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Learn how to use the cosine ratio to calculate the length of either the adjacent side or hypotenuse in a right triangle. Apply your knowledge to triangle problems, including those requiring knowledge of circle theorems to find some of the information.
In this video, we’re going to see how to use the cosine ratio to calculate the length of either the adjacent or the hypotenuse in a right-angled triangle.
So first of all, a reminder of what the cosine ratio is. I have on the screen here a diagram of a right-angled triangle. And I’ve chosen one of the other two angles to label as 𝜃. I’ve then labeled the three sides of this triangle with their names in relation to this angle 𝜃. So we have the hypotenuse. We have the opposite. And we have the adjacent.
The cosine ratio or cos, as it’s often abbreviated to, is the ratio between the adjacent and the hypotenuse of this triangle. So its definition is that cos of the angle 𝜃 is equal to the adjacent divided by the hypotenuse. And if you’re using SOHCAHTOA to help you with trigonometry, then this is the CAH part of that.
We’re now going to see how we can use this ratio in order to calculate the length of missing sides in right-angled triangles. So here is our first problem. We’re given a diagram of a right-angled triangle. We can see we’ve been given the length of one side and one angle. And we’re asked to calculate 𝑥, which is the length of another side.
So for me, the first step for any problem involving trigonometry is to label the three sides of the triangle with their labels in relation to this angle of 28 degrees. So using the first out of each of these words, you can see there I have the hypotenuse, the opposite, and the adjacent.
Now looking at those three sides, you can see I’ve got the six centimeters, which is the hypotenuse. And I want to calculate 𝑥, which is the adjacent. So I have the 𝐴 and 𝐻 involved in this ratio. And if you think about SOHCAHTOA, then 𝐴 and 𝐻 appear together in the cos ratio, which is how we know that it’s the cos ratio that we’re going to need for this particular question.
So I recall the definition of the cos ratio. And it’s this that it’s the adjacent divided by the hypotenuse. What I’m going to do then is that I’m going to write down this ratio. But I’m gonna fill in the information I know. So I’m gonna replace 𝜃, which is the angle, with 28. I’m gonna replace the adjacent with 𝑥 because that’s its letter in the diagram. And I’m gonna replace the hypotenuse with six.
So this gives me cos of 28 is equal to 𝑥 over six. Now this is an equation that I’m looking to solve in order to work out the value of 𝑥. So the first step to do in this, well there’s a six in the denominator. So I need to multiply both sides of this equation by six.
When I do that, I get that 𝑥 is equal to six cos 28. This just means six multiplied by cos 28. But we don’t need the multiplication sign there. Now I’m gonna evaluate this using my calculator, making sure it’s in degree mode because the angle in the question was specified in degrees.
So evaluating this, I get that 𝑥 is equal to 5.29768. The question asked me to find this value to the nearest 10th. So I need to round my answer. And therefore I have that 𝑥 is equal to 5.3.
So a quick summary, in this question then, what we did was we identified that it was the cosine ratio we needed by labeling the sides of the triangle. We then wrote this ratio out, but using the information in this particular question, and then solved the resulting equation in order to work out that missing value 𝑥.
Right, here is the second example we’re going to look at. It says calculate the length of the hypotenuse of triangle 𝐴𝐵𝐶 to two decimal places. And we can see that we’re given a diagram of a right-angled triangle where we’ve got the length of one side and the size of one of the other angles.
So as before, my first step is to label the three sides with the labels in relation to this angle of 65 degrees. So we have their labels here. And again we can see that it’s the cosine ratio we’re going to need because I’ve been given the length of the adjacent side. It’s 5.1 millimeters. And it’s the hypotenuse I want to calculate. So 𝐴 and H remember is the cosine ratio.
So as before, what I’m going to do is I’m gonna write out the cosine ratio. But I’m gonna fill in the information I know. So 𝜃 is gonna become 65. The adjacent is gonna become 5.1. And the hypotenuse I will refer to as 𝐵𝐶 as that’s the notation for this particular triangle.
So I have cos of 65 is equal to 5.1 over 𝐵𝐶. Now again, this is an equation that I’d like to solve in order to work out the value of 𝐵𝐶. It’s slightly more complex because, this time, the unknown length is in the denominator of a fraction. So my first step in solving this equation is going to be to multiply both sides of the equation by 𝐵𝐶.
So doing this gives me 𝐵𝐶 multiplied by cos 65 is equal to 5.1. Now I want to work out 𝐵𝐶. So the next step is to divide both sides of the equation by cos of 65. Cos of 65 is just a number. So it’s perfectly okay for me to do that.
This gives me then that 𝐵𝐶 is equal to 5.1 over cos of 65. This stage here is when I’m gonna use my calculator to evaluate this. And this tells me that 𝐵𝐶 is equal to 12.067 and so on. Now the question asked me to give my answer to two decimal places. So finally, I need to round this answer. And I need to include the units which are going to be millimeters.
So my final answer then is that the length of the hypotenuse 𝐵𝐶 is 12.07 millimeters. So the process for this question was very similar to the previous one. We labeled the three sides, identified the need for the cosine ratio, we wrote it down using the information in the question, and then solved the resulting equation. The equation was just slightly more complex this time because the length we were looking to find was in the denominator of a fraction. So it involved two steps in order to solve it rather than just one.
Right, our next example looks a little bit different. We’re given a circle. And we’re told that 𝐴𝐵 is the diameter of this circle. We’re then asked to calculate the radius of this circle to the nearest 100th.
