In this video, we will learn how to use the cosine and sine law to solve real-world problems. As we’re focusing on the applications of these laws, we’ll assume at this point familiarity with the laws themselves. And we won’t explain the basics of these laws or how to prove them in this video.
Firstly though, a quick reminder of what the sine and cosine laws actually are and when they can be used. Suppose we have a triangle 𝐴𝐵𝐶 labeled as follows. Remember that the convention is to use uppercase letters to represent angles and to use lowercase letters to represent sides, with sides always being opposite the angle with the same letter. None of the angles in this triangle are necessarily right angles. So the first key point we need to recall is that the sine and cosine laws allow us to calculate side lengths and angle measures in nonright triangles.
The law of sines or the sine law first of all then. This tells us that, in any triangle, the ratio between each side length and the sine of its opposite angle is constant, which we can write as 𝑎 over sin 𝐴 equals 𝑏 over sin 𝐵, which is equal to 𝑐 over sin 𝐶. And remember, the lowercase letters represent side lengths and the uppercase letters represent angles.
We don’t need to use all three parts of the sine law, just two parts of this equality. The information needed in order to apply the law of sines is two angles and their opposite sides. So we can recognize the need for the law of sines when we’re working with opposite pairs of information.
This first version of the law of sines is particularly useful when calculating the length of a missing side, as the sides are in the numerators of the fractions, so less rearrangement is required. We do also have a reciprocal version, which is particularly helpful for calculating the measure of an angle. sin 𝐴 over 𝑎 equals sin 𝐵 over 𝑏, which is equal to sin 𝐶 over 𝑐. So that’s the law of sines, and now let’s consider the law of cosines or the cosine law.
The most common form in which you see this written is as follows. 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos of 𝐴. We use it to calculate a side length, in this case the length of side 𝑎, when we know the other two sides of the triangle and the included angle. So we know the sides 𝑏 and 𝑐 and the angle 𝐴.
We can also rearrange the cosine law to a form which enables us to calculate any angle in the triangle if we know all three side lengths. The rearrangement is straightforward, and it gives cos of 𝐴 is equal to 𝑏 squared plus 𝑐 squared minus 𝑎 squared all over two 𝑏𝑐. In this case, as we said, we know all three side lengths and we wish to calculate one of the angles.
Again, the proofs and the basic application of these laws are not going to be considered in this video. Instead, we’re going to focus on applying these laws to some real-life problems. Let’s look at our first question.
A plane travels 800 meters along the runway before taking off at an angle of 10 degrees. It travels a further 1,000 meters at this angle as seen in the figure. Work out the distance of the plane from its starting point. Give your answer to two decimal places.
Looking at the diagram, we can see that we have a triangle. We want to calculate the distance of the plane from its starting point. That’s this length here, which we can refer to as 𝑑 meters. We know the lengths of the other two sides in this triangle. They are 800 meters and 1,000 meters. And using the fact that angles on a straight line sum to 180 degrees, we can work out the size of this angle here. It’s 180 degrees minus 10 degrees, which is 170 degrees.
As this is a non-right-angled triangle, we need to answer this problem using either the law of sines or the law of cosines. So the first step is to decide which of these we need. And that will depend on the specific combination of information we’ve been given and what we want to calculate.
In this triangle, we know two sides and the included angle. And we want to calculate the third side. We recall then that this means we should be using the law of cosines. Let’s recall the law of cosines. It’s 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴. Now, there’s no need to actually label our triangle using the letters 𝐴, 𝐵, and 𝐶. Instead, we just remember that the lowercase letters 𝑏 and 𝑐 represent the two sides we know and the capital letter 𝐴 represents the included angle.
So using 800 and 1,000 as the two side lengths 𝑏 and 𝑐 and 170 degrees as the angle 𝐴, we have the equation 𝑑 squared equals 800 squared plus 1,000 squared minus two times 800 times 1,000 times cos of 170 degrees. We can either type this directly into our calculator or it may be a good idea to break the calculation down into some stages. In either case, we arrive at 𝑑 squared equals 3,215,692.405.
Now, we must remember that this is 𝑑 squared. It isn’t 𝑑, so we aren’t finished. We have to square root in order to find the value of 𝑑. It’s a really common mistake though to forget to do this. Square rooting gives 𝑑 equals 1,793.235178. The question asks us to give our answer to two decimal places. So rounding appropriately, we’ve worked out the distance of the plane from its starting point. It’s 1,793.24 meters to two decimal places.
