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 lesson outcome today is to be able to derive and use the formula for the sine rule.

Our new keywords are sine rule, and it's a formula for used for calculating either an unknown side length or the size of an unknown angle, and you can see it there on the screen.

But we will be using this lesson to derive it and make use of it, so it will make more sense once we come to the end of the lesson.

So the lesson is on the sine rule, and in our first learning cycle, we are looking at deriving the sine rule, and then when we get to the second learning cycle, we'll be applying the sine rule and making use of it.

So let's make a start at looking at where and how we derive the sine rule.

So when we know two sides and one angle in a triangle, the third side is calculable.

So we can see how we would construct that.

So firstly, we've constructed a circle with a radius of 3 centimetres, and that is what maybe we can think of as our base edge.

From the end of that line segment that is in the centre of that circle, so a point on the circumference, we've then constructed a circle with a radius of 4 centimetres.

And any point along that circumference could be the third vertex of a triangle, and you'd join that back to the centre of the first circle.

However, if we know the angle needs to be 30 degrees at that centre point, then there is only one particular point where we can construct our triangle.

If we chose a different point, then the third side length would change but the angle would change as well.

And Sam says, "Can you see that these two triangles have different angles?" And there are a couple of special cases where we might have two possibilities with an angle set at a particular degree.

And so you can see there is this one here, which is still a point on the circumference of the second triangle along the ray line that that angle makes, as well as this one where we have a point on the circumference of that second circle along the ray line.

So there are some special cases where we get two possibilities.

And Sam says, "It's important to note whether the diagram has been drawn accurately." Because on that first one, there is an obtuse angle, whereas on the second, it is an acute angle.

So if it has been drawn accurately, then we can know to expect an acute angle or an obtuse angle.

So we're looking to derive the sine rule, and here we've got our general triangle.

So we've got vertex A, B, and C.

Opposite edges are the lowercase of that letter.

So we can see that opposite the angle at A is an edge length of a.

We've also got a dash line, which is the perpendicular height of this triangle with a length of p.

And that meets, from the vertex, meets a point on the line AC that labelled D.

So we can use, if we focus just on this left hand right-angled triangle that has been formed by the perpendicular height or that altitude, then we can make use of the sine ratio to write the height, p, as an expression.

So just focusing on that purple triangle.

So we know the sine ratio is sine of theta is equal to the opposite divided by the hypotenuse, and the opposite is the side opposite the angle, theta.

So for this right-angled triangle, our angle is angle A, the angle at the vertex A, our opposite is the edge p, and our hypotenuse is the edge c.

So we can write the ratio a sine of A is equal to p over c.

We can rearrange this to make p the subject by multiplying both sides by c.

So on the screen, we have an expression for the length of p, the perpendicular height of this triangle.

If we now move our attention to the right hand side of that altitude, we have another right-angled triangle, and we can make use of the sine ratio once again, but for a different angle.

So our angle is now the angle at vertex C, the opposite is still p, but this time the hypotenuse is a.

So we can write the sine ratio is sine of C equals p over a, and again rearrange it to make p the subject.

So now we have two different expressions for the same length, which means that they are equal to each other, they're equivalent.

That c sine of A is equal to a sine of C.

So we have an equation.

So here is a check: Which of these statements is correct? So pause the video, look at the triangle, refer to the diagram, which of these are correct? Press Play when you're ready to check.

So it's B and D.

So c sine of A and a sine of C.

The left hand triangle and the right hand triangle.

So for the first task, you're gonna complete the missing steps.

So we've seen how we get to this equation.

We had two expressions for the perpendicular height or the altitude of the triangle.

So we equated them because they were expressions for the same thing.

But how do we get from that equation to the form, the equation at the bottom, what steps have taken place? So pause the video whilst you work on question one.

When you press Play, we'll move to question two.

So question two is the last question on task A.

So use the diagram to show that d = c of sine B, d = b of sine C, and lastly, b over sine B = c over sine C.

So pause the video, and when you finish with question two, press on Play, we'll go through our answers to task A.

