WEBVTT mathematics/trigonometry/murray
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We're working on the trigonometry lectures, and today we're talking about finding the area of a triangle.
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We've learned several important formulas over the past few lectures.
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Today, we'll be combining them all and learning a few different methods to find the area of a triangle depending on what kind of initial data you're given.
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I want to remind you about some very important formulas, first of all the master formula that works for right triangles is SOH CAH TOA.
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Let me remind you how that works.
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It only works in a right triangle.
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You have to be very careful of that.
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You have to have one right angle.
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If you talk about one of the other angles, θ, then you label all the sides as the hypotenuse, the side opposite θ and the side adjacent θ.
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SOH CAH TOA stands for the sin(θ) is equal to the length of the opposite side over the hypotenuse, the cos(θ) is the length of the adjacent side over the hypotenuse, the tan(θ) is equal to the opposite side over the adjacent side.
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Remember, we have a little mnemonic to remember that, if you can't remember SOH CAH TOA, you remember Some Old Horse Caught Another Horse Taking Oats Away.
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Now, that'll help you spell out SOH CAH TOA.
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Remember, SOH CAH TOA only works in right triangles, you have to have a right angle to make that work.
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The law of cosines works in any triangle, I'll remind you how that goes.
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We're assuming here that your sides are labeled a, b, and c, little a, little b, and little c, then you label the angles with capital letters opposite the sides with the same letter, that would make this capital A, capital B, and this, capital C.
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The law of cosines relates the lengths of the three sides, little a, b and c, to the measure of one of the angles, c²=a²+b²-2abcos(C).
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This works in any triangle, it does not have to be a right triangle.
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In fact, if it happens to be a right triangle then the cos(C), if C is a right angle, the cosine of C is just 0, that whole term drops out and you end up the Pythagorean theorem for right triangle, c²=a²+b².
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Finally, the important formula that we're going to be using for areas is Heron's formula.
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That's very useful when you know all the three sides, a, b and c, of a triangle ABC.
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First, you work out this quantity s, the semi-perimeter, where you add up a, b and c, that's the perimeter, divide by 2, so you get the semi-perimeter, then you drop that into this area formula, and you drop in the lengths of all three sides.
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Heron's formula gives you a nice expression for the area of a triangle without ever having to look at the angles at all.
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That's very useful as well.
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We'll practice combining all those formulas in different combinations and see how we can calculate the areas of triangles in various ways.
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The first example, we're given a triangle that has two sides of length 8 and 12 with an included angle of 45 degrees.
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Let me set that up.
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That's 8, that's 12, and this is 45 degrees.
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I'll draw in the third side there.
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I'm not going to try to use Heron's formula yet, I'm going to try to use old-fashioned SOH CAH TOA here.
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The thing is, remember, SOH CAH TOA only works in right triangles.
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I don't necessarily have a right triangle here.
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What I'm going to do is draw an altitude, drop a perpendicular from the top corner of that triangle.
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Now I do have a right triangle.
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I'm going to try and find the length of that altitude, I'm going to use SOH CAH TOA.
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Sin(θ) is equal to the opposite over the hypotenuse.
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Sin(45) is equal to the opposite, now the hypotenuse of that little triangle is 8, so root 2 over 2 is equal to the opposite over 8.
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That's because I know the sin(45), that's one of my common values, that's π/4.
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I memorize the sine, cosine and tangent of all the common values way back earlier in the course.
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If you haven't memorized that, you really should commit all those common values 45 and multiples of 30 to memory.
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If I solve for the opposite here, I get that the length of the opposite is equal to 8 root 2 over 2, that's 4 root 2.
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That means that that altitude is 4 root 2.
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Now we can use the old formula from geometry for the area of a triangle, just 1/2 base times height.
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The h stands there for height instead of hypotenuse.
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I know that's a little confusing to be using h for two different things, but we're kind of stuck with that in English that hypotenuse and height both start with the same letter.
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One-half the base here is 12, and the height we figured out was 4 square root of 2.
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We multiply those out, that's 6 times 4 square root of 2, that's 24 square root of 2 for my area there.
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That one came down to drawing an altitude in the triangle and then using SOH CAH TOA.
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We didn't really have to use anything fancy like the law of cosines or Heron's formula, although we could have.
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You'll see some examples later where we use the law of cosines and Heron's formula instead.
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In this one, we just drew this altitude, we used SOH CAH TOA to find the length of the altitude, then we used the old-fashioned geometry formula, 1/2 base times height, to get the formula of the triangle.
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In the next example, we're given a triangle with side lengths 10, 14 and 16.
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Let me draw that.
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10, 14, and 16 ...
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We're asked to find the area.
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If you have all three sides of a triangle and you're asked to find the area, it's kind of a give-away that you're going to use Heron's formula.
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Because Heron's formula works very nicely based on the side lengths of the triangle only.
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You never even have to figure out what the angles are.
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That's what we're going to use.
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Heron, remember, says that you have to start out by finding the semi-perimeter.
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That's (1/2)×(a+b+c), that's (1/2)×(10+14+16).
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(10+14+16) is 40, a half of that is 20.
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Now I know what s is.
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Heron says I plug that into my area formula which is this big square root expression of s times s minus a, s-minus b, and s minus c.
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My s was 20, a is 10, b is 14, and c is 16.
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Now this is just a very easy simplification, this is 20 times 10, times 20-14 is 6, and 20-16 is 4, so 20×10 is 200, times 24 is 4800.
