Lesson video

Hello, I'm Mrs. Lashley and I'm going to work with you as we go through the lesson today.

I really hope you're looking forward to the lesson.

You're willing to give it your best shot and I will be there to support you as we get through it.

So our learning outcome today is to be able to use the sine ratio to find the missing side or angle in a right-angled triangle.

There are some key words on the slide here that you'll have met before, but I would encourage you to pause the video and re-familiarize yourself before we make a start.

So this lesson on checking and securing understanding of sine ratio problems is going to be split into two learning cycles.

The first learning cycle is formalising the sine formula and then we'll move on to the second learning cycle later on in the lesson where we're calculating angles using the formula, let's take a look at formalising the sine formula.

So we know that these two triangles are similar as two of the angles in the triangle are equal to the two angles in the other.

The third angle would also be the same if we worked it out using the angles inside of a triangle add up to 180 degrees.

So we can find the length of the side labelled X by using a ratio table.

If we look at the hypotenuse, the longest edge, the one opposite the right-angle in each of the two triangles, one of them has a hypotenuse of one and the other has a hypotenuse of 12.

And so we can see that the scale factor, the linear scale factor is 12.

We can multiply one by 12 to make it equal to the hypotenuse of 12.

If we look at the opposite edge, the edge opposite the given 47 degree angle, it's 0.

73 centimetres on the triangle that has a hypotenuse of one.

And we can see that the proportion between the two edges within that triangle is that we can multiply the by 0.

73 to give us the 0.

73.

And so this will be true on the similar triangle.

We can multiply the 12 by 0.

73 and that will give us the length of X.

X is calculated by doing 0.

73 multiplied by 12.

And that might be because you're using the ratio within the triangle or that might be that you're using the ratio between the two similar triangles.

Either way, that calculation is equal to 8.

76.

So we now know that the X is worth 8.

76 and we can go on to generalise this process for any right-angled triangle that has an angle theta where the side opposite theta has a length of sine theta.

So if these two triangles are similar because theta and 90 degrees are in both of the triangles, then we can go along with this process of looking at a ratio table, the ratio between the two triangles and the ratio within the triangle to generalise an expression for that side labelled as opposite.

So once again, if we look at the two hypotenuse, one is a unit, one is one and the other is a generalised form of H.

To move between the two triangles, we'd multiply by H.

And so this would be the same for the opposite edges.

The edges that are opposite the angle theta in both of the triangles because of this proportion between the two similar triangles then sine theta on the smaller one of the diagram multiplied by H will give us the edge that is labelled as opposite.

Alternatively, if we look at the ratio within the triangle, we can see that to get from the hypotenuse to the opposite, it's times by sine theta.

And so that will be the case on the larger of the two triangles as well.

And so what's an expression for that edge that is opposite theta on the more generalised form? Will it be H multiplied by sine theta.

So what does this actually mean? Well it means that the length of the hypotenuse of the right-angled triangle multiplied by the sine of the angle theta equals the length of the side opposite that angle.

And that is true for any right-angled triangle with an angle of theta.

This learning cycle is about formalising the sine formula.

So if we take the ratio table and what we have just seen between the similar triangles, we can come up with a formula that shows the relationship between the side opposite the angle theta and the hypotenuse.

And it can be represented in three ways.

So if we take the hypotenuse and multiply it by sine theta this is equal to the length of the side opposite angle theta.

However, if we start with the opposite, the side that is opposite the angle theta and divide by sine theta, then this will be equal to the length of the hypotenuse.

And finally, if we take out opposite, the side that is opposite our angle theta and divide it by our hypotenuse, then this is equal to sine of theta.

And we can also derive two versions of the formula given one by performing a division.

So if we start with that the hypotenuse multiplied by sine of theta is equal to the opposite side and divide both sides of the equation by hypotenuse.

Well, hypotenuse divided by hypotenuse is equal to one and so they will cancel through.

And so we can write this as opposite divided by hypotenuse is equal to sine theta and that is the ratio of sine.

Alternatively, if we start back with hypotenuse multiplied by sine theta equals opposite and this time divide by sine theta, we do that to both sides.

Well sine theta divided by sine theta is equal to one so they cancel through.

And so the hypothesis is equal to the opposite divided by sine theta.

So what's the purpose of having these three forms of the formula? Well, we can substitute the values in a right-angled triangle into each version to represent the relationship between the side opposite the angle and the hypotenuse in different ways.

