Lesson Video: The Sine Rule Physics • 9th Grade

In this video, we will learn how to use the sine rule to find missing sides and angles in different triangles.

12:30

Video Transcript

In this video, our topic is the sine rule. This is a rule that applies to all triangles, and it allows us to solve for interior angles as well as side lengths. In this lesson, we’ll learn what this rule says, as well as how to use it practically.

As we get started talking about the sine rule, also sometimes called the law of sines, we can recall that this word “sine” describes a mathematical function. And specifically, if we take values of 𝑥 ranging from zero up to two 𝜋 and insert them into this function, then out comes a series of values ranging from a maximum value of one to a minimum value of negative one.

An important point here is that these 𝑥-values, the argument of the sine function, are angles. We see here that we’ve written out these angles in units of radians, so they range from zero up to two 𝜋 radians. But we could equivalently write them in degrees. 𝜋 over two radians is 90 degrees, 𝜋 radians is 180 degrees, and so on, up to two 𝜋 radians, which is 360 degrees. So the sine function is a trigonometric function that takes as its argument an angle expressed either in radians or degrees. And it returns a value as high as positive one and as low as negative one.

The sine rule then uses this function, and it uses it specifically with triangles. And one great thing about the sine rule is that it applies to virtually any triangle. It could be a right triangle; an isosceles triangle, where two of the angles are identical; or just generally any three-sided shape. What the sine rule does mathematically is it equates ratios of sines of an angle to corresponding side lengths.

Say we have a triangle, and the interior angles of this triangle are called 𝐴, 𝐵, and 𝐶 like this. When we talk about the sine rule equating ratios of sines of angles to corresponding side lengths, what we mean is that each one of these angles, say 𝐵, for example, has what we can call a corresponding side length of the triangle. And that in the case of 𝐵 is this length right here. For angle 𝐶, that side length would be here. And then for angle 𝐴, it’s the side length opposite that, right here.

What we can do then is name the side lengths of this triangle to correspond with the interior angles they stand opposite to. So we’ll call this side length here lowercase 𝑎. And then the side length opposite angle 𝐶 we’ll call lowercase 𝑐. And that opposite 𝐵 we’ll give lowercase 𝑏 as its name. And just to be clear, the symbol for this angle here is meant to be an uppercase 𝐶.

Our triangle now is completely labeled, all the angles and all the side lengths. And we can go back to our statement about the sine rule, that it describes ratios of sines of an angle to corresponding side lengths. In any triangle, there are always three angles and three side lengths, and so we’ll get three ratios.

The first ratio, we can say, is the sin of the angle capital 𝐴 divided by the side length lowercase 𝑎. The second ratio is the same thing but now for the angle and side we’ve marked 𝐵 and then lastly for the angle and side marked 𝐶. Now that we have these ratios, the last step is to equate them. So we can go and write that the sin of angle 𝐴 divided by side length 𝑎 equals the sin of angle 𝐵 divided by side length 𝑏, which is likewise equal to the sin of angle 𝐶 divided by side length 𝑐.

This overall expression here is the mathematical statement of the sine rule. What it helps us do is solve for interior angles and side lengths in triangles in whatever application we find them. We’ll apply this rule using a few examples in a moment. But before we do, note that while this is perhaps the most common form for the sine rule, there’s another mathematically equivalent way to write it.

If we were to invert all these ratios so that now the side lengths 𝑎, 𝑏, and 𝑐 in lowercase were on top and the sines of all the angles are in the denominators, this is still a mathematically correct way of writing out the sine rule. One way to convince ourselves of this is to start here and then use cross multiplication until we have our expression in this form. With that said, the best way to learn the sine rule is to use it in practice. So let’s now look at an example exercise.

What is the length of side 𝑎 of the triangle shown?

Looking at this triangle, we see that two of the interior angles are given to us. Along with that, the side length corresponding to one of these interior angles is also known. We also see side length 𝑎, which is opposite the angle marked out as 47 degrees. And it’s this length we want to solve for. To do this, we’re going to use the sine rule, also known as the law of sines. This rule says that if we have a triangle, and it could be any triangle, then the ratio of the sine of any one of the triangle’s interior angles to its corresponding side length is equal to that same ratio for any of the other angles and corresponding sides.

Now, we don’t need to know all three of the angles or all three of the sides to use any one of these relations. Because of this, we can apply the sine rule to our triangle over here in order to solve for side length 𝑎. We see that the side length, as we noted, is opposite the interior angle of 47 degrees. So then, as one of our ratios applying the sine rule, we can write the sin of 47 degrees divided by 𝑎. And then by that same rule, this ratio is equal to the sine of any other interior angle in our triangle divided by its corresponding side length. The question is, do we have that information?

Well, we see that we have this other interior angle, 95 degrees, and that opposite that angle we indeed have a known side length, 10 centimeters. So our ratio then is the sin of 95 degrees divided by 10 centimeters. And as we said, the sine rule tells us this is equal to the sin of 47 degrees divided by 𝑎.

And now that we have this equation, all we need to do is rearrange algebraically to solve for 𝑎. If we multiply both sides by 𝑎, that factor cancels out on the left. And then, next, we multiply both sides of the equation by 10 centimeters divided by the sin of 95 degrees, canceling out 10 centimeters and that sine on the right. And finally, we have an expression that we can evaluate to solve for side length 𝑎. Entering this expression on our calculator, to two significant figures, we find a result of 7.3 centimeters. That’s the length of side 𝑎 in the triangle in our diagram.