So you need to complete the missing steps to get from the equation which was equating the expressions for p to a rearrangement of that at the bottom, a different form.

So you may have decided to divide by sine of C, remembering that the expression a sine of C is a product.

So it's actually a multiplied by sine of C.

So dividing by sine of C is the inverse operation, and we are going to do that to both sides to keep things nice and balanced.

And that would give us this equation.

So a is equal to c sine of A, which is what we started with on the right hand side of the equation, divided by sine of C.

Then divide by sine of A, and we finish with the form we were looking for, a divided by sine of A = c divided by sine of C.

Just to note, however, you may have done your division the other way round.

So you may have divided by sine of A as your first step.

So your next line would be, a sine of C divided by sine of A equals c, and then you would've divided by sine of C and finished with the same line.

So those two intermediate steps could have been done in either order.

Your equation at the middle would be different depending on which order you did the division.

On question two, you needed to show and use the diagram to show d = c of sine of B, and that was starting by using the sine ratio for the angle at B, and D is the opposite, and C is the hypotenuse, and then rearranging it to make D the subject.

On part B, you are now looking at the other right-angled triangle on the diagram with the angle of C, D again was the opposite, and B was the hypotenuse, and then rearranging that to get D as the subject.

And then the last one was to equate your answers to part A and part B because they were both equal to D, and rearranging it very similar to what you did in question one.

So we end up with b over sine B = c over sine C.

So we're now onto the second learning cycle where we're going to apply the sine rule.

So Sam and Aisha have said we have shown that a over sine of A = c over sine of C, and also b over sine of B = c over sine of C.

And you did that in learning cycle one within your task.

Question one you showed what Sam has said, and question two is what Aisha has said, they're both equal to c over sine C, if you have a look.

So that means that a over sine A is equal to b over sine B which is equal to c over sine C.

So the ratio of the edge over its sine of its angle is equal in all three.

So that's what we have shown.

And this is our sine rule, so we've now derived the sine rule.

So using any triangle, we can drop the perpendicular, you make use of the sine ratio to find that there is this ratio between the edge opposite the sine of the angle.

So Aisha says, "The sine rule helps us find missing sides and angles when we have three out of four of these.

." So if we have the edge labelled a, the edge labelled b, and the angle at the vertex B, then we can use the sine rule.

So two side lengths and the corresponding angle means we can use the sine rule.

Alternatively, if it was the two sides, a and b, but we had the angle at vertex A, then we still can use the sine rule.

Or if we have two angles and one corresponding side.

So if you have the edge length a and the angle at vertex A and the angle at vertex B, then we can use the sine rule to work out the edge, B, or we might have the other one.

And rearranging the formula helps us find the unknown value.

So if it was a, the side length a, remember that is the side opposite the angle at vertex A, then our rearrangement would be times in by sine of A so that A is the subject, and we'd end up with b sine of A over sine of B.

And that is an expression on the numerator that is a product.

So b multiplied by sine of A over sine of B is equal to a.

If we're trying to work out the angle, then we're going to look at arcsin.

So if we're trying to work out the angle at the vertex A, that means we need two edge lengths and the corresponding angle.

So sine of A is equal to a sine of B over b, and then we can make use of arcsin to get A as the subject.

So Aisha says on this triangle, A, B, C, we have a side and it's corresponding angle and another angle.

So this is the side and it's corresponding angle.

The other angle is the 42 degrees.

And so we can use the sine rule to find the length, AB, by making use of the angle C.

So we're going to be finding the length that is opposite the other angle.

So b over sine of B = c over sine of C.

The sine rule tells us that there is this ratio that is equivalent within a triangle.

So because we've got our side and its corresponding angle, we can get the value that the other side and the corresponding angle are equal to.

So we substitute in our values into the appropriate places.

So 14 divided by sine of 72, that has a numerical value, and that is equal to this unknown edge which is AB over sine of 42.

So we're gonna rearrange this to make c the subject.

So if we times by sine of 42, remember sine of 42 has a numerical value, it's not an exact numerical value.

So you are better off leaving it in our working out a sine of 42.