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I can simplify that a bit.
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I can pull off, let's see, I can pull a 10 out of there, and that's 10 square root of 48.
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Now, I can pull a 16 out and make that 40 square root of 3 for my area.
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Let's recap what we're given there.
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We're given a triangle and we were told the three side lengths.
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If you know the three side lengths and you're going for the area, you almost certainly want to use Heron's formula.
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It's a quick matter of finding the semi-perimeter, and then dropping the semi-perimeter and the three side lenghts into the square root formula.
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Then you just simplify down and you get the area.
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In this third example, we're asked to find the area of a triangle whose side lengths are 7, 9 and 14.
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Ordinarily, I'd say that's a dead give-away that you want to use Heron's formula, but unfortunately, this example they specifically say without using Heron's formula.
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I'd really like to use Heron's formula on that but it has asked me to find the area of the triangle without using Heron's formula.
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I'm going to do a little bit of extra work here.
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I'll start out by drawing my triangle.
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There's 7, 9, and 14.
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I think I'm going to try to find that angle right there.
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I'll call that angle C.
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The reason I called it angle C is because I'm planning to use the law of cosines to find that angle.
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Remember the law of cosines is very useful if you know all three sides.
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You can quickly solve for an angle.
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Let me remind you what the law of cosines is.
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It says c²=a²+b²-2abcos(C).
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If you know little a, b and c, you can just drop them into that formula, and you can solve for capital C.
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That's what we're going to do here.
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I wrote 19 for that side, of course, that's supposed to be 9.
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My c is the side opposite angle C, a is 7 and b is 9.
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I'm going to plug these into the law of cosines, and it says, 14²=7²+9²-2×7×9×cos(C).
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I'll do a little algebra on that.
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14² is 196, 7² is 49, 9² is 81.
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I was looking ahead to this next calculations, 7×9 is 63, 2 times that is 126.
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We get 126cos(C).
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49+81 is 50+80, that's 130.
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If I move that to the other side, I get, let's see, 130 from 196 is 66, is equal to -126cos(C), so cos(C), solving for that part, that's what we don't know, is -66/126.
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If I take the arccosine of that on my calculator, of course I've got to use degree mode if I'm planning to use degrees for this problem that I am, I'll take the arccos(-66/126).
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What I get is 121.6 degrees approximately for angle C.
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Normally, I would just fill that into my drawing, but in my drawing, I've shown it as an acute angle and that really isn't appropriate because 121.6 is bigger than 90 degrees.
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I think I have to modify my drawing based on what C came out to be, and draw that part as an obtuse angle.
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I'll redraw there.
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That's angle C which we figured out was 121.6 degrees.
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That's side c which is 14, that's side a which is 7, and this is side b which is 9.
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What I'm going for is I'm trying to find the area of the triangle ultimately, but I can't use Heron's formula because the problem specifically told me I wasn't allowed to use Heron's formula.
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I want to try to use the old-fashioned 1/2 base times height, but I don't know the height of the triangle.
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I'm going to try to find the height of the triangle, that's why I had to find that missing angle.
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To find the height of the triangle, I'm going to drop an altitude here.
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Since it's an obtuse angle in the triangle, this altitude is actually outside the triangle.
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There's that altitude.
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I want to find the height of that altitude.
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In order to find it, I have to find that angle right there, and that is the supplement of 121.6.
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So θ=180-121.6, which is approximately equal to 59.4 degrees.
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That tells me what angle θ is.
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Based on that and this a=7, I can find the missing height of the triangle.
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I'm going to use SOH CAH TOA for that.
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Sin(θ) is equal to the opposite over the hypotenuse.
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Sin(59.4) times the hypotenuse, is 7 ...
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We don't know what the opposite length is.
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I'll leave that there.
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I get that the opposite is equal to 7sin(59.4).
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I'll work that out on my calculator.
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What I get is that that's approximately equal to 6.03.
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That length right there that we found, 6.03.
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I'm going to need a little more space to work this out, I'm going to go to the next slide here and just redraw my triangle, and remind you what we know about that.
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We were given this triangle with side lengths 7, 9, and 14.
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The first thing we did was we used the law of cosines to find this missing angle.
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That was 121.6 degrees.
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Then we dropped a perpendicular, an altitude, from the top angle there.
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To find that, we had to figure out that θ there was 59.4 degrees.
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Then using that value of θ and the hypotenuse of 7, we figured out that this was 6.03 units long.
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Now we're in good shape to find the area of the triangle using old-fashioned geometry.
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It's just 1/2 times the base times the height.
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The base here is 9.
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That's the base.
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You might be worried a little bit that the top of the triangle sticks out a bit over the edge of the base, that actually doesn't matter.
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That doesn't make the formula invalid.
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It is still ...
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You count the base as being 9, even though the top of the triangle sticks out over the edge of the base.
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The height we've just worked out is 6.03.
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Now I'm going to multiply that by 9 and by 1/2, and I get approximately 27.1 for the area of the triangle.
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Let's recap what happened there.
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I was given three sides for a triangle.
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Normally when you're given three sides and you want to find the area, that's a dead give-away that you want Heron's formula.
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Unfortunately, this example asked us to find it without using Heron's formula.
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That's why we used law of cosines to find that angle, and then we used SOH CAH TOA to find the length of an altitude of a triangle, in other words, the height of a triangle.
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SOH CAH TOA right there.
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Then we used the old-fashioned area formula using the base and the height to give us the area of the triangle.
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We'll try some more examples of these later.