Lesson video

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

By the end of today's lesson, you should be able to use the sine ratio to calculate the missing edges or an angle using it on a right angle triangle.

There are some keywords on the slide here that you might want to re-familiarize yourself before we get going into this lesson.

So pause the video, read through them, make sure you feel confident in all of those, and then we'll get started.

So the lesson's broken into two learning cycles.

The first is to formalise the sine formula.

Then secondly, we're going to calculate angles using the sine formula.

So let's make a start on formalising the sine formula.

Here we have two right angle triangles, and we know that these two triangles are similar.

And we know that because they both have two angles that are equal to each other.

We can find the labelled length X by using a ratio table.

Here, we've got a hypotenuse of 1 and a hypotenuse of 12, the corresponding edges on the two similar triangles.

And therefore, we can think of this as an enlargement in similar triangles, that the scale factor is 12, that the multiplier to get you from the hypotenuse of 1 to the hypotenuse of 12 is 12.

The opposite, 0.

67 centimetres, the side that is opposite the 42 degree angle on the first triangle, would also be multiplied by 12 to give you the value of X to keep them in proportion.

Within the triangle, the proportions are also constant.

The hypotenuse of 1 multiplied by 0.

67 gives you 0.

67, and that will be true on that enlarged similar triangle, that 12, the hypotenuse, multiplied by 0.

67 will give us this value of X.

So, 0.

67 times by 12 is 8.

04.

We now know that the opposite is 8.

04 centimetres.

And this can go further to generalise to any right angle triangle that has an angle of theta where the side opposite it has a length of sine theta.

So hypotenuse of one and an opposite of sine theta.

And on our more general triangle, our opposite, we are just calling opposite, and our hypotenuse were given the variable of h.

So on our ratio table, 1, the hypotenuse, has become h by multiplying it by h.

Our enlargement scale factor would be h.

And sine theta multiplied by h would give you the opposite on the more general triangle.

If we go within the triangle, the ratio between that opposite and the hypotenuse, we're gonna multiply by sine theta, and we do the same on the similar triangle.

So the opposite edge can have an expression of h times sine theta.

So this ratio table shows us the general form of the relationships within a right angle triangle with a hypotenuse and its opposite.

So the length of the hypotenuse multiplied by the sine of an angle theta equals the length of the side opposite of the angle, h multiplied by sine theta.

And we can use the ratio table to generate a formula that shows this relationship between the side that's opposite this angle theta and the hypotenuse, and that formula can be represented in three different ways.

Firstly, the hypotenuse multiplied by sine theta is equal to the opposite.

Or if we start on the edge that is known as the opposite and divide it by sine theta, then we would be left with the hypotenuse.

And so we can hopefully see that that's using inverse operations of the equation to make them equivalent.

Lastly, if we start again with the length of the opposite and divide it by the length of the hypotenuse, this gives us sine theta.

Another way to do this without the ratio table is we can derive two more versions, the final two versions of the formula, using the very first one and doing some division.

So here we have that the hypotenuse multiplied by sine theta is equal to the opposite.

If we divide both sides by the length hypotenuse, then on the left hand side the hypotenuse divided by hypotenuse is equal to one, they cancel through, and the right hand side would still say opposite divided by hypotenuse.

So we have the formula written in this version.

Opposite divided by hypotenuse is equal to sine theta.

We can start with the hypotenuse multiplied by sine theta equals opposite formula again, but this time rather than dividing both sides by hypotenuse, it's a product on the left hand side.

Let's divide by sine theta.

So dividing by sine theta on the left hand side and dividing by sine theta on the right hand side, once again, sine theta divided by sine theta is 1, so they cancel out.

We are left with h only on the left hand side.

And so we have our third and final version of the sine formula that the opposite divided by sine of theta is equal to the length of the hypotenuse.

And we can substitute the values from any right angle triangle into each of these versions of the sine formula to represent the relationship between this side that's opposite the angle and the hypotenuse in different ways.

So here we have a right angle triangle.

The opposite is 15 centimetres, the angle is 66 degrees, and the hypotenuse is b.

So we would say b multiplied by sine 66 is equal to 15, but we could also say that 15 divided by sine 66 is equal to b.

Alternatively, 15 divided by b is equal to sine 66.

These three equations are equivalent to each other, they're just different rearrangements.

So here is a check.

One correct equation showing that relationship between the opposite and the hypotenuse is 20 times sine 38 degrees is equal to the opposite.

