30 is equal to the adjacent over 12.

And then I can add that together, add on my six root two and add on my 12 to get 36.

88 metres for the perimeter.

So to summarise today's lesson, you need to consider the fact that at this point of your learning, you now have multiple strategies available.

So by analysing the problem you can determine which approaches or approach might work.

And that starts with thinking about is it a right-angled triangle or can I have a right-angled triangle? If you do, then you get Pythagoras' theorem as an option and you get your trigonometric ratios.

If it's a non right-angled triangle and there is no way of creating a right-angled triangle with enough information, then you need to start thinking about sine rule, cosine rule, and the sine formula for the area.

The sine rule is the one that I would go to first in terms of can I use it? And that's because if you have a pair of angle and the opposite edge, then you can make use of the sine rule.

Cosine rule can be used if you've got all three edges and you need to work out an interior angle, alternatively, if you've got two adjacent sides and the angle between them.

And that's the same information that you need for the sine formula to find the area.

So keeping the goal in mind, helps you to focus your thinking.

So although it's a good strategy to just make a start at adding anything to the diagram that you can, sometimes you can put too much information and sort of cause yourself to go off track.

So try and keep a focus on what it is that the question, or you are trying to prove or show is.

Consider the information you have and whether you need anything else in order to solve the problem.

So thinking about, for example, if you wanted to work out the area or it asked you to work out the area, then you either need the perpendicular height to the base, or you need two sides and the angle between them.

So that's the information you need.

And then think about how you can get to that information.

But if you are unsure, if you are at of a bit of a blank when you stare at a trigonometric question or a triangle geometry question, begin by calculating or deducing what you can.

And that might be all three angles in the triangle if you were given two, because then it might tell you that there's a right-angled triangle or it might tell you it's an isosceles triangle, and then you can split it into two right-angled triangles and then maybe make use of Pythagoras.

So as you do more questions, as you practise, you'll get more and more confident and familiar with all of your trigonometric rules.

Really well done today.

And I look forward to working with you again in the future.