Learn about Trigonometry word problems - part 1 Video

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Learn about Trigonometry word problems - part 1

Now, we know some Trigonometry, let’s use that Trigonometry to solve some word problems. So, let’s get started with—this is exciting, this is a navigation problem. So, I have a ship and this is to go from point A to point B. So, let me draw that so I’m going from point A which is that point to point B with our nice big, not so big triangle, big enough. There you go, because I need space to do my Math. So, this is point A. I will draw and move point A, point B. So this ship, this is to go from point A to point B and this problem tells us that the distance between point A and point B is 10 kilometers. So, this guy set sail and because of whatever reason, he’s a bad navigator or the water is flowing faster maybe upwards, maybe the water is flowing in this direction, let me draw that in blue to show which direction of the water. Let’s say the current is flowing in that direction and he ends up off track. He actually ends up after five kilometers so he travels for five kilometers, we do that in blue green. It travels for five kilometers and after five kilometers, he realizes that he has gone off track. And somehow I don’t know it maybe using—I don’t know if you know what that is but that’s what people used when they’re on ships to figure out where they are I think. He realizes that he has 15 degrees off course so he is 15 degrees off course. So, what that means is that between the path he took and the path he should have took is 15 degrees. At least that’s my interpretation of it. So, this is 15 degrees. And of course he has traveled five kilometers in the wrong direction, 15 degrees in the wrong direction. So what he needs to know is—so this, let me draw a little ship so this is where the ship is now, that’s my ship. So now, this poor captain he said, “How far do I still have to go to get to point B?” So, he needs to know how far is this distance right here? How far is this distance? This may just involve some Trigonometry. So, how can we figure out this distance? Well, let’s just breakdown what we know and that’s how I do trick problems, I just keep messing around within until I get the right answer. So, let’s see if we can tackle this problem that way. So, what can we figure out? We know that is five kilometers, this side is five kilometers. We know that this is 10 kilometers, we know this angle. Well, one thing I know I can figure out is—let’s just make some right angle so we can start using some Trigonometry so let me draw a right angle here. I just draw up at straight down. So, can I figure out what this side is right here? Can I figure out what this side right here is? Well, this side is going to be what? If we look at this angle, this is adjacent to the angle and this is the hypotenuse. So, if we know the hypotenuse, we know the angle and we want to figure out the adjacent, what trig function should we use? Let me write out down our Socatoa. So, we know the hypotenuse, we want to figure out the adjacent, so what involves adjacent and hypotenuse? Co or cosine. So, we know that the cosine of 15 degrees, so I’m going to write over this just not to space. Cosine of 15 degrees is equal to this brown side so let’s just call this I don’t know, I’m just going to call it x. So, this is equal to x over the hypotenuse, over five. And if we solved for x, we get x is equal to five cosine of 15 degrees. Maybe we made progress maybe we didn’t, let’s keep trying. Now, let’s see if we could figure out this side of this triangle. Well, to this angle this is the opposite and we know the hypotenuse so what trig identity should we order or trig function should we use? Well, if we know the opposite or if we want to figure out the opposite and we know the hypotenuse and we knew the angle, so what trig function deals with opposite and the hypotenuse? Well, that’s sine. So the sine, as we call this y, we know that y is equal to—well, doing it the same way sine, five sine of 15 degrees. I skip the step right because we could say that sine of 15 degrees is equal to the opposite y over the hypotenuse over five. And then that step takes us here, we just multiplied both sides by five and you get y is equal to five sine of 15 degrees. So, that’s pretty cool. We know this y, we know this x, can we figure out what this like this? Let me draw it and yet another color. Can we figure out what the length of this side is? Well, we know from the problems that this whole length is 10. We know this whole length is 10 and we know that this little part of it is x which is we figured out is five cosine of 15 degrees. So, let’s call this as z. We know that z is equal to 10 minus x because this whole thing is 10, this thing is x so it’s using with 10 minus x so we are to figure out what x is equal to, z is equal to 10 minus this, that’s what x is. You don’t even have to use a variable x, we could have just written it like that. So, 10 minus five cosine of 15 degrees, that’s this side, c is equal to 10 minus five cosine of 15 degrees. So, we know z, we know y. This is a right triangle and we’re looking fro this yellow side whatever we want to call it. That’s the answer to the problem. Well, this is trying to look easy; this is just a Pythagorean Theorem. So, y squared plus z squared is going to be equal to our, let’s call this m, picking an arbitrary letter. Let’s call that m, so y2 plus z2 is going to be equal to m2. So, we could just say, let’s just write that down, I don’t know why I picked m really just to confuse you I think. So, m2 is equal to—but what’s y2? Y2 is five sine of 15 degrees so it equals, five sine 15 degrees squared plus 10 minus five cosine of 15 degrees squared. And now, I just have to simplify this. So, if you have a calculator, I mean you could do this without any simplification but I want to get it as simple as I can. So, what’s this squared? Well, this is equal to—let me draw a line here because I don’t want to get too messy but I need all of this space. So, this is equal to—and remember this is m2, we’re going to have to take the square root of all this at the end before you solve for m. This is equal to 25 sine squared 15 degrees or the sine of 15 degrees squared that’s just how you write it. And then we do a little bit of foil here to expand this so plus 100 minus 100 cosine of 15 degrees plus 25 cosine squared of 15 degrees. All I did is I expand it. I said, “Well, this whole expression square is equal to this squared minus two times the two things multiplied out so that’s 100 cosine of 15 degrees and then I just squared this last term which is plus 25 cosine of 15 degrees.” If that confuse you, you might to review the multiplying expressions. So, let’s see if we can simplify this further. So, if we take this term and this term, we could simplify that to 25 times sine squared of 15 degrees plus cosine squared of 15. And you could kind of skip all this and just use a calculator and figure out this exact value but I’m just going to keep working on it just because I like to get it as simple as possible before I use the calculator. So, that term and that term are that and then it’s plus the stuff plus 100 minus 100 cosine of 15 degrees. And then what is the sine squared of 15 times plus the cosine squared of 15? That’s one of our basic identities, right? That’s the Pythagorean identity and sine squared of x plus cosine squared of x or of data is just one, so this term becomes one. And I’m running out of time so I’ll continue in the next video.