So which of these are also correct to show the relationship between the hypotenuse and the opposite side of the triangle on the screen? So pause the video whilst you decide whether it's part A or part B, and then press play to check.

We can divide both sides by 20, and that would leave us with equation A, sine of 38 is equal to the opposite divided by the hypotenuse divided by 20.

Here's another check.

Which of these correctly shows the relationships between the hypotenuse and the side opposite the 82 degree angle for the triangle shown here? Pause the video whilst you make your decisions and then press play to check.

C is a correct statement that shows the relationship between the hypotenuse and the side that's opposite the 82.

That is the hypotenuse multiplied by sine theta equals the opposite, that form of the formula.

But that is true for A and D as well.

You can see from the one that's in the pink box that we have just rearranged it to get A and to get D.

Here's another check.

Which of these correctly shows the relationship between the hypotenuse of the right angle triangle and the side that's opposite the angle? Again, pause the video and then check your answer by pressing play.

So E gives us hypotenuse multiplied by sine theta equals opposite, and B and C are rearrangements of that.

We've divided both sides by sine theta to get B, and we've divided both sides by 116 to get C.

So in each version of the formula, a different part of the triangle is the subject, and this is what makes them useful.

Otherwise, what's the point of having three different versions of the same formula? So it's helpful to use the one with the subject that you want.

So on this triangle, we are trying to calculate the opposite, so we would use the top formula because opposite is the subject.

This way, we can substitute them in and just calculate the opposite.

So our 87 is our hypotenuse, and 55 degrees is our theta.

And we can then calculate this on find the P, is 71.

266 to three decimal places.

If we had used one of the other two versions of the formula, we would've had to do rearranging to make our opposite the subject.

So you may as well start with the one where the opposite is the subject.

So which of these correctly shows the relationship between the hypotenuse and the side opposite the angle of 21 degrees? Pause the video whilst you decide and then press play to continue.

These three, A, B, and D, they all show the formula for sine.

But which of these will help you find q easiest? Press pause so that you can decide which one makes finding q the easiest, and then press play to continue.

So B, because q is the subject.

On A, q is the denominator, and on D, q is part of the product.

B, however, it is the subject, so that's the easiest one to use to find q.

So now find q.

Use your calculator and give your answer to two decimal places.

Press pause whilst you're doing that and then press play to check.

Q is 94.

87 centimetres, which seems a sensible answer because you are calculating the hypotenuse, which should be the greatest length of the sides because it is the longest edge of a right angle triangle.

So now we're up to task A, where question one, you need to substitute the labelled sides and angles into each triangle.

And then for each triangle, choose the most appropriate version of the sine formula to find the length of its missing side.

Press pause whilst you're doing that.

And then when you press play, we'll move on to the next question.

Question two, you need to match the triangle to the equation and then also use the equation to find the length of the side labelled X, giving your answers to two decimal places.

So pause the video whilst you match them up and calculate.

And then when you press play, we'll move on to the next question.

So now we're on to question three, where you're going to use the sine formula to find the length of the missing sides that are labelled with a letter and give your answers to two decimal places.

Pause the video and work through those six parts.

And then when you press play, we've got one more question of the task.

Question four is on the screen.

You need to find the lengths of the distances A, B, C, and D, and round them to the nearest integer.

It may be that you wish to draw the four separate right angle triangles, and when you've worked out A or B or C or D, then you can add that into each of the diagrams where necessary.

So press pause whilst you're working through.

And then when you press play, we're gonna go through the answers to this task.

Here are the answers for question one.

You needed to substitute the values into the appropriate places and calculate A and B.

One version of the formula was more useful than the other.

So on the left hand side, to calculate A, you would use the top one because A is the subject.

So A was 117.

53 centimetres.

On the right hand side of the screen, you were trying to calculate B, so you would use the second version of the formula because B is the subject, and B is 2.

65 centimetres.

Question two, you needed to match the diagram to the equation to find X and then actually calculate X as well.

So A matched with E, and X was 26.

50.

B matched with F, and X was 24.

51.

And lastly, C matched with D, and X was 94.

35.

Question three, the answers are on the screen.

I would suggest you pause the video and take your time to check against your own answers.

Question four, A was 23 centimetres, B was 37 centimetres, C was 44 centimetres, and D was also 44 centimetres.

Unfortunately, with this type of task, there's like a rolling knock on effect.

If you've got one of the earlier answers, so for example B, if you've got B incorrect, then the likelihood is that you may have got C and D incorrect as well.

So do take your time to check your solutions.

