WEBVTT mathematics/trigonometry/murray
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Hi, these are the trigonometry lectures for educator.com, and today we're going to look at some word problems and some applications of triangle trigonometry.
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We're going to be using all the major formulas that we've learned in the previous lectures.
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I hope you remember those very well.
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The master formula which works for right triangles is SOH CAH TOA.
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You can remember that as Some Old Horse Caught Another Horse Taking Oats Away.
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Remember, that only works in a right triangle.
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If you have an angle θ, that relates the sine, cosine and tangent of θ to the hypotenuse, the side opposite θ, and the side adjacent to θ.
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You interpret that as the sin(θ) is equal to opposite over hypotenuse, the cos(θ) is equal to adjacent over hypotenuse, and the tan(θ) is equal to opposite over adjacent.
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The law of sines works in any triangle.
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Let me draw a generic triangle here, doesn't have to be a right triangle for the law of sines to work, so a, b, and c ...
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Generally, you'd label the angles with capital letters, and label the sides with lowercase letters opposite the corner with the same letter.
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That makes this a, this is b, and this is c.
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The law of sines says that sin(A)/a=sin(B)/b=sin(C)/c.
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That's the law of sines.
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Law of cosines also works in any triangle.
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Let me remind you what that one is.
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We had a whole lecture on it earlier, but just to remind you quickly, it says that c²=a²+b²-2abcos(C).
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That's useful when you know all three sides, you can figure out an angle very quickly using the law of cosines.
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Or if you know two sides and the angle in between them, you can figure out that third side using the law of cosines.
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Remember, law of cosines works in any triangle, doesn't have to be a right triangle.
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It still works in right triangles.
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Of course, in right triangles, if C is the right angle, then cos(C) is 0, so the law of cosines just boils down to the Pythagorean formula.
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You can think of the law of cosines as kind of a generalization of the Pythagorean theorem to any triangle, doesn't have to be a right triangle anymore.
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Finally, Heron's formula.
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Heron's formula tells you the area of a triangle when you know the lengths of the three sides.
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Heron says that the area is equal to the square root of s×(s-a)×(s-b)×(s-c).
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The a, b, and c are the lengths of the sides, you're supposed to know what those are before you go into Heron's formula.
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This s I need to explain is the semi-perimeter.
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You add a, b, and c, you get the perimeter of the triangle, then you divide by 2 to get the semi-perimeter.
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That tells you what the s is, then you can drop that into Heron's formula and find the area of the triangle.
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We'll be using all of those, and sometimes it's a little tricky to interpret the words of a problem and figure out which formula you use.
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The real crucial step there is as soon as you get the problem, you want to draw a picture of the triangle involved, and then see which formula works.
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Let's try that out on a few examples and you'll get the hang of it.
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The first example here is a telephone pole that casts a shadow 20 feet long.
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We're told the sun's rays make a 60-degree angle with the ground.
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We're asked how tall is the pole.
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Let me draw that.
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Here's the telephone pole, and we know it casts a shadow, and we know that that shadow is 20 feet long.
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That's a right angle.
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The reasons it casts a shadow is because of these rays coming from the sun.
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There's the sun casting the shadow.
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We want to figure out how tall is the pole.
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We're told that the sun's rays make a 60-degree angle with the ground.
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That means that angle right there is 60 degrees.
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We want to solve for the height of the telephone pole, that's the quantity we want to solve for.
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That's the side opposite the angle that we know.
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We also are given the side adjacent to the angle we know.
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I see opposite and adjacent, and I see a right angle.
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I'm going to use SOH CAH TOA here.
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I know the adjacent side, I'm looking for the opposite side.
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It seems like I should use the tangent formula here.
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Tan(60) is equal to opposite over adjacent.
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Tan(60), 60 is one of those common values, that's π/3.
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I know what the tan(60) is, I've got that memorized and hopefully you do too, square root of 3 is the tan(60).
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If you didn't remember that, I at least hoped that you remember the sin(60) and the cos(60), that tan=sin/cos.
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You can always work out the tangent if you don't remember exactly what the tan(60) is.
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The sin(60) is root 3 over 2, the cos(60) is 1/2, that simplifies down to square root of 3.
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That's what the tan(60) is.
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The opposite, I don't know what that is, I'll just leave that as opposite, but the adjacent side I was given is 20.
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I'll solve this opposite is equal to 20 square root of 3.
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Since this is an applied problem, I'll convert that into a decimal, 20 square root of 3 is approximately 34.6.
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The unit of measurement here is feet, I'll give my answers in terms of feet.
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That tells me that the telephone pole is 34.6 feet tall.
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Let's recap there.
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We were given a word problem, I don't know at first exactly what it's talking about.
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First thing I do is I draw a picture, so I drew a picture of my telephone pole, I drew a picture of the shadow then I noticed that's a right triangle, I could use SOH CAH TOA to solve it.
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I tried to figure out which quantities do I know, which quantities do I not know.
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I knew the adjacent side, I knew the angle, but I didn't know the opposite side and that seemed like it was going to work well with the tangent formula.
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I plugged in what I knew, I solved it down using the common value that I knew, and I got the answer.
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In the next one, we're trying to build a bridge across a lake but we can't figure out how wide the lake is, because we can't just walk across the lake to measure it.
