WEBVTT mathematics/pre-calculus/selhorst-jones
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This is Will Murray for educator.com and we're here today to talk about the law of cosines, which is the second of the two big trigonometric rules.
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Remember, last time we talked about the law of sines.
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You kind of put those together, and together those enable you to find the length of any side and the measure of any angle in a triangle, if you're given enough information to start with.
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Let's start with the formula here.
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The law of cosines is c²=a²+b²-2abcos(C).
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Let me draw you a a triangle so we can see how that applies.
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Remember, the convention is that you use lowercase a, b, and c for the sides of the triangle, and uppercase A, B, and C for the angles.
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You use the same letter for the angle and the side opposite it.
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My uppercase A goes here, and my B goes here because it's opposite of side b, here's angle C.
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The point of the law of cosines is it relates the lengths of the three sides a, b, and c, little a, little b and little c, to the measure of one of the angles which is capital C here.
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The point is that, first of all, you can use this in any triangle.
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It's not just valid in right triangles.
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Remember the big rule we had, SOH CAH TOA, is only valid in right triangles.
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The law of cosines is valid in any triangle.
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It's a generalization of the Pythagorean theorem, in a sense that, remember the old Pythagorean theorem was just c²=a²+b², that only works in a right triangle.
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If you look at the law of cosines, if angle C is a right angle, then the cos(π/2) or the cos(90) is zero.
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If angle C is a right angle, then this term 2ab-cos(c), drops out,the law of cosines just reduces down to the Pythagorean theorem, c²=a²+b².
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You can kind of think of the Pythagorean theorem as just being a consequence of the law of cosines.
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The law of cosines is the more general one that applies to any triangle.
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The Pythagorean theorem is the more specific one that just applied when angle C happens to be a right angle.
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Let's see how it's used.
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The law of cosines is really used in two situations.
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First of all, it's used in a side angle side situation.
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That means where you know two sides of a triangle and the included angle.
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The reason it's useful ...
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Let me write the law of cosines again c²=a²+b²-2abcos(C).
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The point here is that if you label this sides as little a and little b here, that makes this angle capital C and little c is down there.
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If you know the side angle side, in other words, if you know, little a, little b, and capital C, then you know all of the right-hand side of the law of cosines.
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You can then solve for little c.
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That's why the law of cosines is useful for side angle side situations, it's because you can fill in everything you know on one side of the law of cosines, then you can solve for little c.
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It's also useful for side side side situations.
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Let me draw that.
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Side side side means you know all three sides of a triangle, but you don't necessary know any of the angles yet.
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The point is, if you know little, little b, and little c, then you know all of these parts of the law of cosines, so you can solve for the cosine of capital C, and you can figure out what that angle capital C is.
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Then you can figure out what one of the angles is.
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You can just kind of rotate the triangle, and relabel what a, b, and c are to find the other two angles.
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If you know all three sides of a triangle, the law of cosines is very useful for finding the angles, one at a time.
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Remember, there's a couple other ways that you can be given information for triangles.
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You can be given, angle side angle, or side angle angle, or side side angle.
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Those two don't really lend themselves very well to solution by the law of cosines.
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If you're given one of those situations then you really want to use the law of sines which we learned about in the previous lecture.
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There's one more formula we're going to be using in this lecture which is Heron's formula.
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The point here is that, if you know all three lengths of sides of a triangle, I'll call them a, b, and c, as usual, then you have a nice formula for the area.
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It's got one more variable in it, this s.
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S is 1/2 (a+b+c), that's the semi perimeter.
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Remember, the perimeter is the distance around the edge of a triangle, that's a+b+c.
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The semi perimeter is just 1/2 of a+b+c.
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We worked that out ahead of time and we call it s.
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We plugged that into this formula, that's a fairly simple formula just involving s and then a, b, and c, and it spits out the are of the triangle for us.
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That's very useful if you know the lengths of the sides.
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You never really have to look at any angles, and you don't have to get into any sines or cosines, no messy numbers there, hopefully.
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Let's try out some examples here for law of cosines.
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First example, we're given a triangle ABC.
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Let me go ahead and draw that.
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I'm going to put a here, and b here, and c here, which forces the angles, remember the angles go opposite the sides of the same letter.
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We're given the a=3,b=4, and angle c measures 60 degrees.
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We want to first of all determine how many triangles satisfy these conditions and then we want to solve the triangles completely.
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To answer the first question, we have a side angle side situation.
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What we know is that side angle side always has a unique solution assuming the angle is less than 180.
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In this case, the angle is 60 which is less than 180, so there's a unique solution.
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There's exactly one triangle satisfying these conditions.
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That answers the first question, how many triangles satisfy those conditions, exactly one.
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Now we have to solve the triangle completely, this is where the law of cosines is going to be useful.
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Let me copy down the law of cosines c²=a²+b²-2abcos(C).
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This is very useful because we know a and b, and capital C.