So we're now up to the second learning cycle of the lesson, which we're gonna look at calculating angles using the sine formula.

So up 'til now, we've been calculating edges, the hypotenuse, or the opposite.

Now we want to think about finding the angle.

So Izzy says that she's confident to use the sine function to find the length of the hypotenuse or the opposite in a right angle triangle.

And we've got a right angle triangle here on the screen where the opposite is X and the hypotenuse is 79 with an angle of 42 degrees.

We know that the calculation used in our formula would be 79 times sine 42 to give us the value of the opposite, which is 52.

86 centimetres.

So Izzy is confident to do that, but she wants to know is it possible to find the size of the angle using the sine formula.

So in this example, we have the opposite edge and we have the hypotenuse edge, but we don't know the angle between.

And we can use the version of the sine formula, where sine theta is the subject.

This one here.

So opposite divided by hypotenuse is equal to sine of the angle.

And if we substitute the lengths of hypotenuse and the side opposite theta, we end up with sine theta is equal to 7.

07, that's our opposite edge, and 10, our hypotenuse, which equals 0.

707 as a decimal.

And so to find the angle, we can find the 0.

707 in the sine column of our table of trigonometric values.

So, here's the sine column, here's 0.

707, which means that the angle would be 45 degrees.

So one way to find the angle is to use a table of trigonometric values.

Here's a check for you.

So Izzy has found the value of sine theta for this triangle.

By using the lookup table, find the angle theta.

So pause the video whilst you're doing that and then press play to check your answer.

So sine column, 0.

94, 70 degrees.

Theta would be 70 degrees if the opposite was 23.

5 and the hypotenuse was 25.

Another check.

By substituting the lengths of the hypotenuse and the opposite into the sine formula, use the lookup table to find the size of the angle.

So press pause whilst you're doing that, and then when you're ready to check the answer, press play.

So you should have put 173.

2 as the numerator of our fraction because it's the opposite and 200 as the denominator, which when you do the division is 0.

866 as a decimal.

Find that in the sine column, and that means the angle would be 60 degrees.

So theta is 60 degrees.

So Izzy is a bit concerned because what if she gets a value for sine theta that isn't in the table? And there's a high chance that that's gonna happen.

So here, 7 and 20, gives you 0.

35.

0.

35 is not a value in the column for sine.

However, there are two values on either side, which Izzy says, "Well I can see on the table that theta is somewhere between 20 degrees and 25 degrees," but can she be any more precise? So we can say we know the range for theta, we don't know the value of theta.

So we could also use the unit circle starting at the Y axis.

So 0.

35 is our value of sine.

So that's where we would find it on this unit circle graph.

If we go horizontally across and then drop vertically down to create our right angle triangle, that's 0.

35 units, which is what we want it to be with a hypotenuse of one, and we can use the scale to say that's 21 degrees.

But it's another approximation of the angle.

We don't know for certain that's 21 degrees, we just know it's somewhere between 20 and 25, and it's closer to 20 than it is to 25.

So here is a check.

For the given triangle on the screen, which cross shows the starting point for a vertical or horizontal line from which to find the missing angle.

So press pause.

It may be that you want to look back at the example I just went through before you try this yourself, and then press play to check your answer.

So you should go across at B because the value of sine theta is 0.

4.

So we've gone across from 0.

4, we've got our right angle triangle.

So estimate the size of theta using this unit circle.

Pause the video whilst you make a decision on what you think the angle of theta is in this right angle triangle.

Press play when you're ready to check it.

So your answer should be something between 22 degrees and 25 degrees.

Another check.

For the given triangle, which cross shows the starting point for a vertical or horizontal line from which to find the missing angle.

So press pause whilst you calculate that and decide from those six options, and then press play to check your answer.

The answer is A.

Now estimate the size of theta using this unit circle.

Again, press pause whilst you make a decision on your estimation and press play to check.

So your angle should be somewhere between 47 degrees and 50 degrees, using the scale on the arc that we can see.

So Izzy said, "It's better than the table, but it's long-winded, and it's still not precise because you are still estimating.

Is there not a more precise way of finding an angle?" Similar to how we'd use a calculator to find the length of a side.

So on a unit circle, we can find the length of a side opposite the angle of 30 degrees by applying the sine function to that angle.

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

5.

We can read that off of this vertical scale, and that is the length of the side on the unit circle.

The sine function has an inverse function called arcsine, and it may be seen as sine to the power of minus one, especially on a calculator.