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It says that these engineers measure from a point on land that is 280 feet from one end of the bridge, 160 feet from the other.
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The angle between these two lines is 80 degrees.
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From that, we're supposed to figure out what the bridge will be, or how long the bridge will be.
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Lots of words here, it's a little confusing when you first encounter this because there's just so many words here and there's no picture at well.
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Obviously, the first thing we need to do is to draw a picture.
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I had no idea what shape this lake is really but I'm just going to draw a picture like that.
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I know that these engineers are trying to build a bridge across it.
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Let's say that that's one end of the bridge right there and that's the other end.
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They measure from a point on land that is 280 feet from one end of the bridge and 160 feet from the other.
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It says the angle between these two lines measures 80 degrees.
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If you look at this, what I have is a triangle, and more than that I have two sides and the included angle of a triangle.
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I have a side angle side situation, and I know that that gives me a unique solution as long as my angle is less than 180 degrees.
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Of course, 80 is less than 180 degrees.
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I know I have a unique solution.
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I'm trying to find the length of that third side.
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If you have side angle side and you need the third side, that's definitely the law of cosines.
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I'm going to label that third side little c, and call this capital C, label the other sides a and b.
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Now, the law of cosines is my friend here, c²=a²+b²-2abcos(C).
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We know everything there except for little c.
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I'm just going to plug in the quantities that I know and reduce down and solve for little c.
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Let me plug in c², I don't know that yet, a² is 160², plus b² is 280², minus 2×160×280×cos(C), the angle is 80 degrees.
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I don't know exactly what that is but I can find that on my calculator.
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160²=25600, 280²=78400, 2×160×280, I worked that out as 89600, the cos(80), I'll do that on my calculator ...
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Remember to put your calculator in degree mode if that's what you're using here.
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A lot of people have their calculator set in radian mode, and then that gives you strange answers, because your calculator would be interpreting that as 80 radians.
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It's very important to set your calculator to 80 degrees.
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Cos(80)=0.174, let's see, 25600+78400=104000, 89600×0.174=15559, that's approximate of course, if we simplify that, we get 88441.
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That's c², I'll take the square root of that, c is approximately equal to 297.4.
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Our unit of measurement here is feet, so I'll give my answer in terms of feet.
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Let's recap what made that one work.
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We're given this kind of long paragraph full of words and it's a little hard to discern what we're supposed to be doing.
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First thing we see is, okay, it's a lake, so I drew just a random lake, it's a bridge across a lake, so I drew a picture.
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That's really the key ideas to draw a picture.
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I drew my bridge across the lake here.
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That's the bridge right there.
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It says we measure from a point that is 280 feet from one end of the bridge, and 160 feet from the other.
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I drew that point and I filled in the 160 and the 280.
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Then it gave me the angle between those two lines, so I filled that in.
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All of a sudden, I've got a standard triangle problem, and moreover, I've got a triangle problem where I know two sides and the angle between them, and what I want to find is the third side of the triangle.
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That's definitely a law of cosines problem.
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I write down my law of cosines, I filled in all the quantities that I know then I simplified down and I solved for the answer on that.
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We'll try another example here.
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This time, a farmer measures the fences along the edges of a triangular field as 160, 240 and 380 feet.
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The farmer wants to know what the area of the field is.
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Just like all the others, I'll start out right away by drawing a picture.
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It's a triangular field, my picture's probably not scaled, that doesn't really matter, 160, 240 and 380.
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I have a triangle and I want to figure out what the area is.
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If you have three sides of a triangle and you want to find the area that's pretty much a dead give-away that you want to use Heron's formula.
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Let me remind you what Heron's formula is.
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Heron's formula says the area is equal to the square root of s×(s-a)×(s-b)×(s-c).
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The a, b, and c here just the lengths of the three sides but this s is the semi-perimeter of the triangle.
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That's 1/2 of the perimeter a+b+c.
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The perimeter is the distance around if you kind of walk all the way around this triangle.
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That's (1/2)(160+240+300), 160+240=400, plus 300 is 700, 1/2 of that is 350, so s is 350.
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I just drop the s and the three sides into Heron's formula.
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That's 350×(350-160)×(350-240)×(350-300), (350-160)=190, (350-240)=110, and (350-300)=50.
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I can pull 100 out of that immediately, so this is 100 times the square root of 35×19×11×5.
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I'm going to go to my calculator for that, 35×19×11×5=36575, square root of that is 191, times 100 is 19.1, rounds to 25.
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This is the area of a field, the units are squared here, and we were talking in terms of feet, so this must be square feet.
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Let's recap how you approach that problem.
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First of all, you read the words and right away you go to draw a picture, I see that I have a triangular field, the edges are 160, 240 and 300 feet.
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I'm asked for the area of the field.
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Now, once you have the three sides and you want the area, there's pretty much no question that you want to use Heron's formula on that.
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That's the formula that tells you the area based on the three sides very quickly.
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You write down Heron's formula, that's got a, b and c there.
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It's also got this s, the s is the semi-perimeter.
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You figure that out from the three sides.
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You plug that into Heron's formula and you plug in the three sides.
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The a, b and c are 160, 240 and 300.
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Then, it's just a matter of simplifying down the numbers until you get an answer and figuring out that the units have to be square feet, because the original measurements in the problem were in terms of feet, and we're describing an area now, it must be square feet.
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We'll try some more examples later.