WEBVTT mathematics/geometry/pyo
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Welcome back to Educator.com.
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We are going to continue this unit with the Law of Cosines.
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We are still on triangles; now, using Law of Cosines is very similar to when we used Law of Sines.
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We are using both of those for non-right triangles; it is very important to remember that there are different things
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that you can use for right triangles, and then there are things that you can use for non-right triangles.
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So, Law of Sines and Law of Cosines are used for non-right triangles.
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The Law of Cosines consists of three different formulas, depending on what you need--what measure you have to find.
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If you have a triangle ABC with sides a, b, c, and angles A, B, C, angles are capital, and the sides opposite them are the lowercase of that letter.
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So, if this is angle A, then this would be side a; if this is angle B, this is side b, and this is side c.
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It is really important to remember that the angle and the side opposite go together, like a pair: this angle and this side opposite, and so on.
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If we are looking for, let's say, A, you can use this formula.
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If you are given, let's say, a, b, and c, and you are looking for angle B, well, then, you have to use this one.
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Or if you are, let's say, looking for b, and you are given sides c and a, and then you are given the angle C, then you would use this one.
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You would only use one of these, depending on what you need.
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That is a² = b² + c² minus 2 times bc times cos(A).
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If you look at this, this looks kind of like the Pythagorean theorem: a² = b² + c², the other two sides squared.
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But it is minus 2 times bc, and the cosine of A; so then, that is how it is going to work for all of them.
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b² = a² + c² - 2 times a and c, cosine of angle B.
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c² = a² + b² minus 2 times a times b, times the cosine of this angle C.
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And that is the Law of Cosines.
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Now, when you are given a non-right triangle, and you have to determine whether to use Law of Sines or Law of Cosines--
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well, the Law of Sines would probably be the easier option, so if you can, you would use the Law of Sines.
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But remember: the Law of Sines was sin(A) over side a = sin(B)/b and sin(C)/c.
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So, in order for you to be able to use the Law of Sines, you would have to have an angle-side pair.
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If you are given an angle-side pair, then you are always going to be given three measures.
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Out of the six total, the three angles and three sides, you are going to be given three.
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In order to use the Law of Sines for non-right triangles, out of those three, two of them would have to be an angle-side pair.
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When you don't have an angle-side pair, and it is a non-right triangle, then you would have to use the Law of Cosines.
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Again, for non-right triangles, you can use the Law of Sines or the Law of Cosines.
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If you can, you will look to see if you have an angle-side pair; if you do, then you would use the Law of Sines.
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But if all three measures that are given...none of them is an angle-side pair, then you would have to use the Law of Cosines.
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It is either one or the other.
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For example, let's say that the measure of angle B is 27 degrees, and let's say this over here is 5, and this over here is 10.
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Well, here we have no angle-side pair, because angle C is not given; so then, we don't have that angle-side pair;
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we have the measure of angle B, but no side b, so the angle-side pair is not given; and the same thing happens for this.
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So, since we have no angle-side pair, we would have to use the Law of Cosines.
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And see how we are given a and c, and we have angle B; so which one would we use?
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Well, it depends on what we are looking for; if we are looking for this one right here--let's say we are looking for b--
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that went first; then we would have to use this one right here, because this is what you are looking for.
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You are given this; you are given this; and you are given the measure of angle B; so you would have to use this one.
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Let's say we are looking for, let's see...once we find that, if we want to find, let's say...the measure of angle A;
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then after we find B, we will have an angle-side pair; we will have the measure of angle B, and we will have side b.
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So, from there, you can go ahead and use the Law of Sines, using this one and this one that we found, and then sin(A)/a.
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Let's just do a few problems and see if we can get more familiar with this.
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Use the Law of Cosines when both are true: the triangle is a non-right triangle (we know this one),
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and then the measures of an angle and the side opposite are not given, meaning that there is no angle-side pair.
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Here, if I draw this out, and this is a; this is b; and this is c; we know that, for us to use the Law of Cosines, these have to be true.