Let’s look now at a second example.

What is the size of angle 𝐴, in degrees, in the triangle shown?

In this triangle, we see the angle 𝐴 marked out here, as well as the interior angle of 54 degrees. Opposite this 54-degree angle is a side length of 8.4 centimeters. And then opposite angle 𝐴 is a side length of 9.6 centimeters. We see then that this triangle is well set up for us to use the sine rule to solve for this angle 𝐴. The sine rule applies to any triangle. It says that if the interior angles and the corresponding side lengths are marked out as shown, then the sine of any of those interior angles divided by the corresponding side length is equal to that same ratio for any of the other pairs of angles and sides.

So as we think about applying the sine rule to our triangle over here, specifically to solve for angle 𝐴, we remind ourselves that angle 𝐴 corresponds to the side length of 9.6 centimeters and the angle of 54 degrees corresponds to 8.4 centimeters. So then the ratio of the sin of the unknown angle 𝐴 to 9.6 centimeters is equal, by the sine rule, to the sin of 54 degrees divided by 8.4 centimeters.

And now what we want to do is to isolate this angle 𝐴. To do this, we’ll first multiply both sides of the equation by 9.6 centimeters, canceling that factor on the left and then giving us this expression here. Notice that on the right-hand side of our equation, these units of centimeters cancel from numerator and denominator.

And now what we need to do is to invert or undo the application of the sine function on the angle 𝐴 we want to solve for. We do this by applying what’s called the arc sine or the inverse sine to the sin of 𝐴. And then to maintain our equality, we apply the same inverse sine function to the right-hand side of our expression. When we take the inverse sine of the sin of the angle 𝐴, what remains is simply the angle 𝐴.

Our final step is to evaluate the right-hand side of this expression. And just as the sine function is standard on any scientific calculator, so the inverse sine function will be as well. When we evaluate this expression and find our answer in degrees, to two significant figures, it’s 68 degrees. That’s the size of the angle 𝐴 in the triangle shown.

Let’s now consider one last example exercise.

What is the length of side 𝑎 of the triangle shown in the diagram?

In this diagram, we see side 𝑎 right here, as well as two interior angles of this triangle, 64 degrees and 38 degrees. And then opposite this angle of 38 degrees, we have a side length given as 6.1 centimeters. Knowing this, we want to solve for the side length 𝑎. And we’re going to do it by applying what’s called the sine rule. This rule says that if we have a triangle, and it could be any triangle, with its angles and sides marked out like this, then the ratio of the sine of any of these angles to the corresponding side length is equal to that same ratio for the other pairs of sides and angles.

We can see that a key to being able to use the sine rule is to have at least two corresponding angle and side pairs. That way, we can set up an equality and then solve for any one of those four values, given the other three. As we look to apply the sine rule to our diagram in order to solve for the side length 𝑎, we see that we have one such angle and side pair, 38 degrees with 6.1 centimeters, but that that’s the only one. We don’t have a second corresponding pair. If we knew what this angle of the triangle was though, we see that that angle is opposite the side length 𝑎 we want to solve for.

It turns out that we can solve for this angle, but we won’t do it using the sine rule. Instead, we’ll use the fact that the sum of the interior angles of any triangle is always 180 degrees. So therefore, if we call this unknown angle in our triangle capital 𝐴, then we can say that 𝐴 plus 64 degrees plus 38 degrees is equal to 180 degrees. Now, 64 degrees plus 38 degrees is 102 degrees. And if we subtract this amount from the left-hand side, the negative 102 cancels with positive 102. And we find that 𝐴 equals 180 degrees minus 102 degrees, or 78 degrees.

Now that we know this angle in our triangle, we can use the sine rule to solve for the side length 𝑎. We’ll write that the sin of capital 𝐴, which we know is 78 degrees, divided by the side length lowercase 𝑎 is equal to this ratio over here, the sine of this angle in our triangle divided by this side length. Solving for 𝑎 is now just a matter of cross multiplying. We can start by multiplying both sides of the equation by 𝑎, leading that factor to cancel on the left. And then, in our resulting equation, if we want to isolate 𝑎 on the right-hand side, we do this by multiplying both sides by the inverse of this ratio. When we multiply this ratio by its inverse, that product is equal to one, effectively canceling all this out. And finally, we have an expression we can enter in on our calculator to solve for 𝑎. To two significant figures, 𝑎 is 9.7 centimeters. That’s the length of this side of the triangle.

Let’s now summarize what we’ve learned about the sine rule. In this lesson, we learned that the sine rule applies to all triangles. And it equates ratios of sines of angles to corresponding side lengths. This means that given a triangle, with its angles and sides labeled like this, the sine rule tells us that the sin of angle 𝐴 divided by side length 𝑎 is equal to the sin of angle 𝐵 divided by side length 𝑏, which is also equal to the sin of angle 𝐶 divided by side length 𝑐. Lastly, we saw that an equivalent expression for the sine rule is to put the side lengths on top in the numerators and the sines of the angles in denominators. Either way of stating the rule is correct and can be used to solve for side lengths and interior angles of triangles.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.