Lesson Video: Trigonometric Ratios in Right Triangles | Nagwa Lesson Video: Trigonometric Ratios in Right Triangles | Nagwa

# Lesson Video: Trigonometric Ratios in Right Triangles Mathematics • Third Year of Preparatory School

## Join Nagwa Classes

In this video, we will learn how to find and express the values of the three trigonometric ratios—sine, cosine, and tangent—for a given angle in a right triangle.

12:11

### Video Transcript

In this video, we are going to look at the trigonometric ratios: sine, cosine, and tangent. Which are the names given to the ratios that exist between the different pairs of sides in right-angled triangle. So, let’s begin by looking at the right-angled triangle on the screen here. And to start off with, we need to be familiar with the names that are used for the sides in a right-angled triangle.

Now you should already be familiar with the name hypotenuse, which is the name given to the longest side of a right-angled triangle. The side which is opposite the right angle. And you may perhaps have met that before in the Pythagorean theorem. The other two sides of a right-angled triangle also have particular names. And their names are in relation to the angle that we’re interested in. So in this right-angled triangle here, I’ve labeled one of the other two angles, the ones that aren’t right angles. I’ve labeled one of them as 𝜃. So the names of the sides are in relation to this angle 𝜃. The first label is the opposite. And just like the hypotenuse is opposite the right angle, the opposite side is the one opposite this angle 𝜃. So that would be this side here. The name that we give to the final side, the third side, is the adjacent. And that side is adjacent or next to both this angle 𝜃 and the right angle. It’s between those two. That’s how we recognize which of the sides is the adjacent. So you need to be familiar with these three labels: the opposite, the adjacent, and the hypotenuse. And you need to be comfortable identifying which side is which in various different right-angled triangles.

So the three trigonometric ratios that we’re interested in are the ratios between the different pairs of the sides of the right-angled triangle. The first one that we’re going to look at has the name sine. Now for a particular angle 𝜃, this sine ratio is the ratio between the opposite and the hypotenuse. So its definition is that sin of the angle 𝜃 is equal to the opposite divided by the hypotenuse. Whatever the lengths of those two sides are. The second ratio is given the name cosine or often abbreviated to cos. And this is the ratio between the adjacent and the hypotenuse. So it is defined as the cos of the angle 𝜃 is equal to the adjacent divided by the hypotenuse. The final ratio is called tangent, often abbreviated to tan. And this is the ratio between the opposite and the adjacent sides. So its definition is that tan of the angle 𝜃 is equal to opposite divided by adjacent. So that’s how these three ratios are defined.

And what’s important is that these ratios, for a particular angle 𝜃, they are fixed no matter how big I choose to draw the triangle. And therefore, no matter what the lengths of the opposite, the adjacent, and the hypotenuse are. For a particular angle 𝜃, the values of these three ratios will always be the same. So if I were to extend this triangle, but still keeping that angle 𝜃 constant. Then the value of this sine ratio would be the same, whether I use the opposite and the hypotenuse as they’re labeled in that solid triangle. Or whether I use the opposite and the hypotenuse as they are now marked in red in the larger triangle. In both cases, when I do the opposite divided by the hypotenuse, I will get the same value for this ratio. And of course, the same thing is true for cosine and for tangent as well.

Now to help you remember the definitions of sine, cosine, and tangent and which sides are involved, there is a little acronym that you can use to help. And so what we do is we take the first letter of each of those words. So sin 𝜃 is equal to opposite over hypotenuse. So that gives us SOH, S, O, H. Cos is equal to the adjacent over hypotenuse. That gives us CAH, and so on. So SOHCAHTOA. If you remember that, then you’ll be able to remember the definitions of sine, cosine, and tangent more easily. If you wanted to convince yourself that these ratios are in fact constant for a fixed angle 𝜃, you could do your own investigation. So like the one I’ve started on the screen, you could draw out a triangle, measure these sides as accurately as possible, and calculate the ratios. And then continue to a medium triangle and a larger triangle and convince yourself that the ratios do remain the same. So let’s look at how we can use this to answer some questions.

The first question, we’re given a diagram of a right-angled triangle. And we’re asked to write down the value of cos 𝜃.