So here we can see a right-angled triangle where theta is 66 degrees, the hypotenuse is an unknown length labelled B, and the opposite side to the angle 66 degrees is 15.

So if we substitute those into the relevant parts of the formulae, then we would have B multiplied by sine of 66 degrees equals 15, that's the relationship there or 15 divided by sine 66 is equal to B and 15 divided by B is equal to sine of 66 degrees.

They are just rearranged forms of the same formula.

So here's a check, which of these correctly shows the relationship between the hypotenuse and the side opposite the 80 degree angle for this triangle rounded to one decimal place? So pause the video and when you've finished and want to check, press play.

So C, 25 times sine of 80 degrees is equal to 24.

6.

The hypotenuse multiplied by sine of the angle is equal to the side opposite the angle, but you may have said A and D as well because they are rearranged forms of that equation.

24.

6 divided by 25.

So that's the opposite divided by the hypotenuse is equal to sine of 80 degrees.

And on D we've got the opposite, divided by sine of 80 degrees is equal to the hypotenuse.

Here's another check.

Which of these correctly shows the relationship between the hypotenuse and the side opposite the angle theta for this triangle rounded to the nearest integer.

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

E shows us the relationship for the opposite edge to the hypotenuse is 116 multiplied by sine theta equals 89.

And then we can see that B and C are also true and our rearrangements of that.

So dividing both sides by sine theta gives us 89 divided by sine theta equals 116 and divided both sides by 116 gives us C, which is 89 divided by 116 is equal to sine theta.

So in each version of the formula, a different part of the triangle is written as the subject.

So there's sort of three variables.

There's the opposite edge, the hypotenuse and the angle theta.

So it's helpful to use the version of the formula whose subject matches the part of the triangle that you're trying to find the value of.

So in this example, P is the unknown, and P is the side opposite the angle of 55 degrees.

So we're going to take this formula because the opposite is the subject.

So sometimes it's a good idea to label the sides that you are trying to either calculate or have been given.

Substituting in those values into the correct places, then we have 87 times sine of 55 degrees is equal to P.

Where P is our opposite edge.

We can use our calculator to calculate that to three decimal places or whichever degree of accuracy we need.

So P is 71.

266 to three decimal places.

That's a sensible answer because the hypotenuse is the longest edge of a right-angled triangle and we are not trying to calculate the hypotenuse.

So we're expecting our value to be less than 87 and it is.

So here's another check for you.

Which of these show the relationship between the hypotenuse and the side opposite the angle 21 degrees for this triangle rounded to the nearest integer? So pause the video and then when you're ready to check it, press play.

So A, B and D.

A, B and D have correctly shown the relationship between our hypotenuse and the opposite.

So here's A, B and D, the three that were correct, but which is the most helpful to find value of Q? So pause the video and then when you're ready to check, press play.

So B is the most helpful and that's because Q is the subject so there's no further rearrangement necessary.

If you calculate the left hand side of that equation, it will give you the value of Q.

So find that length labelled Q and round your answer to two decimal places.

Pause your video whilst you're using your calculator to do that and then when you're ready to check, press play.

So Q to two decimal places is 68.

84, it's the hypotenuse, so we are expecting it to be greater than 28.

So we're onto the first task of the lesson.

For question one, by using the sine formula, remember that they're all equivalent to each other, but the one with the subject that you are trying to calculate is the most easy one to use.

Find the length of the missing side labelled with a letter for each of the triangles and round your answer to two decimal places.

So pause the video whilst you work through those six questions and when you press play, we'll move on to question two.

So here's question two of task A.

Find the area of the circle with centre O to one decimal place.

So look at the diagram, the centre of the circle is marked as O, you've got a point A on the circumference and a point B that is where the cord meets a radius.

So pause the video and then when you're ready to check your answers to task A and move on, press play.

So here's question one with all six answers on the screen.

So you've got A equals 0.

67, B equals 4.

68, C equals 5.

36, D equals 9.

14, E equals 10.

89 and F equals 410.

28.

Well done if you've got all of those correct.

Moving on to question two, question two, we needed to identify that the cord and the radius would be perpendicular and that makes our right-angled triangle.

And then we would be looking to find the hypotenuse of that right-angled triangle, which is our radius because we would like to work out the area of the circle.

And the formula for the area of a circle is Pi R squared.

So the radius was necessary.

