 Hi, I'm Zor. Welcome to InDesert Education. Well, it's time to solve a few problems related to laws of science and coscience, which we could just learn. And I would like to start with some illustrative examples of these laws. You see, when I first saw the law of science or the law of coscience in the same token, I was quite surprised how nicely they look, especially the law of science, quite frankly. Remember, it looks really interesting, quite cool actually, and the law of coscience is something like this, which is some kind of extension of the Pythagorean theorem, which is beautiful by itself. So, let's just check if it really works. I mean, yes, I know we have proved it, but this exercise would really help to better feel how these laws really work. Okay, so, what I want to do is to illustrate these laws for a couple of triangles, which are quite familiar to everybody, and we definitely know the relationship between their sides and angles, etc. And let's just check for these few very known triangles if these laws really hold. Okay, so my first triangle is the right triangle with 30 and 60 degrees angles. Now, let's check first the law of coscience, for instance. Now, I don't remember the values of coscience of many different angles, except one actually. So, the only thing which I do remember is this, because I know that in this triangle 30, 60, 90 degrees, the side which is opposite to angle 30 degrees is half of the hypotenuse, so A is half of C. So, if that's true, this is 2A, right? And that's what actually needs to be sine of the 30 degree equal to 1 half, because it's A divided by 2A. So, that's actually the only theory which I remember, and that helps me to find this value. Everything else, or other angles, I can basically derive from this one. Okay, now, what's the third side? Well, let's use the Pythagorean theorem, right? Now, if this square plus this square is equal to this square, then B is equal to... Well, let's say B square is equal to square of hypotenuse minus square of these categories, right? Which is 4A square minus A square, which is 3A square, so B is equal to A square root of 3. Okay, now let's check the, let's say, cosine theorem, the law of cosines. Okay, now, obviously, it doesn't really make sense to check the law of cosines for C equal to hypotenuse, because it's really a Pythagorean theorem. Cosine of 90 degrees equal to 0, right? So, that's fine. Now, let's check it for this side, for instance. So, this square plus this square minus 2 product and the cosine of this should be equal to this square. Alright, let's check it out. Square of this is 4A square. Square of this is 3A square minus 2 product of them, which is 2A times A square root of 3 and times cosine of 30. Okay, what is a cosine of 30? Well, actually, you can check it from here. It's adjacent cosines divided by hypotenuse, which is square root of 3 divided by 2. What is it equal to? This is 7A square. Now, this is 2 and 2 can be reduced. Square root of 3 and square root of 3 is 3 and 2 is 6. And as we see, this is a square of this side. So, the law of cosine works fine. Now, how about this side? Alright, that's easy. So, square of this plus square of this and minus 2 product of this by cosine of 60. Okay, A square was 4A square minus 2 times A times 2A and times cosine of 60. Cosine of 60, well, number one, obviously, you can get it from this because it's a adjacent calculus divided by hypotenuse, which is 1 half. Or, in other words, cosine of 60 is the same as sine of 90 minus 60, which is 30. And sine of 30, I do remember, is 1 half. So, anyway, this is 1 half. So, what do we have? This is 5A square minus 2A square, which is 3A square. Now, 3A square is exactly the square of this. So, the law of cosine works fine. Now, let's check the law of sine. So, A divided by sine of 30 degree, which is A divided by 1 half should be equal to this, which is A square root of 3. Divided by sine of 60. Now, sine of 60 is this divided by this, which is square root of 3 over 2, and should be equal to this divided by sine of this. 2A divided by sine of 90 degree, which is 1. Well, let's just check. This is equal to 2A. This is equal to 2A. And this is equal to 2A. So, the law of cosine, the law of sines holds as well. Fine. Finished with this triangle. And let's go to another familiar triangle, which is 45, 45, and 90 degree, right triangle. Now, using the Pythagorean theorem, well, first of all, since these two angles are 45, this is isosceles triangle. So, this is A, then this is A, and this is, using the Pythagorean theorem, A square plus A square is 2A square. So, square root would be A square root of 2. Now, so from here, we can say that sine of 45 degrees equals cosine of 45 degrees equals square root of 2 over 2, right? It's A divided by square root of A, which is 1 over square root of 2, which is square root of 2 divided by square root of 2, square root of 2 over 2, right? I just multiplied numerator and denominator by square root of 2. So, I will have a radical on the top as it's really kind of traditional in mathematics. Alright, so, we've got that. Now, let's check. Again, we don't really have to check the law of cosine for hypotenuse because that's really a Pythagorean theorem. So, let's check it for one of these categories. We just have to check one because they're