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If these are true, then we have to use the Law of Cosines.
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And from the three measures that they give you, there are going to be three different types of problems
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where you are going to use the Law of Cosines, where both of these would be true.
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If I am given a side, a side, and a side--the measure of all three sides of a non-right triangle--then that would be the Law of Cosines.
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Also, if you have side-angle-side (with the one that we just did, that was side-angle side, because that was side b,
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and then angle C, and then side a), that doesn't give you any angle-side pairs.
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So, side-angle-side would be another one; and then, the third one is going to be angle-side-angle.
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So, if you have an angle, a side, and an angle, that is also a case where you are not going to have any angle-side pairs.
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These three are the types of problems, given the measures, where you are going to have to use the Law of Cosines: side-side-side...
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and don't get this confused with the side-side-side postulates and the congruence theorems,
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because that is when you are using two triangles, and you are trying to make them congruent.
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With this, I am just doing this so that you can see that, if you are given the measure of a side, the measure of a side,
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the measure of a side, of one triangle, then you are going to have to use the Law of Cosines, and so on.
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OK, for this triangle ABC, we are going to look for side a, which is this side right here, the side opposite angle A.
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I know that I have no angle-side pair; I have a side, an angle, and this side.
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And so, using one of the formulas (I know that I have to use the Law of Cosines, so), here are the three formulas.
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It is a² = b² + c² - 2bc times the cosine of angle A.
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Remember: it has to be the cosine, because it is the Law of Cosines.
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And remember that all lowercase letters are the sides, and the capital letters are angles.
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The second one is b² = a² + c² (it is the other two sides), minus 2 times ac cosine of B.
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And then, side c² = a² + b² - 2ab cosine of angle C.
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Those are your formulas; now, we are going to see which one we need to use.
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I have side c, side b, and angle A; if I am looking for side a, well, I am given b, c, and angle A, and this is what I am looking for.
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So, I have to use the first one: that is going to be a² = b²...
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our b is 21, so 21², plus...c is 14, squared, minus 2 times 21 times 14, cosine of 60.
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From here, it is all calculator math.
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Now, if your calculator is the type that doesn't perform everything in one screen, then just go ahead and solve it in chunks.
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But keep in mind--remember the order of operations: we have to multiply all of those together.
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And this right here is the cosine of 60, so you would have to punch in cos(60) on your calculator to find that number.
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You cannot separate the cosine and the angle measure.
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It is going to be 2 times 21 times 14 times whatever this number is, so it is four numbers; and cos(60) is actually 1/2,
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so it will be those four, multiplied; you are going to square this; you can add these together,
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because here, multiplying this, this is far away from this right here; so you can solve this, and they won't affect each other.
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Just remember that you have to multiply the 2 by all of those numbers before subtracting the 2.
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So, be careful not to just solve it all in the order of this plus this minus 2; it has to be minus the whole thing.
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I am going to just go ahead and do 21; and I do have a calculator on my screen: 21² is 441,
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plus...14² is 196, minus...and then, let's do all of that, too.
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2 times 21 times 14 times cos(60), .5, equals 294, so a² is going to be 441 + 196 - 294, so 343.
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Then, the square root of that is 18.5, so that is the measure of side a, 18.5.
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All right, we are going to work on some more examples.
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Determine whether the Law of Sines or the Law of Cosines should be used to solve each triangle.
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We are not going to actually solve them; we are just going to determine if we can use the Law of Sines or the Law of Cosines.
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Remember that both the Law of Sines and the Law of Cosines are used for non-right triangles.
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And you want to look to see if you have an angle-side pair.
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If you do, then you can go ahead and use the Law of Sines, because the Law of Sines is a lot easier.
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If you don't, then you would have to use the Law of Cosines.
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Now, if they say to solve the triangle, we went over that before; solving the triangle just means looking for all of the unknown measures.
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So, in this case, they have given you this side, this side, and this angle.
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To solve the triangle, we would have to solve for the measure of angle B, the measure of angle C, and AB, because those are all unknown measures.