So to calculate the radius, which is the hypotenuse, we've got our opposite edge of three centimetres and an angle theta of 25 degrees.

So we'd need to do three divided by sine 25 degrees, which is 7.

1 and then do Pi R squared.

Your answer to one decimal place is 158.

3 centimetres squared.

So we're now at the second learning cycle where we're going to look to calculate angles using the sine formula instead of the sides.

So let's make a start.

So here on the screen we can see a table of values and Izzy says, "I'm confident I can use the sine function to find the length of the hypotenuse or opposite in a right-angled triangle." So here we have a right-angled triangle, X is the opposite edge, opposite that angle of 42 degrees and 79 centimetres is the hypotenuse.

It's opposite the right-angle.

So 79 times sine 42 is the length of the opposite using the relationship between sine of an angle, the hypotenuse and the opposite, which we can calculate as 52.

86 to two decimal places.

But Izzy says, "But is it possible to find the size of an angle using the sine formula?" So we've been able to find the hypothesis, we've been able to find the opposite edge, but can we find the angle? So we can use the version of the sine formula where sine theta is the subject, because theta is that angle and that was when it was the subject is part of this sine of theta.

So this is what it looks like.

sine of theta is equal to the opposite divided by the hypotenuse.

So if we substitute the lengths of the hypotenuse and the side opposite theta into our formula, we'll get sine theta is 7.

07 divided by 10.

The opposite is 7.

07 centimetres and the hypotenuse is 10 centimetres.

7.

07 divided by 10 is 0.

707.

So we now have a value for sine of theta degrees, but we still do not know what the angle actually is.

So to find the angle, we can find the 0.

707 in the sine column and check the corresponding angle.

So because we've used the sine formula, we're gonna go down the sine column and ignore the cosine and tangent for the moment, find 0.

707 and then see which angle gives you that value, which is 45.

So we now know that the angle in this triangle is 45 degrees.

So here's a check, by substituting the length of the hypotenuse and the side of opposite the theta into the sine formula, use the lookup table to find the size of the angle theta, pause the video while you're working through that and press play when you're ready to check.

So eight is our opposite, so that's the numerator.

18.

93 is our denominator because it's the hypotenuse.

If we do eight divided by 18.

93 to get a decimal value, it's north 0.

423.

And if we use our lookup table, that means that the angle theta is 25, so theta degrees is 25 degrees.

Here we have a triangle with a tonne use of 20 centimetres and an opposite of seven centimetres.

If we substitute that into the sine formula, we get a decimal than 0.

35.

Is these recognised that this is an issue because it's not a value of sine theta that's in the table.

If we look in the sine column, there isn't 0.

35, but there are two values either side of it, there's 0.

342 and 0.

423.

And so our angle must be somewhere between 20 and 25 degrees.

And Izzy says, "I can see that it's between 20 and 25, but can I be more precise?" Because in maths we like to be as exact as possible.

So we can also find the angle by reading off of the axis of the unit circle starting at the Y axis.

So if we've got this value for sine theta as 0.

35, we can go to 0.

35 on the Y axis because that is the coordinate, the Y coordinate gives us the sine value.

If we then trace across to our unit circle and draw the radius of our unit circle from the origin to that point on the unit circle circumference, we can see that that is 0.

35 units, the length, that's what we've marked and that is the length of that triangle is our sine of an angle.

And so we can see using this unit circle that it's 21 degrees, we knew it was between 20 and 25, but we weren't too exact, but it's still approximately 21 degrees.

There's no reason that that needs to be an integer.

We are having to use that scale to the best of our ability to approximate theta.

So if here's a check using the unit circle.

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

So pause the video, you might wanna rewind and just check over that last example.

But when you're ready to check, press play.

So we should be using B.

0.

4, the opposite divided by the hypotenuse, which is 1.

5 gives us a decimal value of 0.

27.

Sine is the Y coordinate.

So we're going to go across from 0.

27 on the Y axis.

So what is the estimation of that angle theta for that triangle? Pause the video and then when you're ready to check, press play.

So it's approximately an angle between 15 and 17 degrees.

So if you said 16 degrees or 16.

2 degrees, that's a good approximation for the angle theta.

Here's 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 once again, pause the video and then when you're ready to check, press play.

So we should have been using A.

Hence estimate the size of angle theta using the unit circle, pause the video and then when you're ready to check press play.