the same. Okay, square of this, which is A square plus square of this, which is 2A square minus 2A times A times square root of 2 times square root of 2 over 2. That's the cosine. So, what is it? This is 3A square. This is square root of 2, square root of 2, 2 square A square, which is square of this. So, law of cosine, cosine holds. Now, let's check the sines, law of sines. A divided by sine of 45, which is square root of 2 over 2 should be equal to, well, this is exactly the same. Now, this, A square root of 2 over sine of 9 to the degree, which is 1. And this is true because this thing is equal to 2A over square root of 2, which is square root of 2A, which is the same as this. So, law of cosine, law of cosine and law of sines both hold for this triangle as well. Okay. Now, another triangle, which is not exactly in the category of right triangles, but very much close to it. Here it is. So, let's consider you have isosceles triangle with 120 degrees on the top. Now, it's isosceles, which means these two angles are the same. And since the total sum should be 180, so this is 30 and this is 30 degrees, right? 30 plus 30, 60 plus 120, that's 180. So, that's the triangle which we are given. So, let's check if these laws hold for this particular triangle. Now, to make our life easier, I will draw the altitude, which is the same as the angle bisector and its median as well. So, to check, I will use this is A and this is A. And this is 30, this is H. Now, since this is H, this triangle is 30, 60, 90, right? So, this is to H because H is opposite to 30 degrees angle in the right triangle and this is to H as well. Now, what is this? Actually, we already had this in the first example. This is H times square root of 3 and this is square root of 3. So, by assigning a length H to an altitude, I basically calculated all other sides of this particular triangle. So, I know everything. I know angles and I know sides and all I have to do is check this relationship between them. Alright. So, let's just check it out. For instance, let's just take these two squares, some of them minus two products and cosine of 1 to 1. You should be equal to this one, right? Let's check it out. Square of this, which is 4H square plus square of this minus 2 times 2H times 2H. And what's the cosine of 120? Okay, as I said, I don't remember these numbers. So, cosine of 120. Let's just think about it. Now, this is 120, right? And this is 60. Now, cosine is abscissa. So, it's this piece. Now, obviously symmetrically, if I will do 60 here, I will have exactly the same abscissa, but in this case it's positive, in this case it's negative. So, the cosine of 120 is, by absolute value, the same as cosine of 60, but has an opposite sign. Now, what's the cosine of 60? Same as sine of 30, right? Which is 1, 2, 1 half. So, this is minus 1 half. So, let's multiply it by minus 1 half. So, what is it? First of all, minus and minus, so it will be positive. 2 times 2 times 2 and divided by 2, so it will be 4. So, it will be 4 and 4 and plus another 4, it will be 12H squared. Okay? Now, how about this one? This is H square root of 3 and this is H square root of 3. So, the total length is 2H square root of 3. Now, what if I will square it? What it will be? It will be 4H square and 3, which is 12H squared. Exactly the same as this. So, the cosine, the law of cosines holds in this particular case. For this side as being equal to. Well, let's check this side, for instance. Okay? This square, it's 2H square root of 3 square is 12, as we were talking about. 12H square plus square of this, which is 4H square. Minus 2 times this, which is 2H square root of 3. Times this times 2H and times cosine of 30, which is square root of 3 over 2. Well, let's check what it is. Well, this is 16H square minus 2 times 2 times 2. It's 8 divided by 2 is 4 and multiplied by 3, which is 12. 4H square, which is exactly the square of this. So, the law of cosines holds for the side as well. Alright? Now, let's check the signs. Law of signs. Now, for this, let's just do this. Sign of 30 degrees, we know this is 1 half. Sign of 120 is, again, sign is ordinate. Remember, right? So, this is 120. This is 60. This is 120. So, if I would do 60 here, these points are symmetrical and they have the same ordinate. So, sign of 120 is the same as sign of 60, which is, as we were talking just five minutes ago, square root of 3 over 2. So, now we are ready to check our law of signs. So, this, which is 2H square root of 3 divided by sign of 120, which is square root of 3 over 2. Should be equal to this, which is 2H, divided by sign of 30, which is 1 half. Well, let's check it out. This is reduced, so it's 2H divided by 1 half, and this is 2H divided by 1 half. So, it's fine. Alright, so we have actually checked the validity of the laws of signs and cosines for three different types of triangles. Well, I think it confirms, I mean, it doesn't prove anything obviously, but it does confirm that it all makes sense, and the proof, which was kind of abstract, is really kind of applicable and you understand may be better and you feel better that this is the true statement about the triangles, about laws of cosines and signs. Alright, now let's solve a simple problem. Let's consider you have an isosceles triangle, and you are given the angle at the base, which