In this example, we were given a diagram to use. Often this won’t be the case, and we’ll need to use information from a written description to draw our own diagram to help us answer the question. Let’s now consider an example of this.
A ship is sailing due south with a speed of 36 kilometers per hour. An iceberg lies 24 degrees north of east. After one hour, the ship is 33 degrees south of west of the iceberg. Find the distance between the ship and the iceberg at this time, giving the answer to the nearest kilometer.
A lot of the skill involved in this question is in drawing the diagram. Let’s start with a compass showing the four directions. We’ll then take each statement separately and consider how to represent it. Firstly, we know that the ship is sailing due south. Initially, we’re told that an iceberg lies 24 degrees north of east of the ship’s starting point.
Now, directly east would be directly to the right of the ship on our diagram. 24 degrees north of east would mean the iceberg lies somewhere along the line like this. We’re then told that after one hour, the ship is 33 degrees south of west of the iceberg. Well, west would be the direction directly to the left of our iceberg. And using alternate angles in parallel lines, we know that the angle formed here is 24 degrees.
So the full angle between the horizontal and the position the ship has now moved to is 33 degrees. And we can now see that we have a triangle. We can work out the angles in our triangle. For example, this angle here is the difference between 33 degrees and 24 degrees. It is nine degrees. We could also work out this angle here, it’s 24 degrees, plus the angle between south and east, which is 90 degrees, giving a total of 114 degrees.
The one piece of information we haven’t used yet is that the ship is traveling at a speed of 36 kilometers per hour. And we know it takes one hour for the ship to go from its original position to its new position. The ship will therefore have traveled 36 kilometers in this time. So we also know one side length in our triangle.
What we were asked to calculate is the distance between the ship and the iceberg at this later time. So that’s this side here, which we can refer to as 𝑑 kilometers. We’ve now set up our diagram, and we see that we have a non-right-angled triangle, which means we’re going to apply either the law of sines or the law of cosines. Let’s look at the particular combination of information we’ve got.
We know an angle of nine degrees and the opposite side of 36 kilometers. We also know an angle of 114 degrees. And we wish to calculate the opposite side of 𝑑 kilometers. We therefore have opposite pairs of sides and angles, which tells us that we should be using the law of sines to answer this question.
Remember, this tells us that the ratio between each side length, represented using lowercase letters, and the sine of its opposite angle, represented using capital letters, is constant. 𝑎 over sin 𝐴 equals 𝑏 over sin 𝐵, which is equal to 𝑐 over sin 𝐶. We only need to use two parts of this ratio. And there’s no need to label our triangle using the letters 𝐴, 𝐵, and 𝐶 as long as we’re clear about what they represent.
Our side 𝑑 is opposite the angle of 114 degrees, and the side of 36 kilometers is opposite the angle of nine degrees. So we have 𝑑 over sin of 114 degrees equals 36 over sin of nine degrees. We can solve this equation by multiplying each side by sin of 114 degrees, which is just a value. And it gives 𝑑 equals 36 sin 114 degrees over sin of nine degrees. Evaluating on a calculator, making sure our calculator is in degree mode, and we have 210.23267.
The question asks us to give our answer to the nearest kilometer. So rounding appropriately, we have that the distance between the ship and the iceberg at this time is 210 kilometers.
We’ve now seen one example each of using the law of sines and the law of cosines to calculate a side length. In our next example, we’ll see how we can apply the law of cosines to calculate all the missing angles in a triangle when we know its three side lengths.
Los Angeles is 1,744 miles from Chicago, Chicago is 712 miles from New York, and New York is 2,451 miles from Los Angeles. Find the angles in the triangle with its vertices as the three cities.
Now, whilst a little bit of knowledge of the geography of the United States of America might be useful here, it isn’t essential to answering the problem. We can just draw a triangle using the three lengths given in the question. And if our triangle turned out to be upside down, it isn’t the end of the world. The triangle should look a little something like this, and we can add the three distances.
Now, this triangle certainly doesn’t look as if it’s a right triangle. So we’re going to need to apply either the law of sines or the law of cosines to this problem. We know all three of the side lengths, and we want to calculate each of the angles, which tells us that we should be using the law of cosines. The rearranged version of this, which is useful for calculating angles, is cos of 𝐴 equals 𝑏 squared plus 𝑐 squared minus 𝑎 squared all over two 𝑏𝑐. If you can’t remember this, you’ll have to perform the rearrangement yourself from the law of cosines in its traditional form.