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Now, we know some Trigonometry, let’s use that Trigonometry to solve some word problems. So, let’s get started with—this is exciting, this is a navigation problem. So, I have a ship and this is to go from point A to point B. So, let me draw that so I’m going from point A which is that point to point B with our nice big, not so big triangle, big enough. There you go, because I need space to do my Math. So, this is point A. I will draw and move point A, point B. So this ship, this is to go from point A to point B and this problem tells us that the distance between point A and point B is 10 kilometers.

So, this guy set sail and because of whatever reason, he’s a bad navigator or the water is flowing faster maybe upwards, maybe the water is flowing in this direction, let me draw that in blue to show which direction of the water. Let’s say the current is flowing in that direction and he ends up off track. He actually ends up after five kilometers so he travels for five kilometers, we do that in blue green. It travels for five kilometers and after five kilometers, he realizes that he has gone off track. And somehow I don’t know it maybe using—I don’t know if you know what that is but that’s what people used when they’re on ships to figure out where they are I think. He realizes that he has 15 degrees off course so he is 15 degrees off course. So, what that means is that between the path he took and the path he should have took is 15 degrees. At least that’s my interpretation of it. So, this is 15 degrees. And of course he has traveled five kilometers in the wrong direction, 15 degrees in the wrong direction.

So what he needs to know is—so this, let me draw a little ship so this is where the ship is now, that’s my ship. So now, this poor captain he said, “How far do I still have to go to get to point B?” So, he needs to know how far is this distance right here? How far is this distance? This may just involve some Trigonometry. So, how can we figure out this distance? Well, let’s just breakdown what we know and that’s how I do trick problems, I just keep messing around within until I get the right answer. So, let’s see if we can tackle this problem that way.

So, what can we figure out? We know that is five kilometers, this side is five kilometers. We know that this is 10 kilometers, we know this angle. Well, one thing I know I can figure out is—let’s just make some right angle so we can start using some Trigonometry so let me draw a right angle here. I just draw up at straight down. So, can I figure out what this side is right here? Can I figure out what this side right here is? Well, this side is going to be what? If we look at this angle, this is adjacent to the angle and this is the hypotenuse. So, if we know the hypotenuse, we know the angle and we want to figure out the adjacent, what trig function should we use? Let me write out down our Socatoa.