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I can just plug all those in and solve for little c.
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Let me do that, a² would be 9, b² would be 16 because 4² is 16, minus 2×3×4=24, cos(60) ...
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This is 25-24, now cos(60), that's one of my common values, that's π/3, I remember that the cos(60) is 1/2.
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This is 25-24×1/2, 25-12, c² is 13, c is equal to the square root of 13.
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I can get an approximation for that on my calculator, that's about 3.61.
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That's approximately equal to 3.61.
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I'll fill that in on my triangle.
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Now, I've got the third side of the triangle.
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The only thing that's left is it says to solve the triangles completely.
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I need to find the other two angles A and B.
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To do that, I'm going to use the law of sines.
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Let me write down the law of sines to refresh your memory, sin(A)/a=sin(C)/c.
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I know what side a, side c, and capital C are.
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I'm going to cross multiply this and I get csin(A)=asin(C).
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I'll fill in the values that I know.
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I know little a is 3, sin(C)=sin(60), sin(A), I don't know that yet, and little c, I figured out, is 3.61.
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I'm solving for sin(A)=3sin(60)/3.61.
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I'll just work that out on my calculator.
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What I get is approximately 0.72.
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By the way, it's very important that your calculator be in degree mode if you're using degrees here.
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I gave angle C as 60 degrees.
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It's very important that you set your calculator to degree mode.
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If your calculator's in radian mode, then it will interpret that 60 as a radian measure, so your answers will be way off, so set your calculator to be in degrees before you try this calculation.
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A is arcsin(0.72), I'll work that out on my calculator.
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That tells me that a is just about 46.0 degrees.
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Now, I've got a measure for angle A.
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I'm going to use the law of sines to find the measure for angle B, but I need a little more space.
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Let me redraw my triangle.
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We've got a, b, and c.
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I figured out the c was 3.61, a was given as 3, b was given as 4, C was given as 60.
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I figured out that A was 46 degrees.
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I'm just trying to find the measure of angle B now.
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I'm going to use again the law of sines.
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Sin(B)/b=sin(C)/c, I'll fill in what I know here, I know that little b is 4, sin(B), I don't know, sin(C)=sin(60), little c is 3.61.
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I'll cross multiply that, 3.61sin(B)=4sin(60), sin(B), that's what we're solving for is equal to 4sin(60)/3.61.
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Let me work that out on my calculator, 4sin(60)/3.61=0.96.
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B is arcsin(0.96) and I'll work that out, that's 73, just about 74 degrees, rounds to 74 degrees.
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Now I figured out angle B, 74 degrees.
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Now, I've solved for all three sides of the triangle, and all three angles of the triangle.
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It's nice at this point, even though we're done with the problem to get some kind of check because we've done lots of calculations here, we could have made a mistake.
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What I'm going to do is add up all three angles in the triangle, and make sure that they come up to be 180 degrees.
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To check on my work here, I'll add up 60+46+74, that does indeed come out to be 180 degrees.
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That suggests that we probably didn't make a mistake in solving all those angles.
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Just to recap this problem here, we're given a side angle side situation, that's a definite tip-off that you're going to be using the law of cosines.
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I filled in my side, the included angle, and a side.
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The first thing I did was I used the law of cosines to find the missing side.
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To solve the triangle completely, I still had two angles that I didn't know, I used the law of sines after that to find the two angles that I didn't know based on knowing the other sides and the other side, and angle.
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Let's try another one now.
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In this one, we're given the side lengths of the triangle, a, b and c are 5, 7 and 10.
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Let me draw a possible triangle like that, 5, 7 and 10.
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We want to find out how many triangles satisfy this conditions and solve the triangles completely.
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The first thing to do with this problem is to identify what we're given.
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We're given a side side side configuration.
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That usually gives you a unique triangle.
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What you have to do is check that each side is less than the sum of the other two.
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Let's check that out.
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Unique if each side is less than the sum of the other two.
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That'll be a quick real check.
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We're comparing 5 with 7+10, 5 is certainly less than 17, so that works.
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7 should be less than 5+10, that certainly works, 7 is less than 15.
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10 should be less than 5+7, that certainly works.
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If any of those checks had failed, for example 10 if it had been 13, 5, and 7, instead of 10, 5, and 7, then the last check would have failed because 13 is not less than 5+7.
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At that point, we would have stopped and said, "This is invalid. There is no triangle that satisfies those conditions."
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Since all those three conditions checked, it does mean that there is a unique solution.
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There's exactly one triangle with those three lengths of sides.
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We've found all the side lengths.
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We need to find the angles, this is where the law of cosines is really handy.
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Let me write that down, the law of cosines says c²=a²+b²-2abcos(C).
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You can label the sides and angles whichever way you want.
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I'm going to label the top one c.
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Let me write that outside of the triangle.
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That makes the bottom side c and then the two other sides a and b.