So if we take arcsine to our equation, the left hand side would leave us 30 degrees, and the right hand side we would see written as sine to the power of minus one of our 0.

5.

And 0.

5 is the length of the side on the unit circle, and so that would come back to 30.

So it's doing it in the reverse.

It's the inverse function, and so it outputs the angle.

So we do have a way of using our calculator to find this angle.

So I'm gonna talk you through the buttons that you would need to press.

So to type the inverse sine, or arcsine, of 0.

5, first of all you need to press the up arrow, the shift key.

Then, press the sine button.

That will bring sine to the power of minus one on your screen.

Type 0.

5, close the bracket, and then press the execution key.

30 is the angle.

So the answer to that is the angle that you are looking to calculate.

So here is a check on that.

Which of the following equations is correct when finding the size of angle n in the equation? Pause the video whilst you make your decision.

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

C and D are equivalent correct equations.

So C, you've got arcsine of 0.

24, which leaves you n, 'cause you've taken arcsine on both sides of the equation.

And D is writing the word arcsine as opposed to using the notation that you see on the calculator.

Which calculator display is correct when trying to find the size of angle P in this equation? So, another check for you.

So pause the video, and then when you're ready, press play.

The answer is A.

So finding the size of an angle will require the use of arcsine because we need to try and inverse, we need to undo the use of the sine function, and that will be shown with the sine to the power of minus one on a calculator.

So which calculator display is correct when trying to find the size of angle Y in this equation? Again, pause the video whilst you're making a decision, and then when you're ready to check, press play.

This one's a little bit like spot the difference.

So they both have the arcsine of 0.

9.

They give a different answer.

In terms of what buttons were used to get this, they were the same.

So which one is incorrect? So as I said, it was a little bit like spot the difference.

What was different? There was an R on the calculator screen on part A, and there was a D on the calculator screen on part B.

You need to have a D, which stands for degrees, which means you are measuring your angle in degrees and not radians.

So the answers have been input correctly, but the result would be incorrect.

So you're onto your last task of the lesson.

So question one, using the table of trig values, find the size of the missing angle marked with a letter.

The first couple have got a little bit of support for you, and then you can continue into C and D.

So press pause whilst you're working on question one, and then when you press play, we'll move on to question two.

So question two, by drawing appropriate lines on this part of the unit circle, estimate the size of each angle marked with a letter to the nearest integer.

Part A is equal to 0.

8, and that's why there is a cross at 0.

8.

You need to use that to find your angle, and then continue for B and C.

Press pause whilst you're doing that.

And when you press play, we've got one more question in this task, and then we'll go through the answers.

Question three, you've got six questions where you need to calculate the angle using your calculator and give your answers to one decimal place.

Once again, you've got some support on the first two parts of the question, and then hopefully that will aid you as you move into the final four.

Press pause whilst you're working through question three.

And when you press play, we're gonna go through our answers.

Here are the answers for question one.

So the first triangle had an angle of 20 degrees, the second triangle was 65 degrees, then 60 degrees, and then 5 degrees.

So you were using the lookup table to find the appropriate decimal and then reading off the angle.

Question two, we know that this is not the most accurate method to find an angle, so any angle that's plus or minus two degrees of what is given would be a sensible estimate.

So on that first triangle where there's angle A, you needed to go across from the 0.

8 and then decide where it hits the arc.

And so 53 degrees is approximately the angle.

If you've got 51, 52, 54, or 55, then you would be fine.

On the second and third triangle, 17 degrees and 26 degrees.

So you should have gone across at about 0.

3 and again going across at about 0.

44.

And finally, question three.

This time these were accurate answers.

Yes, we rounded them to one decimal place, but we were using the arcsine function to calculate the angle.

So A, 22.

0 degrees, B, 54.

9 degrees, C, 28.

7 degrees, D, 9.

2 degrees, E, 73.

7 degrees, and F, 28.

7 degrees.

So to summarise today's lesson on using the sine ratio.

So there is a relationship between the hypotenuse of a right angle triangle and the side opposite the angle theta, and this can be written as a formula in three different ways, and you can see those on the screen.

This sine formula can be used to find the length of the hypotenuse, and in that case, you'd use the middle one, the length of the side opposite the angle theta, and you'd use the last one, or the size of the angle theta itself, which we start with the first one.

But in order to find the size of the angle, we need to use the inverse sine function, which is arcsine.

On a calculator, that is usually written as sine to the power of minus one.

Really well done today, and I hope you enjoyed the lesson.

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