We're now going to do a few examples to actually show you why the trig functions are actually useful. So let's get started with a problem. Let's say I have this right triangle. That's my right triangle. There's the right angle. And let's say I know that the measure of this angle is pi over 4 radians. And I'll just write rad for short. If the measure of this angle is pi over 4 radians, and I also know that this side of the triangle-- this side right here-- is 10 square roots of 2. So I know this side of the triangle. I know this angle, which is pi over 4 radians. And now, the question is, what is this side of the triangle? I'm going to highlight that. And let me make it in orange. So let's figure out what we know and what we need to figure out. We know the angle, pi over 4 radians. And actually, turns out if you were to convert that to degrees, it would be 45 degrees. And we know-- what side is this? This is the hypotenuse of the triangle, right? And what are we trying to figure out? Are we trying to figure out the hypotenuse, the adjacent side to the angle, or the opposite side to the angle? Well, this is the hypotenuse, we already know that. This is the opposite side. This is the opposite side. And this yellow side is the adjacent side, right? It's just adjacent to this angle. So we know the angle, we know the hypotenuse, and we want to figure out the adjacent side. So let me ask you a question. What trig function deals with the adjacent side and the hypotenuse? Because we have the adjacent side is what we want to figure out, and we know the hypotenuse. Well, let's write down our mnemonic, just in case you forgot it. SOHCAHTOA. So which one uses adjacent and hypotenuse? Right? It's CAH. And CAH, the c is for what? The c is for cosine. Cosine of an angle-- let's just call it any angle-- is equal to the adjacent over the hypotenuse. So let's use this information to try to solve for this orange side, or this yellow side. So we know that cosine of pi over 4 radians-- so let's say cosine of pi over 4-- must equal this adjacent side right here. Let's just call that a. a for adjacent. The adjacent side divided by the hypotenuse. The hypotenuse is this side. And in the problem, we were given that it's 10 square roots of 2. So we can solve for a by multiplying both sides of this equation by 10 square roots of 2. And we will get-- because, right? If we just multiply times 10 square root of 2, these cancel out. And then you get a 10 square root of 2 here. So you get a is equal to 10 square roots of 2 times the cosine of pi over 4. Now you're probably saying, Sal, this does not look too simple, and I don't know how big the cosine of pi over 4 is. What do I do? Well, no one has the trig functions, or the values of the trig functions memorized. There's a couple of ways to do it. Either I could give you what the cosine of pi over 4 is. That's sometimes given in a problem. Or you can make sure that your calculator is set to radians and you can just type in pi divided by 4-- which is roughly 0.79-- and then press the cosine button. You finally know what it's good for. And you'll get a value. Or-- and this is kind of the old school way of doing it-- there are trig tables where you could look up what cosine of pi over 4 is in a table. Since I don't have any of that at my disposal right now, I'll just tell you what the cosine of pi over 4 is. The cosine of pi over 4 is square root of 2 over 2. So a, which is the adjacent side-- a for adjacent-- is equal to 10 square roots of 2 times square root of 2 over 2. Remember, to get the square root of 2 over 2, you might be a little confused. You're like, how did Sal get that? All I said is, the cosine of pi over 4 is square root of 2 over 2. And that's not something that-- well, actually, this one you might know offhand, because of the 45 degree angle. But this isn't something that people memorize. This is something you would look up, or it's given in the problem, or you'd use a calculator for. And a calculator, of course, wouldn't give you square root of 2 over 2. It'd give you a decimal number that's not obviously square root of 2 over 2. But anyway, I told you that the cosine of pi over 4 is the square root of 2 over 2. And so if we multiply, what's the square root of 2 over 2? What's the square root of 2 times the square root of 2? It's 2. So that's 2, and then that cancels with that 2. And so everything cancels except for the 10. So the adjacent side is equal to 10. Let's do another one. Let me delete this. Give me 1 second. I'm actually-- this is one of the few modules that I'm not generating the problems on the fly, because I need to make sure that I actually have the trig function values before I do the problem. So let's say I have another right triangle. I probably shouldn't have deleted that last one. So let's see, this is my right triangle. How much time do I have-- about 4 minutes left. Should be enough. So this is my right triangle. And I know the angle-- let's call this--. I know this angle right here is 0.54 radians. And I also know that this side right here is 3 units long. And I want to figure out this side. So what do I know? Well, this side is what side relative to the angle? It's the opposite side, right? Because the angle is here, and we go opposite the angle. So this is the opposite side. And what's this side? Is this the adjacent side, or is it the hypotenuse? Well, this is the hypotenuse, right? The long side, and it's opposite the right angle. So this is the adjacent side. So what trig function uses opposite and adjacent? Let's write down SOHCAHTOA again. SOHCAHTOA. TOA uses opposite and adjacent. OA. So T for tangent, right? TOA. So tangent is equal to opposite over adjacent. So let's use that. So let's take the tangent of 0.54 radians. So the tangent of 0.54 will equal the side opposite to it. So that's 3, right? The opposite side is 3. Over the adjacent side. Well, once again, the adjacent side is what we don't know. So we have to solve for a. So if we multiply both sides by a, we get a tan of 0.54-- we could do that because we know it's not 0-- equals 3. Or a is equal to 3 divided by the tangent of 0.54. So once again, I don't have memorized what the tangent of 0.54 is, but I will tell you what it is because you also don't have it memorized. Or you could use a calculator to figure it out if you had a radian function. The tangent of 0.54 is equal to-- let me make sure I have this right. Oh, right. The tangent of 0.54 is 3/5. So then a is equal to 3 over 3/5. Right, the adjacent side-- now, once again, how did I get this 3/5? Well, I just told you. Or you can use a calculator to know that the tangent of 0.54 is 3/5. And of course, I'm using numbers that work out well, just so that the fractions all cancel. So we know that the adjacent side is equal to-- when you divide by fractions, it's like multiplying by the numerator. Multiplying by the inverse. So times 5/3. So the adjacent side is equal to 5. There. There you go. So let's just think about what I always do. I think about what I have, what sides I have, and what side I want to solve for. And in this case, it was the opposite side I had, and I wanted to solve for the adjacent side. And I said, what trig function involves those 2 sides? The opposite and the adjacent. I wrote down SOHCAHTOA. I said, oh, TOA. Opposite and adjacent. That's tan. So I took the tan of the angle. And then I said, the tan of the angle is equal to the opposite side divided by the adjacent side. That's right here. And then I just solved for the adjacent side. And of course, I used a calculator, or I told you what the tangent of 0.54 is. I think I'll do a couple more of these problems in the next module, but I'm out of time for now. Have fun.