 Hi, I'm Zor. Welcome to Unizor Education. It's time to solve a few simple problems related to right triangles. This is probably the most frequently occurring problems in elementary course of trigonometry. So, all these problems are really trivial and they just follow very, very simply from the definition of the trigonometric functions for the right triangle. So, before going into these problems, let me just remind you in general what we are talking about. So, if we have the right triangle with one acute angle phi and two characters A and B, opposite A is adjacent and C is hypogenous, then we can actually talk about slightly different, well, maybe simpler, definition for trigonometric functions. So, if you remember, the general definition is related to the unit circle. In case we have a acute angle for this acute angle, you can actually use this right triangle and define all the trigonometric functions in this way. Sinus phi is equal to opposite characters to hypotenuse, cosine of phi is adjacent to hypotenuse, tangent phi is opposite to adjacent, cotangent phi is opposite A to AB, secant is 1 over cosine, which is C over A, and cosecant is C over B. So, these are definitions which are absolutely equivalent to corresponding definitions for unit circle, which are more universal and applicable to all angles, not only acute angles as these definitions, but for acute angles they are absolutely equivalent and we were talking about this. But now, based on all this, I would like to solve few problems related to simple kind of condition. You have three different characteristics of this triangle. You have some trigonometric function of the acute angle, and you have some two actually linear characteristics, like hypotenuse and a catapult or a catapult and another catapult, etc. And all the problems which are really considered in this particular lecture are, if you know only two elements out of these three, how to find the third one. That's the general approach to everything. And obviously, we are talking about some kind of an equation which we have to solve, like for instance, if I know B and I know sin of phi, how to find C, or if I know A and B, how to find cotangent of phi, etc. So, all these problems which I'm going to talk about are about this particular activity, if you wish. You've got two elements out of three and using the corresponding equations, find the third element. Now, with this introduction, let's just go one by one. So, first what we did here, we have, and this is actually the problem number one, which I have already solved, how to express all the different trigonometric functions of this particular angle phi in terms of other linear elements, sides actually of this triangle. And this is basically the solution to this problem number one, out of whatever I have, eight include. So, this is something which you, well, probably it's one of the few things which you really have to remember, not all of them, obviously. But, you know, at least sine and cosine, because everything else is derived. Tangent is sine over cosine, cotangent is cosine over sine, secant is one over cosine, and cosecant is one over sine. So, everything else is basically followed from these two. But in any case, I would consider this problem number one, which I wanted to offer you, express all the different six trigonometric functions of this acute angle in terms of all other sides. Now, let's go to other problems. And I do suggest if you didn't try it yourself first before listening to this lecture, do this. I mean, it's really kind of very simple problems. All right. So, number one. What if you have a hypotenuse and an angle? So, you have to express a cadgages A in terms of hypotenuse and an angle. Well, let's just think about it. Where are angle A and C participate in this? Well, you can use this one, or you can use this one, because in both cases, we have exactly these three elements which we need. An angle, the cadgages which we have to determine, and the hypotenuse. Well, let's just use this one, but we'll use both. So, in this particular case, how to resolve this equation for A? Well, obviously, A equals to C times cosine of five. Now, how about this one? From this one, A is equal to C over secant of five. And obviously, this is the same thing, because secant by definition, by general definition, is actually one over cosine. So, if you substitute this one over cosine, then you will get exactly the first four nodes. Now, I don't have to worry about denominator being equal to zero, because we're talking about the right triangle, which means five is really an angle which is greater than zero and less than 90. And in this interval, none of these are equal to zero, so everything is fine. That's number two problem. Problem number, I have two problems which are exactly under number two. Another is another cadgages. I have to find a function in terms of hypotenuse and an angle. So, now we have to find, again, the corresponding expression which contains B, C and five. Well, you can use this one, and from this one, B is equal to C times sine of five. Or I can use this one, where B is equal to C over cosine of five. All right? And again, these are completely equivalent because cosecant is one over sine. Okay, let's finish with problem number two, number three. Three and four actually express hypotenuse in terms of catapults A and an angle five. Well, we really have to do the same thing. We have to find the formula which participates A, C and five like this one, but now we have to find C. The C would be equal to A over cosine of five from this formula. And from this formula is C is equal to A times secant of five. Obviously, this is