Otherwise, you're gonna have a decimal to write down that you will inevitably input some rounding error.

At this point, c is the subject.

So now we can input the left hand side into our calculator to get the most accurate answer, and it's 9.

85 to two decimal places.

So here is a check for you: Calculate the length, AB, to two decimal places.

So we have got a side and it's corresponding angle.

So you can see we've got the angle at the vertex B and the edge that is opposite that.

We have another angle and that's the angle at the vertex C, and we are trying to calculate the length of AB, which is the edge opposite our angle at C.

So pause the video, you may wish to go back over the example we just did because this is very similar before you have a go at this yourself.

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

So the answer is 5.

63.

So now, we're gonna look at a different way of applying the sine rule.

So Aisha says, "This time we have an angle to find," and we can see that's labelled as theta, "Do we know three useful values for the sine rule?" So have a look at that, pause the video.

Remember on the sine rule, there are four different values that could be inputted.

You need three of them to be able to work out the fourth.

So do we have three useful values? Press Play when you're ready to move on.

So we have got a side and it's corresponding angle.

We've got the angle at vertex C, which is 42 degrees, and the edge opposite here is 12, so we must have that pair.

And then we have an edge of another pair and it's the angle within that pair that we're trying to calculate.

Do we have three useful values? Yes, so we can substitute those useful values into our sine rule.

12 divided by sine of 42, so that's our corresponding pair, gives us the numerical value of our ratio.

And we know that the edge opposite theta has the same ratio, it's equal to that.

So 14 over sine theta has to equal the numerical value of 12 over sine 42.

Now we need to think about how we rearrange this to calculate theta.

So we can rearrange this to make sine theta the subject.

Pause the video and check that you are happy with how we get from the first line of working to the second.

Press Play when you're ready to move to the next line.

Okay, so hopefully you're now happy at the rearrangement to make sine theta the subject, and now how do we make theta the subject? Well, we make use of the inverse sine function known as arcsin, so we're gonna use our calculator and we're going to apply arcsin to the most exact answer by using sine of 42.

So 14 sine of 42 divided by 12.

Yes that simplifies, 14 and 12 do have a common factor.

So if you wish to make that 7 sine 42 over 6, that's absolutely fine, but really the calculator's gonna do a lot of the work for us, so there's no need to do that.

So we're gonna evaluate that and find that theta is 51.

32 degrees to two decimal places.

Aisha says, "Is there only one possible triangle? What if we knew theta was obtuse?" So if we remember back to when we constructed two sides with an angle, there are some special cases where we find two triangles, they're not unique, and that's why this is not a criteria for congruency if the angle is not between the two sides 'cause it doesn't guarantee two congruent triangles.

So if we are now told that that angle theatre is actually obtuse, we just calculated an acute angle.

But if we now know it's obtuse, how can we work that out? So what if it looked like this? We still got the same three useful values, we still got the corresponding side and the angle, 12 and 42 degrees, and we still have the edge of the other pair with the angle theta, but now we can see that the angle is obtuse.

The calculator told us the angle was 51.

32 degrees to two decimal places.

So how can we calculate the possible obtuse angle? Pause the video and think about this yourself before we go through this.

Press Play when you're ready to move on.

So if we think about the sine curve, so the sine graph, we're only focusing here between 0 and 180 degrees, and that's because we are looking at angles in a triangle, so it's never gonna be an angle greater than 180 degrees because three angles need to sum to 180 degrees.

So we're just restricting our domain to 0 to 180.

So our calculator gave us the value of 51.

32, and that would be our X value along the x-axis.

So when we did arcsin of 14 sine 42 over 12, it gave us the 51.

32.

And the value of 14 sine 42 over 12 is just less than 0.

8.

So if we remember how we read off from the sine graph, we would go across at that value, and where it intersects the graph gives us our X values, and those X values are our solutions.

So we can see that the first solution is at 51.

32, but there is a second solution between 0 and 180 degrees, and that second solution is in the obtuse range of angles because an obtuse angle is greater than 90, but less than 180.

So it's the symmetry of the sine curve that we're gonna make use of that shows that these distances are the same.