So, we know the hypotenuse, we want to figure out the adjacent, so what involves adjacent and hypotenuse? Co or cosine. So, we know that the cosine of 15 degrees, so I’m going to write over this just not to space. Cosine of 15 degrees is equal to this brown side so let’s just call this I don’t know, I’m just going to call it x. So, this is equal to x over the hypotenuse, over five. And if we solved for x, we get x is equal to five cosine of 15 degrees. Maybe we made progress maybe we didn’t, let’s keep trying. Now, let’s see if we could figure out this side of this triangle. Well, to this angle this is the opposite and we know the hypotenuse so what trig identity should we order or trig function should we use? Well, if we know the opposite or if we want to figure out the opposite and we know the hypotenuse and we knew the angle, so what trig function deals with opposite and the hypotenuse? Well, that’s sine. So the sine, as we call this y, we know that y is equal to—well, doing it the same way sine, five sine of 15 degrees. I skip the step right because we could say that sine of 15 degrees is equal to the opposite y over the hypotenuse over five. And then that step takes us here, we just multiplied both sides by five and you get y is equal to five sine of 15 degrees. So, that’s pretty cool. We know this y, we know this x, can we figure out what this like this? Let me draw it and yet another color. Can we figure out what the length of this side is?

Well, we know from the problems that this whole length is 10. We know this whole length is 10 and we know that this little part of it is x which is we figured out is five cosine of 15 degrees. So, let’s call this as z. We know that z is equal to 10 minus x because this whole thing is 10, this thing is x so it’s using with 10 minus x so we are to figure out what x is equal to, z is equal to 10 minus this, that’s what x is. You don’t even have to use a variable x, we could have just written it like that. So, 10 minus five cosine of 15 degrees, that’s this side, c is equal to 10 minus five cosine of 15 degrees. So, we know z, we know y. This is a right triangle and we’re looking fro this yellow side whatever we want to call it. That’s the answer to the problem. Well, this is trying to look easy; this is just a Pythagorean Theorem. So, y squared plus z squared is going to be equal to our, let’s call this m, picking an arbitrary letter. Let’s call that m, so y2 plus z2 is going to be equal to m2. So, we could just say, let’s just write that down, I don’t know why I picked m really just to confuse you I think. So, m2 is equal to—but what’s y2? Y2 is five sine of 15 degrees so it equals, five sine 15 degrees squared plus 10 minus five cosine of 15 degrees squared. And now, I just have to simplify this. So, if you have a calculator, I mean you could do this without any simplification but I want to get it as simple as I can.

So, what’s this squared? Well, this is equal to—let me draw a line here because I don’t want to get too messy but I need all of this space. So, this is equal to—and remember this is m2, we’re going to have to take the square root of all this at the end before you solve for m. This is equal to 25 sine squared 15 degrees or the sine of 15 degrees squared that’s just how you write it. And then we do a little bit of foil here to expand this so plus 100 minus 100 cosine of 15 degrees plus 25 cosine squared of 15 degrees. All I did is I expand it. I said, “Well, this whole expression square is equal to this squared minus two times the two things multiplied out so that’s 100 cosine of 15 degrees and then I just squared this last term which is plus 25 cosine of 15 degrees.” If that confuse you, you might to review the multiplying expressions. So, let’s see if we can simplify this further.

So, if we take this term and this term, we could simplify that to 25 times sine squared of 15 degrees plus cosine squared of 15. And you could kind of skip all this and just use a calculator and figure out this exact value but I’m just going to keep working on it just because I like to get it as simple as possible before I use the calculator. So, that term and that term are that and then it’s plus the stuff plus 100 minus 100 cosine of 15 degrees. And then what is the sine squared of 15 times plus the cosine squared of 15? That’s one of our basic identities, right? That’s the Pythagorean identity and sine squared of x plus cosine squared of x or of data is just one, so this term becomes one. And I’m running out of time so I’ll continue in the next video.

Transcription by: Scribe4you Transcription Services