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The first one: we have a non-right triangle, and let's see how we have the measure of angle A, and we have side a, and then we have side b.
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So, we do have an angle-side pair; since we have an angle-side pair...
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now, I am not going to actually solve it, but I just want you to see that the Law of Sines is this.
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Remember: you don't use all three of these ratios--you would just use two of them to create a proportion, whichever two you need.
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We know that we are going to use the angle A and side a; it is going to be sin(30)/7 =...
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and then, this is side b; so then, we would be looking for angle B; it would be sin(B)/11.
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And then, you would solve it that way; you would have to solve sin(30) on your calculator; you can multiply that by 11, cross-multiplying.
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And then, you are going to divide the 7.
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And then, when you solve sin(B), remember: you have to use the inverse sine, the sin^-1.
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It is probably the 2nd key, and then "sin" on your calculator; use the inverse sine to find the measure of angle B.
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That is the Law of Sines for this one; and then, for this one, we are given all of the side measures, but no angles.
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That means that we have no angle-side pairs, so we would have to use the Law of Cosines--the Law of Sines here, and the Law of Cosines here.
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And since you have three formulas to work with, you would have to choose whichever one you need,
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depending on what angle or what measure...what side...whatever it is that you are looking for.
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That is Example 1; for the next example, we are going to use more Law of Cosines to find missing measures.
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For the first one, I want to look for the measure of angle B.
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I am going to label that as x, just so that you know that that is the missing measure.
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And then, before we start, let's write out all of the formulas again.
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That is a² = b² + c² - 2bc cosine of angle A.
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b² = a² + c² - 2ac cos(B).
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c² = a² + b² - 2ab cos(C); there are the formulas.
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And then, to find the measure of angle B, given all three sides, I want to see what I want to use--which formula I would use.
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So, for this one right here, I would have to use the second formula, because that is the only one that gives me angle B.
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If you are looking for an angle measure, then you would have to pick and choose your formula, depending on this right here.
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These are the only parts of the formulas that give you the angle measures, so it has to be the second one.
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It is going to be b², which is 5², equals a² (is 13 squared), plus 12², minus 2ac cosine of angle B.
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This is going to be 25 = 169 + 144 - 2...here, let's solve this part out on the calculator.
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Remember that you can square this; you can square that; exponents come first, always, always, by the order of operations.
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And once you do that, make sure that you do not subtract 2; it is not this number, subtract 2, because 2 is multiplied to all of this.
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So, it would have to be 2 times 13 times 12...and then, for this one, cos(B)...
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we are looking for angle B, so this is our variable; but we cannot separate the cosine and the B, because they have to go together.
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It has to be the cosine of an angle measure.
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Think of this whole thing as the variable; that is what you are solving for, so think of that whole thing as x.
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Here, it would be minus 2 times 13 times 12; these three numbers, multiplied together, is going to be the coefficient of this variable.
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2 times 13 times 12 is 312, cosine of B; I will circle this again.
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Then, this is going to be 25; it equals this plus this; it is going to be 313, minus 312, cosine of B.
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Be careful, again: do not subtract these two numbers; this is not 25 = 1cos(B); do not subtract them; it is this minus this whole thing.
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But what I can do is subtract this over to the other side.
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If I subtract 313, I am going to get...that is going to be -288, which is equal to negative (don't forget the negative sign) 312 cosine of B.
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Now, from here, again, if I circle this, and that is the variable, then this is the coefficient;
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I would have to divide this number to both sides, so that I can solve for this variable.
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Divide this by -312; divide this by -312 (I'll write that over).
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Your number is going to be -288, divided by -312: 0.923 = cos(B).
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Now, remember, again: we are solving for angle B; that means we have to find the inverse cosine,
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because here, if we have this angle measure, then we can go ahead and punch in the cosine of that angle measure.
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But since that is the actual number that we are looking for, you would have to use the inverse cosine, don't forget--
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because when you punch in a number after sin, cos, or tan, then the calculator is going to think
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that that number is the angle measure, because it is always the cosine, sine, or tangent of an angle measure.