So for me, when I’m answering any problem to do with trigonometry, the first step for me is always to label the three sides of the triangle in the problem with their labels. So the opposite, the adjacent, and the hypotenuse. So I’ve just used the first letter of those words here. Remember the hypotenuse is opposite the right angle. The opposite is opposite the angle 𝜃. And the adjacent is between 𝜃 and the right angle. Now we’re asked about cos of 𝜃. So we need to recall the definition of cos of 𝜃. And if you remember SOHCAHTOA, well CAH is adjacent and hypotenuse. So the definition of cos 𝜃 is that it’s the adjacent divided by the hypotenuse. What I need to do then is just write down what this ratio is for this particular triangle. So I need to replace the adjacent and the hypotenuse with their values in this example.

So looking at the triangle then, the adjacent is 12 centimeters. So I’m gonna have cos 𝜃 is equal to 12 over — and the hypotenuse. Looking at the diagram, that’s 13. So I have that cosine of 𝜃, cos 𝜃, is equal to 12 over 13. So all I needed to do in this question was recall the definition of cos and then write down the ratio using the values that are specific to this question.

Okay, our second question. We’re given another right-angled triangle. And the angle that we’re interested in is labeled 𝛼 this time rather than 𝜃. And the question asked us to write down the value of sin of 𝛼.

So step one, as in the previous question. First of all, I’m just gonna label each of these three sides as the opposite, the adjacent, and the hypotenuse. Next, I need to recall what the definition of sine is. So SOH, sine is opposite divided by hypotenuse. So looking at the diagram, I can see that I know the hypotenuse. It’s 10 millimeters. But I don’t actually know what the opposite is. That length hasn’t been given to me in the diagram. But I need to know in order to write down this sine ratio. So there’s a little bit of work that I need to recall in order to work this out. And it’s the Pythagorean theorem. Because remember, the Pythagorean theorem tells us about the relationship that exists between the three sides in a right-angled triangle. And what the Pythagorean theorem enables me to do is calculate the length of the third side, if I know both of the other two. And in this case I do. I can see they’re 10 millimeters and six millimeters.

Now the Pythagorean theorem, you often see it written as 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared. But remember, what the theorem actually tells us is that if you take the two shorter sides of a right-angled triangle. So that’s 𝑎 and 𝑏. And if you square them and add them together, then it gives you the same result as if you square the longest side, the hypotenuse. So I can use the Pythagorean theorem to work out the length of this third side. Now I’m not gonna call it 𝑂 because that might be confused with zero. So I’m gonna give it a different letter. Let’s call it 𝑏. So in my working for this part of the question, I’m gonna think of that third side as 𝑏. So what I’m gonna do is I’m gonna write out the Pythagorean theorem. But I’m gonna replace 𝑎 with six and keep 𝑏 as it is. And I’ll replace 𝑐 with 10. So I have six squared plus 𝑏 squared is equal to 10 squared.

Next, I’m gonna write down what six squared and 10 squared are. So I’ve 36 plus 𝑏 squared is equal to 100. Now I want to solve this equation for 𝑏. So I’m gonna subtract 36 from both sides. So I have 𝑏 squared is equal to 64. And then in order to work out 𝑏, I need to square root both sides. So I’ll have 𝑏 is equal to the square root of 64, which is eight. So this tells me then that the length of that third side, which is the opposite, must be eight millimeters. Now you may actually have been able to spot that without going through the formal working out because six, eight, 10 is an example of the Pythagorean triple. That is a right-angled triangle, where the lengths of all three sides are integers. And if you knew that and recognized it, then you could cut down on a little bit of the working out here.

Right, the reason we wanted to know the length of this third side is because we wanted to write down the sine ratio. So remember, sine is opposite divided by hypotenuse. And now I know the opposite is eight millimeters. And the hypotenuse is 10 millimeters. So this tells me then that sin of 𝛼 must be eight over 10. Now that could be simplified to four-fifths. Or, we could write it as a decimal. So we have our answer to this question, which is that sin of 𝛼 is equal to 0.8.

Okay, the final example in this video then.

We’re asked to write down the value of tan 𝜃 in this right-angled triangle here.

Now, we’re given the lengths of all three sides in this particular triangle. We’re not going to need them. So first step as always, we need to label the three sides: the hypotenuse, the opposite, and the adjacent. Next, we need to recall what the definition of that tan ratio is. So TOA, tan is the opposite divided by the adjacent. So looking at our diagram, we’re gonna be using the opposite which is root two over two and the adjacent which is a half. So because both of these are fractions, I don’t want to write one over the other and end up with a fraction that has four layers within it. So I’m gonna write tan 𝜃 is root two over two divided by a half, to start off with.