So the distance from 0 to 51.

32 is the same distance from 180 to that second solution.

And therefore we can calculate that the obtuse angle of theta is 128.

68 degrees.

So here's a check: Which of these diagrams shows that two different triangles can share one angle and two side lengths? Pause the video, and when you're ready to check, press Play.

So it's the first one.

They both share two side lengths, which is the radius of the first circle, the line segment between the two vertices, and then there is a radius on that second circle that joined to the first vertex, which is the centre of the first circle has got two points that are being joined back up.

We're up to our final task of the lesson.

And on this question one, you are going to work out the unknown marked length for each of these triangles to two decimal places.

There's an arrow of no label, and that's the edge that you are trying to calculate by making use of the sine rule.

None of these are drawn to scale, so don't try and use any scaling or measuring to calculate that, but instead you should be using the sine rule.

So pause the video, and then when you finish with question one, press Play, and we'll move to question two.

So question two, you're now finding the acute angle.

So although there are special cases where there could be an obtuse angle that your calculator hasn't given you, it'd give you the acute value first, and you'd need to use the symmetry of the sine curve to work out the obtuse angle, all of these, we are only focusing on an acute angle.

So, pause the video and work out the unknown acute angle for each of these triangles, and then when you finish with question two, press Play, and we'll go through our answers.

We're gonna go through the answers one by one.

So on 1 part A, the answer is 9.

34.

You know it's the sine rule because you have three useful values.

You've got a pair of corresponding edge and angle, and that is necessary to be able to use the sine rule.

And then we have the angle opposite the edge that you're trying to work out.

On B, the answer was 11.

00 to two decimal places.

Once again, you've got a pair of corresponding angle and edge, which is necessary for the sine rule, and the angle opposite the edge you're trying to calculate.

On C, the unknown edge was 12.

74.

The 9 centimetre edge was not a necessary value to be able to calculate 12.

74 using the sine rule.

So you had a pair, the 15 and the 52 degrees, were a pair of corresponding edge and angle, and the 42 degree angle is opposite the edge you were trying to calculate.

On D, the answer was 17.

59.

Did you have a pair? Yes, the 58 degree angle and the 15, and then the 84 degree angle is opposite the edge you were trying to calculate, the 11 was not necessary.

E, the answer 24.

54.

We started with the paired, which was the 52 degree and the 20, that was our corresponding pair of angle and edge, and the 26 degrees is opposite the 11.

So we also had a different pair.

So those two pairs, either one could be used.

What we didn't have explicitly was the angle opposite the edge we were trying to calculate, but we could use angles in a triangle sum to 180 degrees to work that angle out, which the two sum together to be 78, so that angle there would be 102.

And as I said, not drawn to scale, so that is drawn as an acute angle but is actually an obtuse.

And on F, we need to work out that missing edge length, 13.

95 centimetres.

We had two pairs again, so it doesn't matter which pair you chose to use, either the 9 with the sine of 40, or the 11 with the sine of 52.

The edge that we're trying to calculate is opposite the unknown angle, but we could calculate that the unknown angle, would be 88 degrees.

On question two, all the answers are on the screen here, you were calculating the angle this time.

So once again you need to make sure that you did have a pair that you could make use of, so that's the corresponding edge and the angle opposite it.

And then we needed the edge that was opposite the angle you were trying to calculate.

So in A, it was 40.

15 degrees.

In B, it was 40.

15 degrees.

In C, it was 28.

22.

In D, it was 38.

46.

In E, it was 25.

68.

And in F, it was 51.

78 degrees.

Really well done on those questions.

So to summarise today's lesson, which was the sine rule.

So the sine rule is a formula used for calculating either an unknown side length or the size of an unknown angle.

A diagram can aid with the understanding.

The sine rule is useful when you know an angle and side pair and wish to find another pair.

In order to do this, you will need either a side length or an angle.

So really important part of the sine rule is having that pair of corresponding angle and edge, because that gives you the value of the ratio, which the sine rule tells us the ratio between the corresponding edge and the sine of the angle are equal in each pair.

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