WEBVTT mathematics/pre-calculus/selhorst-jones
00:00:00.000 --> 00:00:09.000
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.
00:00:09.000 --> 00:00:14.000
We're going to be using all the major formulas that we've learned in the previous lectures.
00:00:14.000 --> 00:00:16.000
I hope you remember those very well.
00:00:16.000 --> 00:00:21.000
The master formula which works for right triangles is SOH CAH TOA.
00:00:21.000 --> 00:00:25.000
You can remember that as Some Old Horse Caught Another Horse Taking Oats Away.
00:00:25.000 --> 00:00:30.000
Remember, that only works in a right triangle.
00:00:30.000 --> 00:00:43.000
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.
00:00:43.000 --> 00:00:56.000
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.
00:00:56.000 --> 00:01:00.000
The law of sines works in any triangle.
00:01:00.000 --> 00:01:11.000
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 ...
00:01:11.000 --> 00:01:22.000
Generally, you'd label the angles with capital letters, and label the sides with lowercase letters opposite the corner with the same letter.
00:01:22.000 --> 00:01:26.000
That makes this a, this is b, and this is c.
00:01:26.000 --> 00:01:46.000
The law of sines says that sin(A)/a=sin(B)/b=sin(C)/c.
00:01:46.000 --> 00:01:48.000
That's the law of sines.
00:01:48.000 --> 00:01:52.000
Law of cosines also works in any triangle.
00:01:52.000 --> 00:01:54.000
Let me remind you what that one is.
00:01:54.000 --> 00:02:09.000
We had a whole lecture on it earlier, but just to remind you quickly, it says that c²=a²+b²-2abcos(C).
00:02:09.000 --> 00:02:14.000
That's useful when you know all three sides, you can figure out an angle very quickly using the law of cosines.
00:02:14.000 --> 00:02:22.000
Or if you know two sides and the angle in between them, you can figure out that third side using the law of cosines.
00:02:22.000 --> 00:02:27.000
Remember, law of cosines works in any triangle, doesn't have to be a right triangle.
00:02:27.000 --> 00:02:29.000
It still works in right triangles.
00:02:29.000 --> 00:02:38.000
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.
00:02:38.000 --> 00:02:46.000
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.
00:02:46.000 --> 00:02:53.000
Finally, Heron's formula.
00:02:53.000 --> 00:02:58.000
Heron's formula tells you the area of a triangle when you know the lengths of the three sides.
00:02:58.000 --> 00:03:13.000
Heron says that the area is equal to the square root of s×(s-a)×(s-b)×(s-c).
00:03:13.000 --> 00:03:19.000
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.
00:03:19.000 --> 00:03:25.000
This s I need to explain is the semi-perimeter.
00:03:25.000 --> 00:03:32.000
You add a, b, and c, you get the perimeter of the triangle, then you divide by 2 to get the semi-perimeter.
00:03:32.000 --> 00:03:38.000
That tells you what the s is, then you can drop that into Heron's formula and find the area of the triangle.
00:03:38.000 --> 00:03:45.000
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.
00:03:45.000 --> 00:03:57.000
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.
00:03:57.000 --> 00:04:01.000
Let's try that out on a few examples and you'll get the hang of it.
00:04:01.000 --> 00:04:07.000
The first example here is a telephone pole that casts a shadow 20 feet long.
00:04:07.000 --> 00:04:10.000
We're told the sun's rays make a 60-degree angle with the ground.
00:04:10.000 --> 00:04:12.000
We're asked how tall is the pole.
00:04:12.000 --> 00:04:14.000
Let me draw that.
00:04:14.000 --> 00:04:22.000
Here's the telephone pole, and we know it casts a shadow, and we know that that shadow is 20 feet long.
00:04:22.000 --> 00:04:25.000
That's a right angle.
00:04:25.000 --> 00:04:34.000
The reasons it casts a shadow is because of these rays coming from the sun.
00:04:34.000 --> 00:04:38.000
There's the sun casting the shadow.
00:04:38.000 --> 00:04:41.000
We want to figure out how tall is the pole.
00:04:41.000 --> 00:04:44.000
We're told that the sun's rays make a 60-degree angle with the ground.
00:04:44.000 --> 00:04:48.000
That means that angle right there is 60 degrees.
00:04:48.000 --> 00:04:55.000
We want to solve for the height of the telephone pole, that's the quantity we want to solve for.
00:04:55.000 --> 00:05:03.000
That's the side opposite the angle that we know.
00:05:03.000 --> 00:05:05.000
We also are given the side adjacent to the angle we know.
