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Video: Finding Angles Using the Cosine Ratio

Lauren McNaughten

Learn how to use the inverse cosine ratio to calculate the measure of an angle in a right triangle given the length of the adjacent side and hypotenuse. Also, explore problems involving shapes that can be split into right triangles.

12:38

Video Transcript

In this video, we’re going to see how to use the cosine ratio in order to calculate the size of the angles in a right-angled triangle.

So first of all, a reminder of what the cosine ratio is. Well, I have here a right-angled triangle, in which I’ve labelled one of the other angles as 𝜃. And then I’ve labelled the three sides of the triangle with their names in relation to that angle 𝜃. So we have the opposite, the adjacent, and the hypotenuse.

The cosine ratio is the ratio of the adjacent and the hypotenuse. So it’s defined as cosine of 𝜃, or cos as it’s usually abbreviated to, is equal to the adjacent divided by the hypotenuse. Now that form of the relationship is very useful, if you’re looking to calculate the length of one of the sides, so either the adjacent or the hypotenuse. But in this video, we’re looking at how to calculate the size of this angle 𝜃.

In order to do that, we need an alternative way of specifying this relationship. And it’s using what’s referred to as the inverse cosine function or cos inverse. So it’s written like this: 𝜃, the angle, is equal to cosine inverse of the adjacent divided by the hypotenuse. And what this means is, if I know the value of that ratio, adjacent over hypotenuse, then this inverse cosine function allows me to work backwards in order to work out the size of the angle 𝜃 that this ratio belongs to. So within this video we’ll be focusing specifically on using this inverse cosine function in order to calculate the size of various angles.

Okay. Our first question, we’re given a diagram of a triangle and we’re asked, what is the value of cos of 𝜃 in this triangle below.

So in this particular example, we’re being asked to find the value of the cosine ratio, but not actually find the angle itself yet. So looking at the diagram, I can see that I’ve got an isosceles triangle. I know that because of the lines that are on those two sides there, which means they’re both sixteen centimeters. There are no right angles marked on this diagram. And in order to do this particular type of trigonometry, we do need a right-angled triangle. So I need to think about how I can create a right-angled triangle, from this one here that currently isn’t right-angled.

So what you’ll hopefully be familiar with, this isosceles triangle, the base is horizontal. So if I mark in a vertical line down the center of this triangle, what it does is it divides the isosceles triangle up into two right-angled triangles because that line gives the perpendicular height of the triangle. So by marking in this line here, I can divide that isosceles triangle up into two congruent, that is identical, right-angled triangles.

So now, in order to answer this problem, I’m just going to think about one of those two triangles. And I’m gonna think about the one on the left because that’s the one that has 𝜃 marked in it. Now because it was an isosceles triangle, this perpendicular height has divided the base exactly in half, which means the base of this right-angled triangle must be half twenty, so it’s ten centimeters. So I therefore have two of the sides in this right-angled triangle. Now because this problem is a problem that I want to use trigonometry for, I’m gonna start off by labelling the three sides of this right-angled triangle in relation to that angle 𝜃, so the opposite, the adjacent, and the hypotenuse.

And having done that, what you can see is that the two sides I know are the adjacent and the hypotenuse, A and H. That’s what tells me that it’s the cosine ratio I’m going to use for this particular question because if you recall SOHCAHTOA, then A and H appear together in the CAH part, which is the cosine ratio. So I need to recall then the definition of the cosine ratio. And it’s this: that cos of 𝜃 is equal to the adjacent divided by the hypotenuse. So what I’m going to do then for this question in order to find cos of 𝜃, is I’m just gonna write this relationship out, but I’m gonna replace the adjacent and the hypotenuse with their values. So I’m replacing the adjacent with ten and the hypotenuse with sixteen.

Therefore, I have that cos of 𝜃 is equal to ten over sixteen. Now that will simplify as a fraction. So in simplified form, it’s five-eighths, or in fact I could convert it to an exact decimal in this instance. So it’s zero point six two five. Either of those, a fraction or the exact decimal, would be appropriate formats to give the answer for this question.

So what we’ve done then, is we’ve recalled that if you draw in the perpendicular height of an isosceles triangle, you create two congruent right-angled triangles. And then we’ve used trigonometry in the right-angled triangle to calculate this value 𝜃. We’ve recalled the cosine ratio and wrote it down for this particular question.

Okay. Now second question, we’re given a diagram and it is a right-angled triangle and we’re asked to calculate the size of angle 𝜃, giving your answer to the nearest degree.

So as always, for a problem involving trigonometry, I’m going to begin by labelling the three sides of the triangle. And as before, we can see that it is the adjacent and the hypotenuse that are the two sides that I know. This tells me then that it’s the cosine ratio that I need in order to answer this question. So I have the definition of the cosine ratio and I’m going to write it down using the information in this question. So I’m gonna have cosine of the angle 𝜃 is equal to eight over twenty-one.