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Just make sure you don't punch in the cosine of this number here, or else your calculator is going to think that that is the angle measure.
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But instead, we are doing the cosine of the unknown angle measure; it equals...and that is the answer.
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You would have to use the inverse cosine; that is cos^-1; on your calculator, you might have to press the 2nd key and cosine.
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It depends on your calculator; it might have its own separate button for it--just look for it.
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See how you just punch that in with this number; and then it is going to be 22.6...let's see, let's just round it to hundredths.
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So, that would be the measure of angle B, 22.6 degrees.
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The next one: we are given angle, side, and angle; we have no angle-side pairs.
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So, let's say that I want to look for two unknown measures for this one.
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If I have this as the unknown measure--this is the first unknown measure--remember...
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now, you can use the Law of Cosines, but it is a lot of work, and it does take quite some time.
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Here, we actually don't have to use the Law of Cosines, because this is the third angle of a triangle.
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So, try to remember that, if you are looking for the third angle of a triangle, then you can use the angle sum theorem,
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where all three angles of a triangle add up to 180; so it is just going to be 180 = 40 degrees + 59 + x.
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So, all three angles together add up to 180.
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This is 99 + x; if we subtract it, that is going to be 81; it equals x, so the measure of angle C is 81 degrees.
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Now, I also want to look for, let's say, side b.
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So, in this case, we would have to use the Law of Cosines; which one would I use?
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If I am looking for this here, for this one, do I have all the sides--what sides do I have?
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I have this side right here; so here, I have 14; this is unknown; and then, for this one, let's see...
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I have all three angles, so I don't have to worry about that; I have side c; I have this; I have this; the first one I have...
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And then, this is what I am looking for; and then, I have the measure of angle A, and I have this, again.
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I wouldn't be able to use that one, because I only have that side right there.
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Now, if you notice that all three of these don't work, it is because here, now that I found this angle measure, I have an angle-side pair.
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I would use Law of Sines, instead of Law of Cosines.
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Let's save ourselves a lot of work; and then here, since the measure of angle C is now 81,
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in order for me to look for side b, I can use the Law of Sines, because of this angle-side pair.
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Here, I am going to do this with the sine of angle B.
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If you have something like this, you wouldn't have to use the Law of Cosines, because you can just find the third angle measure.
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Now, if you want, you could; but again, this would be a lot easier.
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This here, the sine of C, the sine of 81, over 14, equals the sine of 59, over b.
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So, the sine of 81...we are going to cross-multiply, so bsin(81) = 14sin(59).
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I can divide the sine of 81 (remember: don't forget that you can't separate them)...so b would equal all of this, over sin(81).
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I am going to put 12 over 0.988...so b is going to be 12.5.
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Again, here we have a side and a side and a side given; and so, we are going to have to use this one, because the unknown measure would be angle B.
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For this one, when you are given angle-side-angle, then just solve it by finding the third angle measure,
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and then using the Law of Sines, using the angle-side pair.
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For this next one, we are going to solve the triangle.
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Remember: to solve the triangle means to solve for all unknown measures, angles and sides.
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Here, I want to list out all of the different measures that I am going to look for: the measure of angle B, the measure of angle C, and then side a.
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And that is just so that I don't forget to solve for something.
00:31:20.300 --> 00:31:28.300
Now, we have no angle-side pairs; we have the measure of angle A, but we don't have side a...and here, and here.
00:31:28.300 --> 00:31:35.600
So, there are no angle-side pairs; that means that I am going to have to use the Law of Cosines.
00:31:35.600 --> 00:31:49.000
The three formulas, again, would be a² = b² + c² - 2 times b times c times the cosine of angle A.
00:31:49.000 --> 00:32:01.300
Then, b² = a² + c² - 2 times a times c times the cosine of angle B.
00:32:01.300 --> 00:32:15.400
c² = a² + b² - 2 times a times b cosine of angle C.