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What I can do here is I can plug in little a, little b and little c, and then I can solve for the cosine of capital C, then in turn solve for what angle capital C is.
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Let me do that.
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If c is 10, that's 10² equals, a and b is 5²+7²-2×5×7×cos(C).
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Now it's just a little bit of arithmetic, 100=25+49-70cos(C).
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25+49=74, if we pull that over to the other side, we get 26, -70cos(C).
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Cos(C)=-26/70.
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I'm going to figure out that C is arccosine or inverse cosine of -26/70, I'll do that part on my calculator.
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Remember, you have to be in degree mode for this.
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What I get is that C is approximately equal to 111.8 degrees.
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That's 111.8 in that corner.
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Now I'm going to go over in the next page.
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I'm going to keep going to find out the other two angles.
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We'll find them exactly the same way.
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Let me go ahead and redraw my triangle, 5, 7, 10.
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We've already figured out that that angle is 111.8.
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What I'm going to do is relabel the sides and the angles because I still want to use that law of cosines c²=a²+b²-2abcos(C).
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I'm going to relabel everything here so that I can label C as a new angle, and I can solve for a new angle.
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Relabel that angle as capital C, then my a and b will be 7 and 10.
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Sorry, small C will be 7, and a and b will be 5 and 10.
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We'll go through and we'll work that one out.
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Little c², that's 49 is equal to a², that's 100, plus b² is 25, minus 2ab, that's 2×10×5×cos(C).
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Let me work this out, this is 125, 49 equals 125 minus 2×10×5, that's 100, cos(C).
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If I subtract 125 from both sides, I get -76 is -100cos(C), divide both sides by -100, we get 76/100=cos(C).
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C=arccos(76/100), I'll plug that into my calculator.
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It tells me that that's approximately 40.5 degrees for that angle right there.
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Now, there's one angle left to find.
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Again, I'm going to relabel the sides and the angles so that I can continue to use the law of cosines in its standard form.
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I don't have to change around what a, b and c are in the law of cosines.
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I'll do this in red.
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In red, I'm going to call this angle capital C, and that means I have to relabel my sides, that means little c is equal to 5, a and b are 7 and 10.
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Now I'm going to plug those values into the law of cosines.
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c² is 25, equals 49, 7², plus b² is 100, minus 2×a×b, 7×10, times cos(C).
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Now it's a matter of solving that for capital C again.
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I get 25 equals 149, minus 2×7×10, that's 140, cos(C).
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If I subtract 149 from both sides, I get -124, is equal to -140cos(C).
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Cos(C)=124/140, C=arccos(124/140), I'll go to the calculator on that.
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I get 27.7 degrees.
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Let me fill that in, 27.7 degrees.
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Now we've solved the triangle completely.
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We started out with all three side lengths and we found all three angles in the triangle.
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We're really done but it's always good to find a way to check your work.
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Let me check my work in blue here.
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Again, I found all three angles.
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I'm going to add them together and see if I get 180 to check that out.
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I'm going to add up 111.8+40.5+27.7, what I get is exactly 180 degrees.
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That tells me that I must have been right in getting those three angles for the triangle.
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Just to recap here, what we were given was three sides of a three angle.
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We were given three side lengths, that was a side side side situation.
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We had to check that the three lengths, that we didn't have a situation where two sides added up to be less than the third side.
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We had to check that each side was less than the sum of the other two.
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Once we did that, we knew we have exactly one solution.
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Then we filled in the side lengths of the triangle, and we used the law of cosines.
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The law of cosines lets you fill in three side lengths and then solve for the cosine of one of the angles.
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You work it through, solve for the cosine, then you get each one of the angles by taking the arccosine, then you just go through a separate procedure like that for each one of the angles.
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In our third problem, we're trying to find the area of a triangle whose side lengths are 5, 7 and 10.
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Now, this one is really a set up for Heron's formula, because Heron's formula works perfectly when you know the three sides of a triangle.
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We're going to use Heron here.
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Heron says that the area is equal to the square root of s times s minus a, s minus b, and s minus c.
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You've got to figure out what s is.
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S is the semi-perimeter of a triangle which means 1/2 of the perimeter, 1/2 of the sum of the three side lengths.
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That's 1/2 of 5+7+10, which is 1/2 of 22, which is 11.
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I'm going to plug that in to Heron's formula, wherever I see an s, that's 11×(11-5)×(11-7)×(11-10).
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Then I'll just work on simplifying that, that's 11×6×4×1, that's the square root of 264, 11×24.
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That problem was really pretty quick if you remember Heron's formula.
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Heron's formula is very useful.
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If you know three side lengths of a triangle, then what you do is you work out the semi-perimeter, you just drop the side lengths into this formula for the semi-perimeter, then you drop the semi-perimeter and the three side lengths for the Heron's formula for the area.
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It simplifies down pretty quickly to give you the area of the triangle.
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We'll try some more examples later on.