So an angle between 47 degrees and 50 degrees, we can see that it has passed the point for 45 degrees, so your answer should be within 45 and 50.

And it definitely looks to be closer to 50 than 45.

Izzy says using the unit circle is better than the table, but it's so long-winded and still not that precise.

There was still an element of estimation.

Is there not a more precise way of finding an angle similar to how we can use a calculator to find the length of the side? Well, on a unit circle we can find the length of the side opposite the angle by applying the sine function to that angle.

So if our angle is 30 degrees, that's our angle of rotation of the radius from the X axis.

And then we apply sine, the function of sine.

Sine of 30 degrees is 0.

5.

So Y coordinate of the point where the radius meets the circumference of the unit circle.

The sine function has an inverse function called the arcsine function and you may see it as sine minus one on a calculator.

So have a look at that notation.

That notation stands for arcsine, which is the inverse function of sine.

So if we apply arcsine to both sides of our equation, then 30 degrees would be equal.

So that notation sine minus one of 0.

5.

So that's 0.

5 is the length of side on the unit circle, that Y coordinate.

And when we apply the inverse sine to it, arcsine, it gives us 30 degrees.

So by using the inverse sine function onto the value, it will return the angle theta.

And our calculator can do this for us or we can use the arcsine function on the calculator.

So if we are trying to type arcsine of 0.

5, then this is the steps you should be taking.

So you're gonna press the shift key or the second function key and then sine, that button will then use inverse sine.

Because we've asked for the second function, we're gonna type our length, our ratio, 0.

5, and close the bracket and then we'll press equals or execute.

And the value that it returns is angle theta, so we can use our calculator to find the angle theta.

So here's a check, which of the following equations is correct when finding the size of angle N in this equation? So pause the video and then when you're ready to check, press play.

So we wanted to remove the function of sine on N.

So we need to use the inverse sine.

And there's two ways that you can see that written down.

So sine with the power of minus one but actually means inverse sine and you'll note that from inverse function notation or arcsine, which means the inverse of sine.

Here's another check, which calculator display is correct when trying to find the size of angle Y in the equation sine of Y equals 0.

9.

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

So both answers have been inputted into the calculator correctly.

If you look at the screen, it says the same thing except from the answer value and the difference is the radiance and the degrees.

So the calculator is not in the correct unit for angles, we need it to be in degrees because our equation was in degrees.

It says sine of Y degrees equals north 0.

9.

So its result has been given for radians and not for degrees on A, and so B is the correct one.

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

And so for question one, you're gonna use the table of trig values to find the size of the missing angle marked with a letter in each of those four triangles.

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

So here we're up to question two.

And this time you're gonna use your calculator rather than a table of values to find the size of each angle marked with a letter and round your answer to one decimal place.

So pause the video and then when you're ready for the answers to task B, press play.

So on question one, you needed to use the table of values.

So there was a little bit of support on the first couple.

So on the first question where you were trying to work out angle A, you need to do the opposite divided by the hypotenuse to give yourself the decimal to look up in the lookup table.

And that was 0.

259.

And if you look in the sine column, 0.

259 is for the angle of 15 degrees.

On the second triangle where the angle was B, you needed to use the opposite and the hypotenuse to calculate, once again the decimal value 0.

906, look it up on the sine table and you see that that's 65 degrees.

on C, 0.

906 divided by one would still be north 0.

906.

So actually that one's 65 degrees as well.

So triangle with the angle of B and the triangle of the angle of C are actually similar triangles.

And then on the last one, you need to do 17.

4 divided by 200 to give you the value of 0.

087, and so that means that the angle theta was five degrees.

On question two, you were using your calculator.

Once again, triangle with the angle A and triangle with the angle B had a little bit of support to get you going.

So we're gonna be using the arcsine function of the calculator, making sure we're in the degrees setting.

And so our answers were 22.

0 degrees to one decimal place.

B was 24.

6 degrees, C was 24.

6 degrees, D was 19.

5 degrees, E was 30 degrees, and F was 28.

7 degrees.

So to summarise today's lesson on checking and securing understanding of sine ratio problems, the sine ratio involves the hypotenuse, the opposite and the angle.

If you know the length of the hypotenuse and the size of the angle, then you can use the sine ratio.

If you know the length of the opposite and the size of the angle, you can use the sine ratio.

And if you know the lengths of the hypotenuse and the opposite, you can use the sine ratio.

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