00:32:15.400 --> 00:32:21.400
First, in order for me to solve for this angle measure, I want this sine here.
00:32:21.400 --> 00:32:30.700
If I want to solve for angle B, then I am going to have to know all of the measures of these sides.
00:32:30.700 --> 00:32:42.400
Since we are given the two sides, side c and side b, I am going to look for side a.
00:32:42.400 --> 00:32:51.100
Then, since I have the measure of angle A, I want to use this first one, because that is what goes in there.
00:32:51.100 --> 00:33:14.000
It is going to be a² = 20² + 24² - 2(b)(c)(cos(47)).
00:33:14.000 --> 00:33:31.800
a² = 400 + 576 -...don't forget that you cannot subtract this number with this;
00:33:31.800 --> 00:33:37.000
it is going to be 2 times 20 times 24 times the cosine of 47.
00:33:37.000 --> 00:34:07.100
You are going to multiply all four of those numbers together...that number, and this...so this is 654.72.
00:34:07.100 --> 00:34:20.700
And then, now that you have multiplied all of that, you are just going to take this, add that, and subtract that.
00:34:20.700 --> 00:34:44.100
400...add the 576, and subtract 654.72; you are going to get 321.28.
00:34:44.100 --> 00:34:55.100
And then, to find a, we are going to have to take the square root of that; you are going to get 17.92.
00:34:55.100 --> 00:35:04.200
So, I am going to write that answer here, for this one...92; and then I need to solve for these two.
00:35:04.200 --> 00:35:15.200
So then, the measure of angle B...now that I have this, if I write this in here...
00:35:15.200 --> 00:35:22.000
Keep in mind that, for any given problem where you have to use the Law of Cosines, you only have to use it once--you don't have to keep using it.
00:35:22.000 --> 00:35:28.500
So, here, if I am solving the triangle, then I have to find the measures of three unknowns.
00:35:28.500 --> 00:35:36.000
I don't have to use the Law of Cosines for each one of these; I only have to use it once.
00:35:36.000 --> 00:35:41.200
And then, once you use it once, you are going to end up with an angle-side pair.
00:35:41.200 --> 00:35:50.200
So, since I solved for a, I have side a, and I have angle A; there is my angle-side pair.
00:35:50.200 --> 00:35:58.400
If you have an angle-side pair, then you know to use the Law of Sines: so sin(A)/a
00:35:58.400 --> 00:36:07.700
equals the sine of angle B, over side b, which is equal to sin(C) over side c.
00:36:07.700 --> 00:36:13.600
Since we have the angle-side pair of angle A and side a, we are going to use this one.
00:36:13.600 --> 00:36:24.400
And then, if we solve for angle B first, then I am going to use that one.
00:36:24.400 --> 00:36:43.500
I will put it down here: sin(47)/17.92 = sin(B)/20.
00:36:43.500 --> 00:36:55.800
So, you are going to cross-multiply; that will be 20sin(47) = 17.92sin(B).
00:36:55.800 --> 00:37:18.900
This is going to be, using your calculator...remember that this is sine, not cosine...it is this number, and then divided by this number;
00:37:18.900 --> 00:37:42.700
this is your variable that you are solving for; that is the coefficient; so I need to divide this number...17.92.
00:37:42.700 --> 00:37:52.500
I get 0.816; I am just going to double-check that really quickly.
00:37:52.500 --> 00:38:14.100
47...the sine of 47 degrees...multiply that by 20 (you can do 20 times sin(47); it depends on what your calculator can do); divide it by 17.92.
00:38:14.100 --> 00:38:29.300
So, .816 = the sine of angle B; now again, here we are looking for the angle measure, so we are going to have to use the inverse sine.
00:38:29.300 --> 00:38:55.600
That means that you are going to hit the 2nd button; and that is going to give us B; the measure of angle B is 54.71.
00:38:55.600 --> 00:39:04.600
We still have one more to solve for, the measure of angle C.