00:05:05.000 --> 00:05:09.000
I see opposite and adjacent, and I see a right angle.
00:05:09.000 --> 00:05:15.000
I'm going to use SOH CAH TOA here.
00:05:15.000 --> 00:05:25.000
I know the adjacent side, I'm looking for the opposite side.
00:05:25.000 --> 00:05:30.000
It seems like I should use the tangent formula here.
00:05:30.000 --> 00:05:39.000
Tan(60) is equal to opposite over adjacent.
00:05:39.000 --> 00:05:46.000
Tan(60), 60 is one of those common values, that's π/3.
00:05:46.000 --> 00:05:53.000
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).
00:05:53.000 --> 00:06:04.000
If you didn't remember that, I at least hoped that you remember the sin(60) and the cos(60), that tan=sin/cos.
00:06:04.000 --> 00:06:08.000
You can always work out the tangent if you don't remember exactly what the tan(60) is.
00:06:08.000 --> 00:06:17.000
The sin(60) is root 3 over 2, the cos(60) is 1/2, that simplifies down to square root of 3.
00:06:17.000 --> 00:06:20.000
That's what the tan(60) is.
00:06:20.000 --> 00:06:27.000
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.
00:06:27.000 --> 00:06:34.000
I'll solve this opposite is equal to 20 square root of 3.
00:06:34.000 --> 00:06:51.000
Since this is an applied problem, I'll convert that into a decimal, 20 square root of 3 is approximately 34.6.
00:06:51.000 --> 00:07:00.000
The unit of measurement here is feet, I'll give my answers in terms of feet.
00:07:00.000 --> 00:07:08.000
That tells me that the telephone pole is 34.6 feet tall.
00:07:08.000 --> 00:07:10.000
Let's recap there.
00:07:10.000 --> 00:07:13.000
We were given a word problem, I don't know at first exactly what it's talking about.
00:07:13.000 --> 00:07:27.000
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.
00:07:27.000 --> 00:07:31.000
I tried to figure out which quantities do I know, which quantities do I not know.
00:07:31.000 --> 00:07:41.000
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.
00:07:41.000 --> 00:07:49.000
I plugged in what I knew, I solved it down using the common value that I knew, and I got the answer.
00:07:49.000 --> 00:08:01.000
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.
00:08:01.000 --> 00:08:10.000
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.
00:08:10.000 --> 00:08:14.000
The angle between these two lines is 80 degrees.
00:08:14.000 --> 00:08:18.000
From that, we're supposed to figure out what the bridge will be, or how long the bridge will be.
00:08:18.000 --> 00:08:24.000
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.
00:08:24.000 --> 00:08:33.000
Obviously, the first thing we need to do is to draw a picture.
00:08:33.000 --> 00:08:43.000
I had no idea what shape this lake is really but I'm just going to draw a picture like that.
00:08:43.000 --> 00:08:49.000
I know that these engineers are trying to build a bridge across it.
00:08:49.000 --> 00:08:52.000
Let's say that that's one end of the bridge right there and that's the other end.
00:08:52.000 --> 00:09:11.000
They measure from a point on land that is 280 feet from one end of the bridge and 160 feet from the other.
00:09:11.000 --> 00:09:17.000
It says the angle between these two lines measures 80 degrees.
00:09:17.000 --> 00:09:26.000
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.
00:09:26.000 --> 00:09:36.000
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.
00:09:36.000 --> 00:09:40.000
Of course, 80 is less than 180 degrees.
00:09:40.000 --> 00:09:45.000
I know I have a unique solution.
00:09:45.000 --> 00:09:50.000
I'm trying to find the length of that third side.
00:09:50.000 --> 00:09:55.000
If you have side angle side and you need the third side, that's definitely the law of cosines.
00:09:55.000 --> 00:10:05.000
I'm going to label that third side little c, and call this capital C, label the other sides a and b.
00:10:05.000 --> 00:10:23.000
Now, the law of cosines is my friend here, c²=a²+b²-2abcos(C).
00:10:23.000 --> 00:10:26.000
We know everything there except for little c.
00:10:26.000 --> 00:10:34.000
I'm just going to plug in the quantities that I know and reduce down and solve for little c.
00:10:34.000 --> 00:10:58.000
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.
00:10:58.000 --> 00:11:04.000
I don't know exactly what that is but I can find that on my calculator.
00:11:04.000 --> 00:11:32.000
160²=25600, 280²=78400, 2×160×280, I worked that out as 89600, the cos(80), I'll do that on my calculator ...