is too far, and same thing here, obviously, because it's isosceles. Now, there is a bisector of this angle. Well, this is bisector, so this is obviously the phi. And my question is, given the lengths of the base and the angle 2 phi, or phi actually, half of this angle, and the angle at the base, I would like to determine the lengths of this angle bisector AL. Well, let's consider triangle ADL. We know about this triangle, well, basically enough to determine everything. We know the base and we know 2 angles, right? Since this is a bisector, this angle is phi. This angle is 2 phi. Angles were given as, angles were congruent. Now, now let's use the law of signs. If this is x, then x divided by sin of 2 phi is equal to c divided by alpha, sin alpha. Now, we know what alpha is, right? Alpha is equal to 1h minus 2 phi and minus phi minus 3 phi, right? So, sin of 1h minus 3 phi, just a second ago, we were talking about 120 degree angle. So, the 60 degree, which is supposed to be added to 120 to get 180, has the same sign because it's the same ordinate, right? So, if this is 3 phi, this is also 3 phi, and this is 180 minus 3 phi. So, the ordinance of these are the same. So, this is equal to sin of 3 phi. So, we basically have solved everything because since x divided by 2 phi is equal to c divided by sin of 3 phi, from here, we find x which is equal to c times sin of 2 phi divided by sin of 3 phi. That's the answer, and we were using the law of cosines. Well, and a little calculation of this angle, which is really very simple, this one. So, that's the solution. So, now we have effectively used the law of sin to determine one side of the triangle by another side and two angles at it. That's a simple problem which can be solved with law of sines. Okay, the last one. Let's say that you have a triangle. These are three angles, and let's say you have one side, the c is given, and alpha, beta and gamma are not given. However, what's given is alpha to beta to gamma is 1 to 3 to 8. So, I have a ratio between the size, the measurements of these angles. So, I don't have angles, but I do have the ratio they are in. So, I have a side, a b, which is equal to c, and I have a ratio. Now, I have to determine other sides, this and this, a b. Well, if I knew angles, then I know how to solve this problem. It was just the previous one. I'll use the law of sines. So, how can I find the angles if I know just the ratio between angles? Well, hey, I do know something else. I know that the sum of these angles is equal to 180 degrees. So, that's the second condition. So, now, basically, we have three unknowns, if you wish, and three equations. This is one, a to b, this is another, alpha to beta, then beta to gamma is another, and sum of alpha plus beta plus gamma is the third one. So, how can I find the values of three different angles knowing these two conditions? Well, very simple. If alpha to beta is one to three, it means that the beta is equal to three alpha. Now, if alpha to gamma is one to eight, it means gamma is equal to eight alpha, right? So, I substitute it to this, and I have alpha plus three alpha plus eight alpha equals to one eight. This is what? Twelve. Twelve alpha is equal to one eight. Alpha is equal to one eight divided by twelve is fifteen. So, I've got this. So, alpha is fifteen degrees, beta is three times as much, which is forty-five degrees, and gamma is eight times as much, which is one twenty degrees. Okay, so we know all the angles, and since we know the angles, now we can use the law of sines to get to all the sides, because A divided by sine of fifteen degrees is equal to C divided by sine of sixty, sorry, one point of degrees. Now, this is equal to C divided by, again, the same thing. Sine of one twenty is equal to sine of one eighty minus one twenty, which is sixty, and the sine of sixty we're talking many times before is square root of three divided by two. From here, I get A is equal to, let's multiply by square root of three. So, two goes there, so it's two square root of three C over three. Am I right? I think I'm right. Multiply by square root of three, I will have three here, and square root of three here, and two goes there, yes, times sine of fifteen degrees. That's my A. Now, B over sine of forty-five degrees. So, B is equal to two square root of three C over three times sine of forty-five degrees is this. So, we can reduce this, and I will have C square root of six over three. So, that's the answer. Actually, I can determine what is a sine of fifteen degrees, but let's just not talk about this right now. This is sufficient for now. It's still a number, I mean, obviously, so this is an answer. By the way, if you have in your problem something like a given angle, well, then you can consider the functions of this angle also given like sine or cosine or tangent or whatever. Well, in as much as if you have a length of a square, then you also have basically its area, which is the length squared. So, function of a known value is considered to be given as well. All right, so that's the answer, and that's how this particular triangle looks like. Okay, that's it for today. I'm sure I will have much more difficult problems in the future, but as I illustrated examples to the laws of sines and cosines, I hope they will do. Thanks very much, and good luck.