00:39:04.600 --> 00:39:07.800
Now, we want to look for the easiest possible way to solve for this.
00:39:07.800 --> 00:39:15.600
If you want, you can go ahead and use the Law of Sines again, because you have two different angle-side pairs.
00:39:15.600 --> 00:39:23.600
But for this one, since I have two angle measures, I just want to look for the third one using the angle sum theorem.
00:39:23.600 --> 00:39:30.500
Keep that in mind: if you have two angles, that would be the easiest way to look for a third angle.
00:39:30.500 --> 00:39:50.200
The measure of angle C is going to be 180 minus 47 added to 54.71.
00:39:50.200 --> 00:40:21.200
180 minus...and then, what are these two?...that will be 101.71; that means that the measure of angle C is going to be 78.29.
00:40:21.200 --> 00:40:36.100
180 minus 101.71...the measure of angle C is 78.29.
00:40:36.100 --> 00:40:44.600
All three of these are your answers, because these are the unknown measures of the triangle.
00:40:44.600 --> 00:40:51.600
When it says to solve the triangle, then you are going to be looking for all of the unknown measures.
00:40:51.600 --> 00:40:59.800
The easiest way would be to just first write out what you have to find, and then do one at a time.
00:40:59.800 --> 00:41:04.500
First, always look to see if you can use the Law of Sines or the Law of Cosines.
00:41:04.500 --> 00:41:09.800
In our case, we had to use the Law of Cosines, because we had no angle-side pair.
00:41:09.800 --> 00:41:20.600
We are going to look for that side; and then, from there, you only have to use Law of Cosines once in a single problem.
00:41:20.600 --> 00:41:29.500
We use the Law of Cosines; and then, since that gave us an angle-side pair, we went ahead and used the Law of Sines to find one of the angle measures.
00:41:29.500 --> 00:41:40.900
Then, since you are given two angle measures, you can solve for the third angle measure by the angle sum theorem--that is solving the triangle.
00:41:40.900 --> 00:41:55.100
And the fourth example: In parallelogram ABCD, AD is 8; AB is 11; and the measure of angle A is 110; find the measure of each diagonal of the parallelogram.
00:41:55.100 --> 00:42:09.100
First, I am going to draw a parallelogram and label it ABCD.
00:42:09.100 --> 00:42:27.800
AD is 8; AB is 11; the measure of angle A is 110; and then, find the measure of each diagonal.
00:42:27.800 --> 00:42:36.900
That is the first diagonal; by drawing this diagonal in, we now have this triangle here.
00:42:36.900 --> 00:42:46.000
If this is 11, and this is 8, then this is what we are looking for; this would be side a.
00:42:46.000 --> 00:42:56.100
And then, in order to solve for that, I know it is a non-right triangle...do I have an angle-side pair?
00:42:56.100 --> 00:43:00.700
No, I don't; so then, I have to use the Law of Cosines.
00:43:00.700 --> 00:43:22.100
And remember: it is a² = b² + c² - 2 bc cosine of angle A.
00:43:22.100 --> 00:43:32.400
Here, a² is what we are looking for; and even though it is not sides b and c, it would just be the other two sides.
00:43:32.400 --> 00:43:36.500
There are three sides total; it is just all three, written out here.
00:43:36.500 --> 00:43:43.800
So, a² = b²...we are going to say 8²...
00:43:43.800 --> 00:43:57.800
+ c², just the third side, 11², minus 2 times those two sides cosine of angle A.
00:43:57.800 --> 00:44:05.700
It would be the angle with that side.
00:44:05.700 --> 00:44:24.200
8² is 64, plus 11² is 121, minus...this would be 176cos(A).
00:44:24.200 --> 00:44:31.500
Then, from here, remember that this is what we are solving for.
00:44:31.500 --> 00:44:59.200
a² = 185 - 176...OK, we are solving for side a; this would be 110; let me circle the variable right here;
00:44:59.200 --> 00:45:05.000
then this is the cosine of 110; make sure you do not subtract these two numbers.