equivalent because secant, again, by general definition, is one over cosine. So that's the same thing. Now, how about B, another catapult which I have given, and opposites to five. Well, in this case, I suggest to use either this one or this one, right? In the first case, C from this equation is equal to B over sine of five. And from this one, B is equal to, sorry, C. C is equal to B times cosecant of five. Okay, next. Now, it's a little bit more precise. Express hypotenuse in terms of the catapult A and not just anything related to five, but precisely tangent of five. Okay, well, let's just think about it. There is no formula here, right? However, what can be done is knowing A and knowing tangent, I can find B, right? Knowing this catapult and the angle I can find B because B over A is equal to tangent. So, let me do it first. B is equal to A times tangent of five from this one. Now, since I know B and A, I can calculate C using the theory, right? I know that C squared is equal to A squared plus B squared from which C is equal to square root of A squared plus, and instead of B squared, I'll use this one, which is the same thing as A outside of the square root. And here, I have one plus tangent squared five. Tangent squared five is tangent five squared. It's just more traditional usage of this particular way of how to multiply the trigonometric function by itself. Usually, this power is written between the function name and the function argument. All right? So, that's it now. We don't have to worry about positive value of the square root or negative value because we're talking about a triangle which means everything is positive. So, this is a arithmetic value, the real main value of the square root, the positive one. And A is obviously decided. All right? So, that's how it is. Very similarly, what if my side, my tangent is different instead of A, I have B? Well, we can do the same thing. From B now, we can take A, so what's A? A is equal to, I think this is more convenient. A is equal to B times cotangent five. And now, again, I'm using the Pythagorean theorem. So, C squared is equal to A squared plus B squared. So, C is equal to square root of A squared, which is this, B squared cotangent squared five plus B squared, which is equal to B squared root of cotangent squared five plus one. So, that's the formula for C. What else? This is just one little, little complication. So, we have to do two steps instead of one. Instead of just straight formula, we have to do first some intermediary calculation to calculate the second calculus and then use the Pythagorean theorem. Alright, next is express A in terms of C and tangent five. Well, basically it's more or less the same thing. What can we do about this? Now, tangent from C and tangent five, what can be done? Well, what we can do actually is, let me just think about what's the best way of doing this. Since I have tangent, I have B over A, right? So, on one hand, B over A is tangent. So, that's the known thing, right? On another hand, A squared plus B squared is equal to C squared, right? That's the Pythagorean theorem. What we can do right now is, we can solve this system of equations for A. Now, how can we do it? Well, we can express B from here. It would be B is equal to A tangent five. Substitute B into this equation. We will have A squared plus A squared tangent squared five equals C squared. Or A squared times one plus tangent squared five is equal to C squared. Well, from which A is equal to, we can have square root from both sides and divide by one plus tangent squared. That's what we will have. So, that stands. Now, very similarly, that's the last one. If I want to find B in terms of C and five, I have to do very similarly. Did I do it right with A? Was it tangent or cotangent I had to use? All right. Well, in case of B, if I made a mistake with A, it's exactly the same for B would be. So, you will figure it out. So, this is the expression from here and I can substitute it. No, I have to do it vice versa. I have to do the A in terms of B now. Which means what I should probably do better, I will use A over B, which is cotangent five. Now, I will express A in terms of B. So, everything is right from this, right? So, A over B is cotangent. And this is Pythagorean theory. I express A in terms of B and substitute it here. And I will have B square one plus cotangent square five is equal to C square. From which B is equal to C over square root one plus cotangent square five. That's the answer. So, these are really as simple as it can be. Most of these problems are just derived from the equivalent definition of trigonometric functions for right triangle. And again, let me just emphasize again and again, these are not true definitions for any angle. These are different definitions for the Q-tangle which are exactly equivalent to definitions which we have used for any angle using the unit circle. So, from these six equalities, identitudes, whatever you want to call them, follows immediately practically all of these gathers except a couple of the last ones where you really have to do just very, very little manipulation using some intermediary steps or something. Alright, so that's it. These are simple problems. Do not think that trigonometry is that simple. I will have a few very important lectures about different trigonometric identitudes and equologies and I will present much more complicated problems related to trigonometry which is not really easy like one step and you will get a solution even two steps. So be prepared. Anyway, this is simple and I would suggest you to repeat again just using the notes to this lecture on Unisor.com. That's it. Thank you very much.