00:11:32.000 --> 00:11:34.000
Remember to put your calculator in degree mode if that's what you're using here.
00:11:34.000 --> 00:11:44.000
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.
00:11:44.000 --> 00:11:50.000
It's very important to set your calculator to 80 degrees.
00:11:50.000 --> 00:12:41.000
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.
00:12:41.000 --> 00:12:58.000
That's c², I'll take the square root of that, c is approximately equal to 297.4.
00:12:58.000 --> 00:13:11.000
Our unit of measurement here is feet, so I'll give my answer in terms of feet.
00:13:11.000 --> 00:13:14.000
Let's recap what made that one work.
00:13:14.000 --> 00:13:20.000
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.
00:13:20.000 --> 00:13:28.000
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.
00:13:28.000 --> 00:13:30.000
That's really the key ideas to draw a picture.
00:13:30.000 --> 00:13:33.000
I drew my bridge across the lake here.
00:13:33.000 --> 00:13:37.000
That's the bridge right there.
00:13:37.000 --> 00:13:45.000
It says we measure from a point that is 280 feet from one end of the bridge, and 160 feet from the other.
00:13:45.000 --> 00:13:51.000
I drew that point and I filled in the 160 and the 280.
00:13:51.000 --> 00:13:55.000
Then it gave me the angle between those two lines, so I filled that in.
00:13:55.000 --> 00:14:07.000
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.
00:14:07.000 --> 00:14:11.000
That's definitely a law of cosines problem.
00:14:11.000 --> 00:14:22.000
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.
00:14:22.000 --> 00:14:24.000
We'll try another example here.
00:14:24.000 --> 00:14:33.000
This time, a farmer measures the fences along the edges of a triangular field as 160, 240 and 380 feet.
00:14:33.000 --> 00:14:36.000
The farmer wants to know what the area of the field is.
00:14:36.000 --> 00:14:45.000
Just like all the others, I'll start out right away by drawing a picture.
00:14:45.000 --> 00:14:55.000
It's a triangular field, my picture's probably not scaled, that doesn't really matter, 160, 240 and 380.
00:14:55.000 --> 00:15:00.000
I have a triangle and I want to figure out what the area is.
00:15:00.000 --> 00:15:07.000
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.
00:15:07.000 --> 00:15:10.000
Let me remind you what Heron's formula is.
00:15:10.000 --> 00:15:29.000
Heron's formula says the area is equal to the square root of s×(s-a)×(s-b)×(s-c).
00:15:29.000 --> 00:15:39.000
The a, b, and c here just the lengths of the three sides but this s is the semi-perimeter of the triangle.
00:15:39.000 --> 00:15:45.000
That's 1/2 of the perimeter a+b+c.
00:15:45.000 --> 00:15:49.000
The perimeter is the distance around if you kind of walk all the way around this triangle.
00:15:49.000 --> 00:16:08.000
That's (1/2)(160+240+300), 160+240=400, plus 300 is 700, 1/2 of that is 350, so s is 350.
00:16:08.000 --> 00:16:16.000
I just drop the s and the three sides into Heron's formula.
00:16:16.000 --> 00:16:54.000
That's 350×(350-160)×(350-240)×(350-300), (350-160)=190, (350-240)=110, and (350-300)=50.
00:16:54.000 --> 00:17:06.000
I can pull 100 out of that immediately, so this is 100 times the square root of 35×19×11×5.
00:17:06.000 --> 00:17:40.000
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.
00:17:40.000 --> 00:18:01.000
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.
00:18:01.000 --> 00:18:06.000
Let's recap how you approach that problem.
00:18:06.000 --> 00:18:19.000
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.
00:18:19.000 --> 00:18:21.000
I'm asked for the area of the field.
00:18:21.000 --> 00:18:29.000
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.
00:18:29.000 --> 00:18:35.000
That's the formula that tells you the area based on the three sides very quickly.
00:18:35.000 --> 00:18:38.000
You write down Heron's formula, that's got a, b and c there.
00:18:38.000 --> 00:18:42.000
It's also got this s, the s is the semi-perimeter.
00:18:42.000 --> 00:18:46.000
You figure that out from the three sides.
00:18:46.000 --> 00:18:53.000
You plug that into Heron's formula and you plug in the three sides.
00:18:53.000 --> 00:18:57.000
The a, b and c are 160, 240 and 300.
00:18:57.000 --> 00:19:13.000
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.
00:19:13.000 --> 00:34:25.000
We'll try some more examples later.