Now we can see that, within this circle, we have a triangle. And we’ve got one of the angles of the triangle which is 40 degrees and one side length which is four centimeters. The question is, well is this a right-angled triangle? Because if it is, then we can apply trigonometry to this problem.
Now in order to see that this is in fact a right-angled triangle, you need to make a connection with some work in another area of mathematics. And that is the area of circle theorems.
One of the key circle theorems tells us that if you have a scenario like this, then the angle that is at the circumference of the circle is half the angle that’s at the center. Now here the angle at the center is a straight line. So it’s 180 degrees. And therefore the angle at the circumference, the angle 𝐴𝐶𝐵, is half of that. So it’s 90 degrees, which means that this angle here is in fact a right angle.
If you’ve perhaps studied topics in a different order and you haven’t met circle theorems yet, then you’d expect that right angle to be marked on the diagram for you. But ultimately once you’ve met circle theorems, you would be able to deduce that for yourself.
So what this tells us is that we do have a right-angled triangle. And therefore we are able to use trigonometry in order to answer this problem. Therefore, my first step is gonna be to label the three sides of this triangle with their names, the opposite, the adjacent, and the hypotenuse, in relation to the angle of 40 degrees.
And so what we can see is we know the length of the adjacent. If we want to calculate the radius of this circle, then we need to calculate the length of 𝐴𝐵, which is the diameter, and then halve it. So again we need to know the hypotenuse. So A and 𝐻 are involved, which tells us it’s the cosine ratio that we need.
So as in the previous questions, we’re going to write the cosine ratio right again but filling in the information in this particular question. So we have that cos of the angle which is 40 is equal to four, the adjacent of 𝐴𝐵, the hypotenuse.
Now what we’re looking to do is solve this equation in order to work out the value of 𝐴𝐵. So this is very similar to what we saw in the previous example. 𝐴𝐵 is in the denominator of this fraction. So first of all we need to multiply both sides of the equation by 𝐴𝐵.
And doing so gives me 𝐴𝐵 multiplied by cos 40 is equal to four. The next step then is I need to divide both sides of this equation by cos of 40. So I have 𝐴𝐵 is equal to four over cos 40. Now I’m not going to evaluate this quite yet because the question asked me to calculate the radius of the circle. And 𝐴𝐵 is the diameter. So I need to halve this in order to answer the question.
So 𝑟, which I’m using to represent the radius of the circle, is four over cos 40 multiplied by a half. And I’ll use my calculator now to evaluate this, which tells me that 𝑟 is equal to 2.6108 and so on. Now the question asked me to find this radius to the nearest 100th. So I’ll round my answer.
So we have then that the radius of the circle is equal to 2.61 centimeters. So in this question, the actual process is very similar to what we’ve already seen previously. The only difference was that, this time, we were combining our knowledge of trigonometry with our knowledge of circle theorems as well and using circle theorems in order to identify that the angle 𝐴𝐶𝐵 was 90 degrees. And therefore we could apply trigonometry to the problem.
Right, the final question we’re going to look at is a non-calculator question. So we have a diagram of a right-angled triangle. And we’re told to use the information in the table in order to calculate 𝐵𝐶. And in the table, we can see that we’ve been given the values of the sine, cosine, and tangent ratios for this angle of 35 degrees.
So when doing trigonometry, you almost always have access to a calculator because you need your calculator in order to tell you the values of these sine, cosine, and tangent ratios. There are some particular angles, so 30 degrees, 45 degrees, and 60 degrees, for which these values are quite simple ratios in terms of surds. And so you can do trigonometry without a calculator. But the majority of the time we do need it.
However, this question has got round the need for a calculator by giving us the values of sin, cos, and tan within the question itself. And so we’ll see how to use those in a minute. First step then as it’s probably involving trigonometry is to label the three sides in relation to this angle of 35 degrees.
So we have the three labels here. And again as with all the examples in this video, we can see that it’s the cosine ratio we’re going to need because we have the length of the hypotenuse 𝐻. And we’re looking to calculate 𝐵𝐶 which is the adjacent. So 𝐴 and 𝐻 tells us it’s the cosine ratio we need.
So as in the previous questions, we are going to write out the cosine ratio but filling in the information for this question, so replacing the angle with 35, replacing the adjacent with 𝐵𝐶, and replacing the hypotenuse with 10.
We have then that cos of 35 is equal to 𝐵𝐶 over 10. Now I need to solve this equation for 𝐵𝐶. So I’m going to multiply both sides of the equation by 10. And I have that 𝐵𝐶 is equal to 10 cos 35. Now this is where the information in the question comes in because remember we haven’t got a calculator. But by looking at the table, we can see that we have the value of cos 35 here. It’s 0.819. So I can use that value at this stage in the question.
So this tells me that 𝐵𝐶 is equal to 10 multiplied by 0.819. And that’s a straightforward multiplication to do without a calculator. So we have that the length of 𝐵𝐶 is 8.19 centimeters. Now that’s not its exact value because cos of 35 isn’t exactly 0.819. It’s a longer decimal. But we’ve only been given it to three decimal places. So this value for 𝐵𝐶, this is exact to two decimal places or to the nearest 100th.
So this question then just gives you one example of how you might use trigonometry when you don’t have access to a calculator. In summary then, we’ve recalled the definition of the cosine ratio as the adjacent divided by the hypotenuse. And we’ve seen how to apply this ratio to four different questions where we’re looking to calculate either the adjacent or the hypotenuse in a right-angled triangle.
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