00:45:05.000 --> 00:45:56.300
So here, we are going to just solve this out on the calculator...and then...a² = 185 - -60.196.
00:45:56.300 --> 00:46:45.500
a² = 245.196; and then, a would be √245.196...15.67.
00:46:45.500 --> 00:46:56.500
So, if I were to round to the hundredths, it would be 15.67; that is BD.
00:46:56.500 --> 00:47:03.200
Now, they want us to find the measure of the other diagonal.
00:47:03.200 --> 00:47:18.400
Now, this, obviously, doesn't look like it is 110; if you want to draw it so that it would fit better, you can draw it the other way, like this.
00:47:18.400 --> 00:47:30.700
This was A, B, C, D; and it is the same thing, but just if you want to have it more scaled, this would be 110;
00:47:30.700 --> 00:47:40.900
since this side is a little bit shorter...8...11...that is 110, and then we found that this right here is 15.67.
00:47:40.900 --> 00:48:03.300
BD, one of the diagonals, is 15.67; now, if I erase that, and I look for the other diagonal right here,
00:48:03.300 --> 00:48:15.400
well, I know that this is 11; and this is the side that I am looking for.
00:48:15.400 --> 00:48:29.100
Here, this angle measure is...we have 8 and 11; and then, do we know any of the angle measures?
00:48:29.100 --> 00:48:34.600
Well, what do we know about parallelograms?
00:48:34.600 --> 00:48:46.700
If this is 110 (let me erase this diagonal, so that it is easier to see), we know that this angle right here,
00:48:46.700 --> 00:48:54.200
with this angle, are supplementary, because if you were to draw these a little bit longer--
00:48:54.200 --> 00:49:04.800
extend these sides and the side right here--then remember: we have same-side interior angles, or consecutive interior angles.
00:49:04.800 --> 00:49:12.900
So then, with parallel lines and a transversal, if this is 110, then this angle right here would be that.
00:49:12.900 --> 00:49:21.000
So, using that, let me erase this, and then redraw the diagonal right there.
00:49:21.000 --> 00:49:31.500
I am going to be looking for the side AC; and it doesn't matter what we label it.
00:49:31.500 --> 00:49:44.000
I can redraw just that triangle...70...and then I label this a, b, and c...8, 11, and this is what we are looking for.
00:49:44.000 --> 00:49:53.100
Just for the sake of our formulas, I can use this same formula again, because this would be side a.
00:49:53.100 --> 00:50:13.600
Then a² would be b², which is 11², plus 8², minus 2(11)(8) cosine of angle A.
00:50:13.600 --> 00:50:32.000
So, a² = 121 +...8² is 64, minus 176 cosine of 70 degrees.
00:50:32.000 --> 00:50:44.500
a² = 185 - 176cos(70); and then, you can solve this part right here.
00:50:44.500 --> 00:50:52.500
Make sure that you multiply this number by this number on the calculator, instead of subtracting it first.
00:50:52.500 --> 00:51:53.100
The cosine of 70, times 176, is minus 60.196; and then, take that and subtract the 180; a² = 124.80; so a is going to equal 11.17.
00:51:53.100 --> 00:52:08.500
That side is the other diagonal, and that is AC; that would be 11.17; those are the measures of the two diagonals of this parallelogram.
00:52:08.500 --> 00:52:12.500
And remember how we initially drew this parallelogram.
00:52:12.500 --> 00:52:18.900
Now, this is just so you have a visual; if you need to draw it more to scale, you can redraw it,
00:52:18.900 --> 00:52:24.600
depending on what sides you have and what is bigger and what is smaller.
00:52:24.600 --> 00:52:32.300
And then, we know that the measure of angle A was 110, so that would be an obtuse angle; so you can draw it in that way.
00:52:32.300 --> 00:52:40.900
And here, I just redrew this triangle right here, so that it would be easier to see what we are dealing with, what we are using.
00:52:40.900 --> 00:52:43.000
That is it for this lesson; thank you for watching Educator.com.