Now in this question, I’m not just asked to write down this ratio, but I’m asked to go a step further and actually work out the angle 𝜃. So this is where I need to use the inverse cosine function. Therefore, this tells me that 𝜃 is equal to inverse cosine of this ratio, eight over twenty one. Now I’m going to need my calculator to evaluate this. And if you look on your calculator, you’ll see usually that above the cos button there is cosine inverse or cos inverse. And so you’re often have to press shift in order to get to it. Now that would depend on the particular make and model of calculator you have, but that’s usually where that button is located. So typing this into my calculator, I get that 𝜃 is equal to sixty-seven point six zero seven three. Now the question asked me to calculate 𝜃 to the nearest degree, so I need to round my answer. And therefore, I have that 𝜃 is sixty-eight degrees.

Now it’s just worth mentioning that you do need to make sure your calculator is in the correct mode when you’re performing trigonometry. This question asked for 𝜃 in degrees, so I need to make sure that my calculator is in degree mode. There are other ways of measuring angles, which you may or may not have met, but you need to make sure that your calculator is in degrees for this particular question.

Okay. The next question says: 𝐴𝐵𝐶𝐷 is an isosceles trapezium in which 𝐴𝐵 is equal to 𝐵𝐶 is equal to 𝐴𝐷; they’re all ten centimeters. And the final side, 𝐶𝐷, is equal to twenty-six centimeters. We’re asked to calculate the measure of angle 𝐴𝐷𝐶 to the nearest second.

So first of all, angle 𝐴𝐷𝐶, that’s the angle formed when I go from 𝐴 to 𝐷 to 𝐶. And therefore, it is this angle here that we’re being asked to find. Now looking at the diagram, we haven’t got any right-angled triangles. In fact, we haven’t got any triangles at all, but we can create one. And what I’m gonna do is, I’m gonna draw a perpendicular height in from 𝐴 down to the base 𝐶𝐷. And by doing that, you can see that I’ve now created a right-angled triangle on this left side of the diagram here. I’m actually going to do just the same thing on the other side of the diagram as well.

Now I want to work within this right-angled triangle. So I’m gonna just label up its three sides in relation to the angle 𝜃. So we have the opposite, the adjacent, and the hypotenuse as usual. Now at the moment, it appears that I only know one of the lengths in this right-angled triangle. I know that the hypotenuse is ten centimeters. But currently, I haven’t got either of the other two lengths marked on. So we just need to think about this a little bit. Now I can see that 𝐶𝐷 is equal to twenty-six centimeters. And I can see that 𝐶𝐷 is composed of three parts. It is the base of this right-angled triangle, and then another straight portion, and then the base of this right-angled triangle again. Because these two triangles are congruent to each other, as this is an isosceles trapezium, it’s therefore symmetrical. And so these two triangles are identical.

Therefore, I can use all the information in this question in order to work out what the length of the base of those right-angles triangles is. So this part here is equal to ten centimeters because it’s vertically below 𝐴𝐵, and therefore it must be the same length as 𝐴𝐵. That means that out of the twenty-six centimeters, which is the total for 𝐶𝐷, I therefore got sixteen centimeters left over. And as these two right-angled triangles are identical, their bases are the same length. So their bases must both be half of that sixteen centimeters. Therefore, I’ve got eight centimeters for each of these bases.

Right. Now I have enough information in order to calculate this angle. So looking at the right-angled triangle, I can say that I know the adjacent now; it’s eight centimeters. And I know the hypotenuse. Therefore, I need the cosine ratio in order to answer this question. So I have the definition of the cosine ratio and now I write it down for this question specifically. So I have that cos of 𝜃 is equal to eight over ten. And of course, that will simplify. Or, I can just write it as a decimal in this case. So we have nought point eight. Now this is when I need to use that inverse cosine function in order to work out the value of 𝜃. So I have that 𝜃 is equal to cosine inverse of nought point eight.

Evaluating this then, tells me that 𝜃 is equal to thirty-six point eight six nine eight nine degrees. Now if I’ve been asked for my answer in degrees, I could stop here or I could just round it to the nearest degree. But I’ve been asked for my answer to the nearest second, so I need to recall how to convert from an answer in degrees to an answer in seconds. So looking at this, I can see that I have thirty-six full degrees, and then I have a decimal of nought point eight six nine eight nine leftover. So this needs to be converted, first of all, into minutes and then anything that’s leftover into seconds. So remember then that a minute is one sixtieth of a degree. And therefore, in order to work out how many minutes I have, I need to multiply this decimal by sixty. When I do that, I get fifty-two point one nine three eight. What this means then is, I have fifty-two full minutes and then I have a decimal of zero point one nine three eight leftover, which needs to be converted into seconds. Now again, a second is one sixtieth of a minute. So in order to work out how many seconds this decimal represents, I need to multiply this by sixty. And when I do that, I get eleven point six three one five. So I’m asked for this to the nearest second. So I’ll round that then to twelve seconds.

So finally, I need to pull all these three parts together. And this gives me a final answer then of thirty-six degrees fifty-two minutes and twelve seconds, for the measure of angle 𝐴𝐷𝐶.

So in this question, we didn’t have a right-angled triangle initially. We had to create one within the isosceles trapezium by drawing in that perpendicular height. And once we had that, it was a case of recalling the cosine ratio using the inverse cosine function in order to work out the angle, and then converting it to seconds because that’s how the answer was asked for.

In summary then, we’ve recalled the cosine ratio as the adjacent divided by the hypotenuse. We’ve seen the inverse cosine function which can be used to work backwards from knowing this ratio, in order to working out the size of an angle. And then we’ve applied it to a couple of problems.