In this question then, let’s use 𝐴 to represent Los Angeles, 𝐶 to represent Chicago, and 𝐵 to represent New York. We’ll use the corresponding lowercase letters to represent the opposite sides. To calculate our first angle then, that’s this angle here, we substitute the relevant values. Giving cos of 𝐴 equals 1,744 squared plus 2,451 squared minus 712 squared all over two multiplied by 1,744 multiplied by 2,451.
We can evaluate this on a calculator. And then to find the value of 𝐴, we need to use the inverse cosine function. Doing so gives 𝐴 equals 2.334 degrees. So we’ve found the first angle in the triangle. And we’ll give our answer to two decimal places.
To calculate the next angle in this triangle, which this time we’ll use angle 𝐶, we don’t need to relabel our triangle. We just need to remember that the letters 𝑏 and 𝑐 represent the two sides which enclose the angle and the letter 𝑎 represents the opposite side. So we use 1,744 and 712 for the two sides which enclose the angle and 2,451 for the side which is opposite. This gives cos of 𝐶 equals negative 0.9901. And again, applying the inverse cosine function, we find that angle 𝐶 equals 171.939 degrees.
So we found two angles in the triangle. And in fact, to find the third, we could subtract the two angles we’ve calculated from 180 degrees. But if we don’t use that method, that will be a useful check. In exactly the same way then, but this time using 712 and 2,451 as the two sides which enclose the angle and 1,744 as the opposite side, we find the measure of angle 𝐵 is 5.726 degrees.
Adding the three angles we’ve found, now each rounded to two decimal places, does indeed give 180 degrees. So we can have some confidence in our answer. The measures of the three angles in the triangle formed by these three cities, each to two decimal places then, are 2.33 degrees, 5.73 degrees, and 171.94 degrees.
In our final example, let’s see how we can use the law of sines and the law of cosines to solve problems in other mathematical contexts.
𝑀 is the center of a circle, and 𝐴, 𝐵, and 𝐶 are points on the circumference. If 𝐵𝐶 equals 13 centimeters and the measure of angle 𝐶𝑀𝐵 is 84 degrees, find the area of the circle 𝑀, giving the answer to the nearest square centimeter.
We know that the area of a circle is 𝜋𝑟 squared. So really, this problem is about finding the radius of this circle. Let’s begin by putting the information we’ve been given on the diagram. 𝐵𝐶 is 13 centimeters, and the measure of angle 𝐶𝑀𝐵 is 84 degrees. We don’t know the lengths of 𝑀𝐶 or 𝑀𝐵, but they’re each the radius of the circle.
Now, there are numerous different approaches we could take. But one approach is to apply the law of cosines in the triangle 𝐶𝑀𝐵. This states that 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴, where 𝑏 and 𝑐 represent two sides of a triangle and 𝐴 represents the included angle. In our triangle, 𝑎 is 13 centimeters. The angle 𝐴 is 84 degrees. And the two sides which enclose this angle 𝐴 are each the radius of the circle 𝑟.
We can therefore form an equation. 13 squared equals 𝑟 squared plus 𝑟 squared minus two 𝑟 squared cos of 84 degrees. We can solve this equation to find the value of 𝑟 squared, which we’ll then be able to substitute directly into our area formula. Factorizing the right-hand side of our equation by 𝑟 squared, we have 169 equals 𝑟 squared multiplied by two minus two cos of 84 degrees. Dividing through, we have that 𝑟 squared is equal to 169 over two minus two cos 84 degrees. And we’ll keep our value for 𝑟 squared in this exact form.
We can then substitute this value of 𝑟 squared into the area formula and evaluate on a calculator. Rounding our answer, and we have that the area of circle 𝑀 to the nearest square centimeter is 296 square centimeters.
As I mentioned, there are in fact numerous approaches to this problem, which you can try out yourself if you wish. We could’ve applied the law of sines in triangle 𝐶𝑀𝐵. Or we could’ve divided it in half to form two right triangles and then used right-angle trigonometry.
Let’s now summarize the key points in this video. Firstly, the law of sines and the law of cosines can be used to calculate side lengths and angle measures in nonright triangles. The law of sines in either of its two forms can be used to calculate either a side or an angle when we’re working with opposite pairs of information. The law of cosines in its first form can be used to calculate a side length when we know the other two sides and the included angle. And in its rearranged form, it can be used to calculate any angle when we know all three sides.
We can apply the law of sines and the law of cosines to many problems involving triangles. And although we haven’t seen an example of this, we can also apply both rules within the same problem.